# Cryptanalyzing two image encryption algorithms based on a first-order time-
delay system

Sheng Liu  Chengqing Li<EMAIL_ADDRESS>Qiao Hu  School of Computer
Science and Electronic Engineering, Hunan University, Changsha 410082, Hunan,
China  School of Computer Science, Xiangtan University, Xiangtan 411105,
Hunan, China

###### Abstract

Security is a key problem for the transmission, interchange and storage
process of multimedia systems and applications. In 2018, M. Li et al. proposed
in-depth security analysis on an image encryption algorithm based on a first-
order time-delay system (IEATD) and gave a specific chosen-plaintext attack on
it. Moreover, an enhanced version called as IEACD was designed to fix the
reported security defects. This paper analyzes the essential structures of the
two algorithms and evaluates their real security performances: 1) no efficient
nonlinear operations are adopted to assure the sensibility of keystream; 2)
the equivalent secret key of IEATD can be efficiently recovered from one known
plain-image and the corresponding cipher-image; 3) IEACD can still be
efficiently cracked with a chosen-plaintext attack. Both rigorous theoretical
analyses and detailed experimental results are provided to demonstrate
effectiveness of the advanced cryptanalytic methods.

###### keywords:

Chosen-plaintext attack , cryptanalysis , image encryption , chaotic
cryptography , image privacy.

††journal: IEEE Multimedia

##  1 Introduction

Social media not only drive product discovery and purchase, but also incur
serious concern on the security and privacy of the images shared by the
Internet users. Due to the special properties of multimedia information, the
modern text encryption standards, such as AES and Triple DES, cannot
efficiently protect them in general. To cope with the challenge, a number of
special image encryption algorithms, e.g. joint encryption and compression
together, were proposed every year [1, 2, 3]. It is well known that
cryptography (designing encryption algorithm) and cryptanalysis (security
analysis of a given encryption algorithm) are two integral parts of
cryptology. The cryptanalysis results facilitate the designers strengthen or
replace flawed algorithms. Cryptanalysis of a given image encryption scheme
also provides a special perspective for promoting some multimedia processing
techniques, e.g. image recovery. Some image encryption algorithms like that
proposed in [4, 2] are found to be insecure to different extents from the
viewpoint of modern cryptology [5, 6, 7, 8].

The complex dynamics of a chaotic system demonstrated in an infinite-precision
domain is very similar to the expected properties of a secure encryption
scheme outlined by Shannon in [9]. So, a large number of chaos-based
encryption schemes were proposed in the past three decades [10, 11]. In [12],
Ikeda adopted a one-variable differential-difference equation to model light
going around a ring cavity containing a nonlinear dielectric medium and found
“chaotic” phenomena in the transmitted field. In [4], Mannai et al. introduced
the equation’s variant

$\frac{dx(t)}{dt}=-\alpha x(t)+m\sin(x(t-T))$ (1)

as a chaos-based pseudorandom number generator (PRNG), where $\alpha$ and $m$
are coefficients, and $T$ is the positive delay time. The evolution of the
dynamics is dependent on not only the present value $x(t)$ but also earlier
one $x(t-T)$. To solve equation (1), it is discretized with the following way:
1) each interval $T$ is divided into $N$ subintervals $h$ and each subinterval
$h$ is approximated with a scalar value, where $T=N\times h$; 2) the $N$
samples of each interval are considered as an $N$-dimension vector. In [4], an
image encryption algorithm based on a time-delay Ikeda system (IEATD) was
proposed. The designers of IEATD believed that utilizing the rich dynamics of
a discretized Ikeda system and a new keystream generation mechanism associated
with the average of all pixels of the plain-image can provide sufficient
capacity to withstand known/chosen-plaintext attacks.

In reality, the security strength of IEATD is very weak as its equivalent
secret key can be obtained with only two chosen plain-images [13]. Meanwhile,
M. Li et al. pointed out that two security defects exist in IEATD: 1) the
regularity of the keystream and absence of position scrambling; 2) incapacity
to resist differential attack [13]. To remedy the defects, they adopted much
more complex encryption operations: permutation and crossover diffusion
phases. In short, we call the enhanced image encryption algorithm using the
crossover diffusion as IEACD. This paper focuses on security analysis of the
two image encryption algorithms, IEATD [4] and IEACD [13]. We found that the
authors of [13] did not notice a fatal drawback of the keystream generation
mechanism: insensibility to minor change of a pixel. This leads to that IEACD
still cannot withstand chosen-plaintext attack. Furthermore, there is improper
keystream configuration in diffusion that almost discloses the whole
keystream. The essential structures of the two algorithms cause that the
equivalent secret key of IEATD and IEACD can be recovered with known-plaintext
attack and chosen-plaintext attack, respectively.

The rest of this paper is organized as follows. Section 2 concisely describes
the encryption procedures of the encryption algorithm IEATD and its enhanced
version IEACD. Then, Sec. 3 and Sec. 4 present the detailed cryptanalysis
results on the two encryption algorithms, respectively. The last section
concludes the paper.

##  2 Description of two analyzed image encryption algorithms

The input of algorithm IEATD is an 8-bit gray-scale image of size $H\times W$.
The plain-image is scanned in the raster order and then can be represented as
a sequence $\mathbf{I}=\\{I(i)\\}_{i=0}^{HW-1}$. The corresponding cipher-
image is denoted by $\mathbf{I}^{\prime}=\\{I^{\prime}(i)\\}_{i=0}^{HW-1}$.
Then, IEATD and its enhanced version IEACD can be described in Sec. 2.1 and
Sec. 2.2, respectively.

###  2.1 The framework of IEATD

  * •

The secret key: a positive integer $N_{0}$, three control parameters of
discretized Ikeda chaotic system

$\left\\{\begin{aligned} X_{1}(k+1)&=X_{2}(k)\\\ X_{2}(k+1)&=X_{3}(k)\\\
&\vdots\\\ X_{M}(k+1)&=X_{M}(k)+h\cdot(-\alpha\cdot
X_{M}(k)+m\cdot\sin(X_{1}(k)))\end{aligned}\right.$ (2)

and its initial condition $\mathbf{X}(0)=\\{X_{i}(0)\\}_{i=0}^{M-1}$, where
$h\in[0.1,1]$, $\alpha\in[1,6]$, and $m\in[18,20]$.

  * •

The confusion procedure:

    * –

Step 1: Divide $\\{I(i)\\}_{i=0}^{HW-1}$ into $s$ vectors, where the length of
$k$-th vector $I^{(k)}$ is $N(k)$, and the length of subsequent vector depends
on the previous one:

$N(k)=N(k-1)+\lfloor\text{mean}(I^{(k-1)})\rfloor,$ (3)

where $k=1\sim(s-2)$, $N(0)=N_{0}$. Finally, assign the actual length of the
last vector $I^{(s-1)}$ to $N(s-1)$. Obviously, the longest vector is either
the last vector $I^{(s-1)}$ or the penultimate one $I^{(s-2)}$.

    * –

Step 2: As for the $k$-th vector $I^{(k)}$, iterate Eq. (2) $N(k)$ times from
the initial condition and generate a chaotic sequence
$S^{(k)}=\\{S^{(k)}(i)\\}_{i=0}^{N(k)-1}$, and then obtain the quantized
sequence $Y^{(k)}=\\{Y^{(k)}(i)\\}_{i=0}^{N(k)-1}$ via

$Y^{(k)}=\left(S^{(k)}\cdot 10^{14}\right)\bmod 256,$

where $k=0\sim(s-1)$. Finally, concatenate $\\{Y^{(k)}\\}_{k=0}^{s-1}$ further
into a sequence $\mathbf{Y}=\\{Y(i)\\}_{i=0}^{HW-1}$, where

$Y\left(i+\sum_{j=0}^{k-1}N(j)\right)=Y^{(k)}(i).$

The longest vector among $\\{Y^{(k)}\\}_{k=0}^{s-1}$ is either $Y^{(s-1)}$ or
$Y^{(s-2)}$, and let $Y^{\rm L}$ denote it. Note that every element of
$\\{Y^{(k)}\\}_{k=0}^{s-1}$ is a subsequence of $Y^{\rm L}$.

    * –

Step 3: Perform confusion operations on sequence $\mathbf{I}$ by

$I^{\prime}(i)=I(i)\oplus Y(i)$ (4)

for $i=0\sim(HW-1)$, where $\oplus$ donates the bitwise XOR operation.

###  2.2 The enhanced elements of IEACD compared with IEATD

To enhance the security level of IEATD, some extra operations were appended to
withstand the chosen-plaintext attack reported in [13].

  * •

The added secret sub-keys: a positive integer $C$, the initial condition
$q(0)\in(0,1)$ and the control parameter $\beta\in(3.5699456,4]$ of Logistic
map

$q(i+1)=\beta\cdot q(i)\cdot(1-q(i)).$ (5)

  * •

The modified encryption procedures:

Step 1: Iterate Eq. (5) $HW$ steps from $q(0)$, and obtain a chaotic sequence
$\mathbf{Q}=\\{q(i)\\}_{i=0}^{HW-1}$, which is used to produce permutation
vector $\mathbf{Z}=\\{z(i)\\}_{i=0}^{HW-1}$, where $q(i)$ is the $z(i)$-th
largest element in the sequence $\mathbf{Q}$. Then, permute $\mathbf{I}$ with
the permutation vector $\mathbf{Z}$ and obtain a permuted intermediate image
$\mathbf{I}^{\star}=\\{I^{\star}(i)\\}_{i=0}^{HW-1}$ by performing

$I^{\star}(i)=I(z(i))$ (6)

for $i=0\sim(HW-1)$.

Step 2: Divide $\\{I^{\star}(i)\\}_{i=0}^{HW-1}$ into $s$ vectors
$\\{I^{\star(k)}\\}_{k=0}^{s-1}$ and obtain a sequence $\mathbf{Y}$ like Step
1, 2 of IEATD, where the division size is determined by

$N(k)=N(k-1)+\lfloor\text{mean}(I^{\star(k-1)})\rfloor.$ (7)

Then perform the confusion operation on $\\{I^{\star}(i)\\}_{i=0}^{HW-1}$ via

$I^{\star\star}(i)=I^{\star}(i)\oplus Y(i),$ (8)

where $i=0,1,\cdots,(HW-1)$.

Step 3: Crossover diffusion:

    * –

Step 3a: Generate index array $\mathbf{U}=\\{u(i)\\}_{i=0}^{HW-1}$, where

$u(i)=\begin{cases}\lfloor\frac{i}{2}\rfloor&\text{if }i\bmod 2=0;\\\
\lfloor\frac{i}{2}\rfloor+\lfloor\frac{HW}{2}\rfloor&\text{if }i\bmod
2=1,\end{cases}$ (9)

$i=0,1,\cdots,(HW-1)$.

    * –

Step 3b: Conduct the first round crossover diffusion with $C$ and sequence
$\mathbf{Z}$ by

$I^{*}(u(i))=I^{\star\star}(u(i))\oplus\left(I^{*}(u(i-1))\boxplus
z(u(i))\right)$ (10)

for $i=1\sim(HW-1)$, where $a\boxplus b=(a+b)\bmod 256$ and
$I^{*}(u(0))=I^{\star\star}(u(0))\oplus\left(C\boxplus z(u(0))\right)$.

    * –

Step 3c: Perform the second round of crossover diffusion via

$I^{**}(u(i))=I^{*}(u(i))\oplus\left(I^{**}(u(i-1))\boxplus z(u(i))\right)$
(11)

for $i=1\sim(HW-1)$, where
$I^{**}(u(0))=I^{*}(u(0))\oplus\left(I^{*}(u(HW-1))\boxplus z(u(0))\right)$.

Step 4: Permute $\\{I^{**}(i)\\}_{i=0}^{HW-1}$ and obtain the ciphertext
$\\{I^{\prime}(i)\\}_{i=0}^{HW-1}$ by performing

$I^{\prime}(z(i))=I^{**}(i)$ (12)

for $i=0\sim(HW-1)$.

##  3 Cryptanalysis of IEATD

In [4], the authors claimed that the adopted intermediate keystream is
dependent on the plaintext, so it can withstand the classic attacks, such as
plain/chosen-plaintext attack and chosen-ciphertext attack. However, we argue
that the statement is not always correct. In this section, the weak keys about
$N_{0}$ are discussed. After briefly describing the chosen-plaintext attack on
IEATD proposed in [13], we present a known-plaintext attack on it.

###  3.1 Weak keys with respect to $N_{0}$

In IEATD, the plaintext is first divided into some vectors, which is dependent
on a given key $N_{0}$ and the plaintext itself. The scope of $N_{0}$ is not
specifically given in the algorithm. However, $N_{0}$ should be less than $L$
from the security standpoint, where $L=\lfloor\frac{HW}{2}\rfloor$. If
$N_{0}\geq L$, it is observed that the keystream generation mechanics is
futile since sequence $\mathbf{Y}$ generated from any plain-image is the same.
In such case, a mask image, generated by XORing a plain-image and the
corresponding cipher-image pixel by pixel, can be directly used as the
equivalent secret key. As shown in Fig. 1, a cipher-image is fully decrypted
with the mask image. As this is contrary to the original intention of the
designers, it is assumed that $N_{0}<L$ in the subsequent analysis.

a)

b)

c)

Figure 1: The decrypted result using a mask image when $N_{0}\geq L$: a) the
mask image; b) a cipher-image; c) the decrypted result.

###  3.2 The chosen-plaintext attack proposed by M. Li et al.

To make the cryptanalysis of IEATD more complete, we briefly introduce the
chosen-plaintext attack on IEATD proposed by M. Li et al. in [13] and comment
its performance:

  * •

Determining $N_{0}$: Referring to Eq. (3), one can see that $N(i)=N_{0}$ and
$Y^{(i)}=Y^{(0)}$ for any $i\in\\{1,2,\cdots,s-2\\}$ if
$\lfloor\text{mean}(I^{(k)})\rfloor=0$ for $k=0\sim(s-2)$. To ensure this
condition exists, one can choose a plain-image of fixed value zero. Then,
sequence $\mathbf{Y}$ can be obtained by
$\mathbf{Y}=\mathbf{I}\oplus\mathbf{I}^{\prime}$. If one calculates the
autocorrelation coefficients $R(t)$ of $\mathbf{Y}$, the maximum should be
$R(N_{0})$, where

$R(t)=\frac{\sum_{i=0}^{HW-t-1}(Y(i)-\bar{\mu})\cdot(Y(i+t)-\bar{\mu})}{\sum_{i=0}^{HW-1}(Y(i)-\bar{\mu})^{2}},$

$\bar{\mu}$ is the average of $\mathbf{Y}$, and $t\in[1,L-1]$ is lag.

  * •

Obtaining $Y^{\rm L}_{255}$: Choose a plain-image $\mathbf{I}_{255}$ of fixed
value 255 and get the corresponding longest sequence $Y^{\rm L}_{255}$ from
$\mathbf{Y}_{255}$, where $\mathbf{Y}_{255}$ is the result by XORing
$\mathbf{I}_{255}$ and its cipher-image $\mathbf{I}_{255}^{\prime}$ pixel by
pixel.

Set $Y^{\rm L}=Y^{\rm L}_{255}$ and denote the length of $Y^{\rm L}$ by
$N_{Y^{\rm L}}$. The decryption procedure can be described as follows:

  * •

Step 1: Set $N(0)=N_{0}$, $N_{\rm sum}=0$, $k=0$.

  * •

Step 2: Do $I(N_{\rm sum}+i)=I^{\prime}(N_{\rm sum}+i)\oplus Y^{\rm L}(i)$ for
$i=0\sim(N(k)-1)$, and then set $k=k+1$.

  * •

Step 3: Set

$N_{\rm sum}=N_{\rm sum}+N(k-1)$

and calculate $N(k)$ via Eq. (3). If

$N(k)\leq N_{Y^{\rm L}}$ (13)

and $N_{\rm sum}\leq HW-N(k)$, go back to Step 2.

  * •

Step 4: Assign $\min((HW-N_{\rm sum}),N_{Y^{\rm L}})$ to $N(k)$, where
$\min(a,b)$ returns the smaller element between $a$ and $b$. Then do

$I(N_{\rm sum}+i)=I^{\prime}(N_{\rm sum}+i)\oplus Y^{\rm L}(i)$

for $i=0\sim(N(k)-1)$.

In the above decryption process, the intermediate keystream
$\mathbf{Y}=\\{Y(i)\\}_{i=0}^{HW-1}$ corresponding to a cipher-image is
gradually recovered. In other words, one can reconstruct the specific
$\mathbf{Y}$ belonging to a cipher-image from $Y^{\rm L}_{255}$ and $N_{0}$.
Thus, they can be regarded as the equivalent secret key. The time complexity
of the attack is $O((WH)^{2})$ instead of that claimed in [13], $O(WH)$.

###  3.3 Known-plaintext attack on IEATD

Known-plaintext attack can be considered as a stronger version of the chosen-
plaintext attack as the former can recover the information with the secret key
from some given plaintexts, instead of that specially constructed or selected.
As for algorithm IEATD, even if only one plain-image and the corresponding
cipher-image are available, one can obtain $N_{0}$ effortlessly and then
derive a counterpart of $Y^{\rm L}$, $Y^{\rm L^{\prime}}$. They can be used to
disclose some visual information of the other cipher-images encrypted with the
same secret key.

According to Eq. (3), one can get $\\{Y^{(k)}\\}_{k=0}^{s-1}$ after obtaining
$N_{0}$ and $\mathbf{Y}$. Then, as for two adjacent vectors $Y^{(k)}$ and
$Y^{(k+1)}$, one has $Y^{(k)}(i)=Y^{(k+1)}(i)$ for $i=0\sim(N(k)-1)$, where
$k=0\sim(s-2)$. Hence, the condition can be used to verify the search of
$N_{0}$. Since the scope of $N_{0}$ is relatively small as mentioned in Sec.
3.1, the confirmation of $N_{0}$ is feasible through brute-force searching:

  * •

Step 1: Produce sequence $\mathbf{Y}$ by XORing the plain-image and its
corresponding cipher-image pixel by pixel.

  * •

Step 2: For $j=0\sim(L-1)$, do the following operations:

    * –

Step 2a: If condition $\\{Y(i)\\}_{i=0}^{j-1}=\\{Y(i)\\}_{i=j}^{2j-1}$
satisfies, set $N(0)=j$, $N_{\rm sum}=0$, $k=1$; otherwise, go to the next
loop.

    * –

Step 2b: Set $N_{\rm sum}=N_{\rm sum}+N(k-1)$ and calculate $N(k)$ using Eq.
(3). If $N_{\rm sum}>HW-2N(k)$, set $N_{0}=j$ and terminate the attack.

    * –

Step 2c: If condition

$\left\\{Y(i)\right\\}_{i=N_{\rm sum}}^{N_{\rm
sum}+N(k)-1}=\\{Y(i)\\}_{i=N_{\rm sum}+N(k)}^{N_{\rm sum}+2N(k)-1}$

satisfies, set $k=k+1$ and go back to Step 2b.

After confirming $N_{0}$, one can easily obtain $Y^{\rm L^{\prime}}$ from
$\mathbf{Y}$. Set $Y^{\rm L}=Y^{\rm L^{\prime}}$ and the decryption procedure
is the same as that in Sec. 3.2. Once condition (13) does not exist during the
decryption process, the following cipher-pixels are all decrypted incorrectly.
Referring to Eq. (3), one can know a simple rule: the brighter a plain-image
is, the fewer the divided vectors become (the corresponding $Y^{\rm L}$ gets
longer). When the brightness of the plain-image corresponding to a cipher-
image to be decrypted is higher than that of the known plain-image, condition
(13) is not satisfied for smaller index $k$. This means that more portion of
the cipher-image cannot be decrypted correctly. As shown in Fig. 2b), the
image with lower brightness than that in Fig. 2a) is even completely
decrypted. By contrast, many consecutive pixels of two brighter plain-images
cannot be recovered correctly (See Fig. 2c) and d)). As for the same plain-
image to be decrypted, if brightness of the available known plain-image is
lower, more consecutive pixels cannot be decrypted correctly. This point can
be verified by comparing Fig. 2b), c) and d) with Fig. 2f), g) and h),
respectively.

a)

b)

c)

d)

e)

f)

g)

h)

Figure 2: The result of known-plaintext attack on IEATD: a) a known plain-
image; b)-d) the decrypted images with the equivalent secret key derived from
Fig. a); e) a known plain-image with half brightness of that in Fig. 2a);
f)-h) the decrypted images with the key obtained from Fig. 2e).

##  4 Cryptanalysis of IEACD

To cope with the insecurity problems of IEATD reported in [13], multiple
confusion and diffusion operations are appended, making the algorithm become
another algorithm IEACD. In fact, the original keystream generation mechanism
indeed exists a serious pitfall, which leads to that the patched algorithm
IEACD still cannot withstand chosen-plaintext attack. In this section, three
weaknesses of IEACD are first analyzed to facilitate description of the
following chosen-plaintext attack.

###  4.1 Three weaknesses of IEACD

  * •

The real size of key space is much smaller than the expected one

Due to the limitation of finite-precision presentation, dynamics of any
chaotic system is definitely degraded when it is implemented in a digital
device. As investigated in [14, 15, 16], the structure of the state-mapping
network (SMN) of a digitized chaotic system implemented with fixed-point
precision $e+1$ is largely dominated by that with precision $e$. The short
period problems of PRNG based on Logistic map (5) implemented in a digital
device (with fixed-point arithmetic or floating-point arithmetic) were
comprehensively discussed in [15]. As shown in Fig. 3, discretized Ikeda
system obeys this rule also. No matter what the precision is, the SMN of
discretized Ikeda system follows the following rules: 1) an SMN is composed of
some weakly connected components; 2) there are some self-loops (an edge
connecting a node to itself); 3) As for each connected component, there is one
and only one cycle (including special cycle, self-loop), and every node
evolves to it via a transient process; 4) Many nodes have two and only two
parent nodes. Generating a pseudo-random number sequence by the orbits
determined by a chaotic map is actually walking along a path of an SMN. Now,
one can see that the period of a sequence by solving the discretized Ikeda
system may be very short (even only one). So, there are a number of equivalent
secret keys and invalid secret keys as for the function of IEATD and IEACD.
Note that such pitfall always exists no matter how large the precision $e$
gets.

a)

b)

Figure 3: The State-Mapping Networks of the Ikeda map with
$\alpha=\frac{48}{2^{3}},h=\frac{1}{2^{3}},m=\frac{156}{2^{3}}$ under $e$-bit
fixed-point precision: a) $e=3$; b) $e=4$.

  * •

Insensibility of keystream generation mechanics

Although the permutation operations are performed before the confusion step to
frustrate the predictability of keystream $\mathbf{Y}$, the keystream is still
insensitive to minor changes of some pixels of the plain-image. Referring to
Eq. (7), one can see that the possibility that $N(k+1)$ change is roughly
$\frac{1}{N(k)}$ when a pixel in vector $I^{\star(k)}$ is slightly changed
with variation $\Delta=1$. More generally, when a randomly chosen pixel in one
plain-image is slightly altered with variation $\Delta=1$, the probability
that $\mathbf{Y}$ generated by the altered plain-image is different from the
previous version is

$P_{\rm
c}=\sum_{k=0}^{s-1}\left(\frac{1}{N(k)}\cdot\frac{N(k)}{HW}\right)=\frac{s}{HW}.$

Taking a nature image of size $256\times 256$ as an example, $s\approx 40$ and
$P_{\rm c}\approx 6.103\times 10^{-4}$. Note that the more the average of the
pixels of the image approaches 255, the smaller the probability is.

  * •

Improper configuration of keystream

The two-round crossover diffusion is performed to resist chosen-plaintext
attack and differential attack, but the keystream $\mathbf{Z}$ used in
permutation is wrongly reused in the diffusion part, which makes the algorithm
more insecure. From Eq. (6), (8) and (10), one has

$\displaystyle I^{*}(u(i))$
$\displaystyle=I^{\star\star}(u(i))\oplus\left(I^{*}(u(i-1))\boxplus
z(u(i))\right)$ $\displaystyle=I^{\star}(u(i))\oplus
Y(u(i))\oplus\left(I^{*}(u(i-1))\boxplus z(u(i))\right)$
$\displaystyle=I(z(u(i)))\oplus Y(u(i))\oplus\left(I^{*}(u(i-1))\boxplus
z(u(i))\right),$ (14)

where $i>0$ and $I^{*}(u(0))=I(z(u(0))\oplus Y(u(0))\oplus\left(C\boxplus
z(u(0))\right)$. Also, incorporating $i=u(i)$ and Eq. (12) into Eq. (11), one
has

$\displaystyle I^{\prime}(z(u(i)))$ $\displaystyle=I^{**}(u(i))$
$\displaystyle=I^{*}(u(i))\oplus\left(I^{**}(u(i-1))\boxplus z(u(i))\right)$
$\displaystyle=I^{*}(u(i))\oplus\left(I^{\prime}(z(u(i-1)))\boxplus
z(u(i))\right),$ (15)

where $i>0$ and
$I^{\prime}(z(u(0)))=I^{*}(u(0))\oplus\left(I^{*}(u(HW-1))\boxplus
z(u(0))\right)$. Obviously, the keystream used in the diffusion is not
private. Specifically, as for the plain-pixel in position $z(u(i))$, its
corresponding random integer used in the modulo addition is actually
$z(u(i))$.

###  4.2 Chosen-plaintext attack on IEACD

To conceal the relationship between the plain-image and keystream
$\mathbf{Y}$, one can generate a pair of plain-images
$(\mathbf{I}_{0},\mathbf{I}_{1})$, where $\mathbf{I}_{0}$ is a nature image,
$|I_{1}(a)-I_{0}(a)|=1$, $I_{1}(i)=I_{0}(i)$ for $i\in[0,a)$ and $i\in(a,HW)$,
and the index $a$ is any given integer. Denote the intermediate keystreams and
cipher-images corresponding to $(\mathbf{I}_{0},\mathbf{I}_{1})$ by
$(\mathbf{Y}_{0},\mathbf{Y}_{1})$ and
$(\mathbf{I}^{\prime}_{0},\mathbf{I}^{\prime}_{1})$, respectively. Define the
bitwise XOR operation of two plain-images $\mathbf{I}_{0}$ and
$\mathbf{I}_{1}$ as $\mathbf{I}_{0\oplus
1}=\mathbf{I}_{0}\oplus\mathbf{I}_{1}$, where $\mathbf{I}_{0}$ and
$\mathbf{I}_{1}$ are encrypted by the same secret key. For simplicity, define
a sequence $\mathbf{Z}^{\prime}=\\{z^{\prime}(i)\\}_{i=0}^{HW-1}$, where
$z^{\prime}(i)=z(u(i))$ for $i=0\sim(HW-1)$.

####  4.2.1 Determining $\mathbf{Z}$

As $\mathbf{U}$ is known, one can obtain $\mathbf{Z}$ once
$\mathbf{Z}^{\prime}$ is recovered. So, one can attempt to determine
$\mathbf{Z}^{\prime}$ first. Assume $z^{\prime}(b)=a$ and $b>1$ in the
following analysis. Let $I^{*}_{0\oplus 1\oplus z^{\prime}}(u(i-1))$ denote
$(I^{*}_{0}(u(i-1))\boxplus z^{\prime}(i))\oplus(I^{*}_{1}(u(i-1))\boxplus
z^{\prime}(i))$. Based on the analysis in Sec. 4.1, one can assume that
$\mathbf{Y}_{0}=\mathbf{Y}_{1}$. From Eq. (• ‣ 4.1), one has

$I^{*}_{0\oplus 1}(u(i))=I^{*}_{0\oplus 1\oplus z^{\prime}}(u(i-1))\oplus\\\
\begin{cases}0,&\mbox{when $0<i\leq(HW-1)$, $i\neq b$};\\\ I_{0\oplus
1}(z^{\prime}(i-1)),&\mbox{when $i=b$},\end{cases}$ (16)

According to Eq. (16), if $I_{0}^{*}(u(i-1))=I_{1}^{*}(u(i-1))$ and $i\neq b$,
then $I_{0}^{*}(u(i))=I_{1}^{*}(u(i))$. From
$I^{*}(u(0))=I(z^{\prime}(0))\oplus Y(u(0))\oplus\left(C\boxplus
z^{\prime}(0)\right)$, one can deduce $I_{0}^{*}(u(0))=I_{1}^{*}(u(0))$, and
then get $I_{0}^{*}(u(i))=I_{1}^{*}(u(i))$ for $i=1\sim(b-1)$. Consequently,
Eq. (16) can be represented as

$I^{*}_{0\oplus 1}(u(i))=\begin{cases}0,&\mbox{when $0\leq i<b$};\\\
I_{0\oplus 1}(z^{\prime}(i)),&\mbox{when $i=b$};\\\ I^{*}_{0\oplus 1\oplus
z^{\prime}}(u(i-1)),&\mbox{when $b<i\leq HW-1$}.\end{cases}$ (17)

Referring to Eq. (• ‣ 4.1) and XORing the two cipher-images, one has

$I^{\prime}_{0\oplus
1}(z^{\prime}(i))=(I_{0}^{\prime}(z^{\prime}(i-1))\boxplus
z^{\prime}(i))\oplus(I^{\prime}_{1}(z^{\prime}(i-1))\boxplus
z^{\prime}(i))\oplus I_{0\oplus 1}^{*}(u(i)).$

Then, incorporating Eq. (17) into the above equation, one can get

$\displaystyle I^{\prime}_{0\oplus
1}(z^{\prime}(i))=(I_{0}^{\prime}(z^{\prime}(i-1))\boxplus
z^{\prime}(i))\oplus(I^{\prime}_{1}(z^{\prime}(i-1))\boxplus z^{\prime}(i))$
$\displaystyle\oplus$ $\displaystyle I_{0\oplus 1}(z^{\prime}(i)),$ when
$i=b$; (18a) $\displaystyle\oplus$ $\displaystyle 0,$ when $0<i<b$; (18b)
$\displaystyle\oplus$ $\displaystyle I^{*}_{0\oplus 1\oplus
z^{\prime}}(u(i-1)),$ when $b<i\leq(HW-1)$, (18c)

which is the key equation for the attack. As the above equation has three
cases, sequence $\mathbf{Z}^{\prime}$ is likewise divided into three recovery
parts: $z^{\prime}(b-1)$, $\\{z^{\prime}(i)\\}_{i=0}^{b-2}$, and
$\\{z^{\prime}(i)\\}_{i=b+1}^{HW-1}$, which are discussed separately in the
following:

  * •

Determining $z^{\prime}(b-1)$

Incorporating $z^{\prime}(b)=a$ into Eq. (18a), one has

$I^{\prime}_{0\oplus 1}(a)=(I_{0}^{\prime}(z^{\prime}(b-1))\boxplus
a)\oplus(I^{\prime}_{1}(z^{\prime}(b-1))\boxplus a)\oplus I_{0\oplus 1}(a).$
(19)

Enumerating $z^{\prime}(b-1)=j$, one can obtain a set containing all possible
values of $z^{\prime}(b-1)$ via Eq. (19), where $j=0\sim(HW-1)$ and $j\neq a$.
Adopting more known plain-images and the corresponding cipher-images, one can
get more different sets and intersect them, which makes the probability
$z^{\prime}(b-1)$ is correctly determined approach one. Ideally, the
probability is one if and only if the cardinality of the intersection of these
sets is equal to one.

  * •

Determining $\\{z^{\prime}(i)\\}_{i=0}^{b-2}$

Substituting $i$ with $b-1$ in Eq. (18b), one can get

$I^{\prime}_{0\oplus
1}(z^{\prime}(b-1))=(I_{0}^{\prime}(z^{\prime}(b-2))\boxplus
z^{\prime}(b-1))\oplus\\\ (I^{\prime}_{1}(z^{\prime}(b-2))\boxplus
z^{\prime}(b-1)).$ (20)

In the above equation, $z^{\prime}(b-1)$ is determined and only
$z^{\prime}(b-2)$ is unknown. Similar to the recovery of $z^{\prime}(b-1)$,
one can enumerate $z^{\prime}(b-2)=j$ and verify it via Eq. (20), where
$j\in\\{0,1,\cdots,HW-1\\}$ and $j\neq z^{\prime}(b-1)$. As every element in
$\mathbf{Z}^{\prime}$ is unique, the derived element should be recorded and
not used in the following enumeration. Apparently, the elements before
$z^{\prime}(b-2)$ in $\mathbf{Z}^{\prime}$ can be likewise determined via Eq.
(18b). Since $z^{\prime}(0)$ is the first element, this process is naturally
finished when no element can be found using Eq. (18b). After
$\\{z^{\prime}(i)\\}_{i=0}^{b-2}$ is obtained, $b$ is also determined by the
way. In case of $b=HW-1$, $\mathbf{Z}^{\prime}$ is completely recovered. But
in the other cases, the elements after $z^{\prime}(b)$ in
$\mathbf{Z}^{\prime}$ remain undetermined at this moment.

  * •

Determining $\\{z^{\prime}(i)\\}_{i=b+1}^{HW-1}$

Now, determine $z^{\prime}(b+1)$ via Eq. (18c). When $i=b+1$, Eq. (18c)
becomes

$I^{\prime}_{0\oplus
1}(z^{\prime}(b+1))=(I_{0}^{\prime}(z^{\prime}(b))\boxplus
z^{\prime}(b+1))\oplus(I^{\prime}_{1}(z^{\prime}(b))\boxplus
z^{\prime}(b+1))\\\ \oplus(I^{*}_{0}(u(b))\boxplus
z^{\prime}(b+1))\oplus(I^{*}_{1}(u(b))\boxplus z^{\prime}(b+1)).$ (21)

Besides $z^{\prime}(b+1)$, $I^{*}_{0}(u(b))$ and $I^{*}_{1}(u(b))$ are still
unknown in Eq. (21). Since $z^{\prime}(b-1)$ is obtained, they can be
calculated via

$I^{*}(u(b))=I^{\prime}(z^{\prime}(b))\oplus\left(I^{\prime}(z^{\prime}(b-1))\boxplus
z^{\prime}(b)\right),$ (22)

which is derived from Eq. (• ‣ 4.1). Similarly,
$\\{I^{*}_{0}(u(i))\\}_{i=1}^{b-1}$ and $\\{I^{*}_{1}(u(i))\\}_{i=1}^{b-1}$
also can be calculated. Just as determining $\\{z^{\prime}(i)\\}_{i=0}^{b-1}$,
$z^{\prime}(b+1)$ can be confirmed through enumeration and verification via
Eq. (21). Again, one can calculate $I^{*}_{0}(u(b+1))$ and $I^{*}_{1}(u(b+1))$
via Eq. (22), and then determine $z^{\prime}(b+2)$ by Eq. (18c). By this way,
the rest can also be determined one by one in turn.

In the process of determining $\mathbf{Z}^{\prime}$, as for a known or given
element $z^{\prime}(i)$, one attempts to find its neighbor by verifying
whether the corresponding equation holds. Therefore, $\mathbf{Z}^{\prime}$ is
reconstructed by seeking the relative positions of elements. As mentioned
before, after constructing $\mathbf{Z}^{\prime}$, the permutation vector
$\mathbf{Z}$ can be restored via $z(u(i))=z^{\prime}(i)$ for $i=0\sim(HW-1)$.

Now, the attack in case of $b\leq 1$ is discussed. In fact, the special cases
of $b$ can be identified through Eq. (18a) and (18b). If $b=1$, since
$z^{\prime}(b-1)$ is the first element in $\mathbf{Z}^{\prime}$, no element
can be found via Eq. (18b). Hence, one should determine $z^{\prime}(0)$ via
Eq. (18a) and then find the remainder of $\mathbf{Z}^{\prime}$ through Eq.
(18c). If $b=0$, the attack is failed. Since Eq. (18a) and (18b) both no
longer hold, one would directly attempt to determine $z^{\prime}(1)$ through

$I^{\prime}_{0\oplus 1}(z^{\prime}(1))=(I_{0}^{\prime}(z^{\prime}(0))\boxplus
z^{\prime}(1))\oplus(I^{\prime}_{1}(z^{\prime}(0))\boxplus
z^{\prime}(1))\oplus(I^{*}_{0}(u(0))\\\ \boxplus
z^{\prime}(1))\oplus(I^{*}_{1}(u(0))\boxplus z^{\prime}(1)),$

which is derived from Eq. (18c). Here, $I^{*}_{0}(u(0))$ and $I^{*}_{1}(u(0))$
are still unknown, and only $I^{*}_{0\oplus 1}(u(0))=I_{0\oplus
1}(z^{\prime}(0))$ can be obtained from Eq. (17). Apparently, the available
information is insufficient to obtain $I^{*}_{0}(u(0))$ and $I^{*}_{1}(u(0))$,
and then the attack cannot proceed. Such case occurs with a low probability
$\frac{1}{HW}$, which is $1.52\times 10^{-5}$ when $H=W=256$. So it does not
impact the attack much. If it occurs, one just needs to choose a different
index $a^{\prime}$ and generate the corresponding plain-images again.

Next, let us investigate how many plain-images are sufficient to recover
$\mathbf{Z}^{\prime}$ exactly. As there is a strong correlation between
$\mathbf{I}_{0}^{\prime}$ and $\mathbf{I}_{1}^{\prime}$, it is intractable to
estimate it. Therefore, assume that each pixel in $M_{\rm p}$ pairs of cipher-
images follows independently identical distribution in subsequent discussion.
The retrieve process of $\mathbf{Z}^{\prime}$ is similar to the attack method
on permutation-only ciphers given in [17, 18], which attempts to find the sole
exact permutation position from a set containing all possible positions. As
discussed in [17, 18], some minor error elements in permutation matrix have no
much influence on the decryption performance. However, due to the two-round
crossover diffusion, any wrong element in $\mathbf{Z}^{\prime}$ can incur that
the decryption result has no any visual information. In other words, the
error-tolerant rate of $\mathbf{Z}^{\prime}$ for decryption performance is
zero.

When one attempts to determine $z^{\prime}(i-1)$ via Eq. (18b) and
$z^{\prime}(i)$ is known, among 65536 combinations of
$I_{0}^{\prime}(z^{\prime}(i-1))$ and $I_{1}^{\prime}(z^{\prime}(i-1))$, only
256 ones satisfy the equation. And Eq. (18b) should satisfy for $M_{\rm p}$
pairs of cipher-images. Thus, the possibility deriving a wrong value as the
neighbor of a given element is $\frac{1}{256^{M_{\rm p}}}$. As for Eq. (18a)
and (18c), the analysis is similar and the corresponding possibility is the
same. Then, the probability that an element only has sole exact candidate
after enumeration and verification is roughly

$P_{\rm s}=\left(1-\frac{1}{256^{M_{\rm p}}}\right)^{HW-2}.$

Determining $\mathbf{Z}^{\prime}$ exactly relies on three conditions:
$\mathbf{Y}_{0}=\mathbf{Y}_{1}$, $b\neq 0$, and the exact neighbor for each
element can be obtained. Hence, the probability of recovering
$\mathbf{Z}^{\prime}$ exactly can be calculated by
$P_{z}=(1-\frac{1}{HW})\cdot(1-P_{\rm c})^{M_{\rm p}}\cdot(P_{\rm s})^{HW}$.
When $M_{\rm p}=5$ and $H=W=256$, the probability is about 0.993. So it is
expected that $\mathbf{Z}^{\prime}$ can be always recovered exactly when
$M_{\rm p}\geq 5$. The possibility $P_{z}$ can also be regarded as the
attacking success rate. Apparently, the time complexity of the whole
recovering process is $O(M_{\rm p}\cdot(HW)^{2})$.

####  4.2.2 Determining $N_{0}$, $C$, and $Y^{\rm L}$

Now $\mathbf{Z}$, $\\{I^{*}_{0}(u(i))\\}_{i=1}^{HW-1}$, and
$\\{I^{*}_{1}(u(i))\\}_{i=1}^{HW-1}$ are known, only the confusion part is
left. The unknown elements of $\mathbf{I}^{*}_{0}$ and $\mathbf{I}^{*}_{1}$,
$I_{0}^{*}(u(0))$ and $I_{1}^{*}(u(0))$, can be calculated via

$I^{*}(u(0))=I^{\prime}(z^{\prime}(0))\oplus\left(I^{*}(u(HW-1))\boxplus
z^{\prime}(0)\right).$

And the elements of $\mathbf{Y}$ can be recovered through

$Y(u(i))=I^{*}(u(i))\oplus I(z^{\prime}(i))\oplus(I^{*}(u(i-1))\boxplus
z^{\prime}(i))$

except $Y(0)$ ($u(0)=0$). Similar to Sec. 3.3, one can guess $N_{0}$ through
brute-force searching and ignoring the influence of the unrecovered value in
comparison (assuming $Y(0)=Y(N(0))$). Then, one can calculate

$C=\left(I^{*}(u(0))\oplus I(z^{\prime}(0))\oplus Y(u(0))\right)\boxminus
z^{\prime}(0),$

where $a\boxminus b=(a-b+256)\bmod 256$. To decrypt cipher-images completely,
choose a plain-image of fixed value 255 as [13]. Thus, one can calculate the
corresponding sequence $\mathbf{Y}_{255}$ and extract the longest vector
$Y^{\rm L}_{255}$ from the sequence. Finally, the equivalent secret key
$(N_{0},C,\mathbf{Z},Y^{\rm L}_{255})$ can be obtained.

$\mathbf{I}_{0}$

$\mathbf{I}_{0}^{\prime}$

$\mathbf{I}_{1}$

$\mathbf{I}_{1}^{\prime}$

Figure 4: Two similar plain-images and the corresponding cipher-images.

To test the real performance of the preceding chosen-plaintext attack, some
experiments were performed. As [4, 13], the typical secret key is set as
$N_{0}=100$, $C=216$, $\alpha=6$, $m=19.5$, $h=0.1$, $q(0)=0.75$, and
$\beta=3.7488464$. The initial condition $\mathbf{X}(0)$ of the discretized
Ikeda system is a randomly generated vector of length 50. Figure 4
demonstrates two plain-images used for determining $\mathbf{Z}$ and the
corresponding cipher-images, where the index of the changed pixel is 46240. It
is found that the sequence $\mathbf{Z}$ can be recovered with five pairs of
plain-images and the corresponding cipher-images. After $N_{0}$ and $C$ are
determined, $Y^{\rm L}_{255}$ is derived using a plain-image of fixed value
255. Two corresponding intermediate images and the final result decrypted
using the equivalent secret key are shown in Fig. 5. To show the attack
vividly, Table. 1 and 2 list the encryption process of a sample image of size
$4\times 4$ and the corresponding attacking results, respectively.

a)

b)

c)

Figure 5: Two intermediate images and a decrypted image: a) the image removed crossover diffusion; b) the image removed confusion; c) the decrypted result. Table 1: Encryption process of an image of size $4\times 4$ by IEACD. Item | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---
$\mathbf{I}_{0}$ | 7 | 5 | 7 | 3 | 3 | 10 | 0 | 1 | 6 | 3 | 3 | 2 | 3 | 6 | 5 | 6
$\mathbf{Z}$ | 14 | 6 | 4 | 12 | 2 | 15 | 10 | 0 | 8 | 7 | 9 | 1 | 11 | 3 | 5 | 13
$\mathbf{I}_{0}^{\star}$ | 5 | 0 | 3 | 3 | 7 | 6 | 3 | 7 | 6 | 1 | 3 | 5 | 2 | 3 | 10 | 6
$\mathbf{N}$ | 4 | 6 | 6 |  |  |  |  |  |  |  |  |  |  |  |  |
$\mathbf{Y}_{0}$ | 233 | 33 | 101 | 80 | 233 | 33 | 101 | 80 | 187 | 24 | 233 | 33 | 101 | 80 | 187 | 24
$\mathbf{I}_{0}^{\star\star}$ | 236 | 33 | 102 | 83 | 238 | 39 | 102 | 87 | 189 | 25 | 234 | 36 | 103 | 83 | 177 | 30
$\mathbf{I}_{0}^{*}$ | 10 | 148 | 224 | 92 | 149 | 241 | 215 | 58 | 175 | 130 | 3 | 121 | 199 | 167 | 109 | 89
$\mathbf{I}_{0}^{**}$ | 109 | 116 | 29 | 109 | 140 | 174 | 247 | 171 | 218 | 249 | 37 | 23 | 80 | 22 | 145 | 225
$\mathbf{I}^{\prime}_{0}$ | 171 | 23 | 140 | 22 | 29 | 145 | 116 | 249 | 218 | 37 | 247 | 80 | 109 | 225 | 109 | 174
Table 2: The items obtained in an attack. Items | The corresponding value
---|---
$a$ | 7
$z^{\prime}(b-1)$ | 6
$\\{z^{\prime}(i)\\}_{i=0}^{b-2}$ | $14,8$
$b$ | 3
$\\{z^{\prime}(i)\\}_{i=b+1}^{HW-1}$ | $4,9,12,1,2,11,15,3,10,5,0,13$
$\mathbf{I}_{0}^{*}$ | $10,148,227,65,134,34,230,13,175,129,6,102,246,118,90,4$
$\\{Y_{0}(i)\\}_{i=1}^{HW-1}$ | $33,101,80,233,33,101,80,187,24,233,33,101,80,187,24$
$N_{0}$ | 4
$C$ | 216
$Y^{\rm L}$ | $233,33,101,80,187,24,172,246,41,29,12,24$

##  5 Conclusion

This paper analyzed security performance of an image encryption algorithm
based on a first-order time-delay system IEATD and the enhanced version IEACD.
Although another research group proposed a chosen-plaintext attack on IEATD,
we presented an enhanced attack using the correlation between adjacent vectors
of one plain-image and the corresponding cipher-image. Although IEACD is
designed by the attacking group with intention to fix the security defects of
IEATD, there still exist some security pitfalls, such as invalid secret keys,
insensibility of keystream generation mechanics, and improper configuration of
keystream. Based on these, we designed an efficient chosen-plaintext attack
and verified it with extensive experiments. The serious insecurity of the two
algorithms cannot be improved by simple modifications. They can work as
typical counterexamples to remind us to recast scenario-oriented image
encryption algorithms following the guidelines and lessons summarized in [9,
7, 3].

## Acknowledgements

This work was supported by the National Natural Science Foundation of China
(no. 61772447), Scientific Research Fund of Hunan Provincial Education
Department (no. 20C1759), and Science and Technology Program of Changsha (no.
kq2004021).

## References

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###

Sheng Liu is a graduate student in computer science at the School of Computer
Science and Electronic Engineering, Hunan University. His research interests
include image privacy protection and image forensics. Contact him at
<EMAIL_ADDRESS>

Chengqing Li is a professor with the College of Information Engineering,
Xiangtan University, China. His research interests include image privacy
protection and multimedia cryptanalysis. Li received a PhD in electronic
engineering from City University of Hong Kong. He is the corresponding author
of this article. Contact him at<EMAIL_ADDRESS>

Qiao Hu is an assistant professor with the College of Computer Science and
Electronic Engineering, Hunan University, China. His research interests
include RFID security and privacy, cloud computing, and wireless communication
security. Hu received a PhD in information security from City University of
Hong Kong. Contact him at<EMAIL_ADDRESS>


Slipping and rolling on a rough accelerating surface

M. Khorrami1,***Corresponding Author E-mail<EMAIL_ADDRESS>A.
Aghamohammadi1, C. Aghamohammadi2

1 Department of Fundamental Physics, Faculty of Physics, Alzahra University,
Tehran, Iran

2 Cold Spring Harbor Laboratory, Cold Spring Harbor, NY, USA

Keywords: slipping, rolling, friction, inertial force

###### Abstract

The two-dimensional motion of an object on a moving rough horizontal plane is
investigated. Two cases are studied: the plane having a translational
acceleration, and a rotating plane. For the first case, the motions of a point
particle and a sphere are studied, and it is shown that the solution to the
latter problem can be expressed in terms of the solution to the former one.
Examples of constant acceleration and periodic acceleration along a fixed
line, and specifically sinusoidal acceleration along a fixed line, are studied
in more detail. Also a situation is investigated where the friction is
anisotropic, that is the friction coefficient depends on the direction of the
velocity. In this situation, there may be stick-slip motions, and these are
investigated in detail. For the second case, the motion of a point particle on
a rough turntable is investigated.

##  1 Introduction

The one-dimensional motion of a sliding point particle on a rough horizontal
or an inclined plane is among the standard pedagogical problems in mechanics
[1, 2]. In [3, 4], the two-dimensional motion of a particle sliding on an
inclined plane has been studied for a special choice of the friction
coefficient. Although friction theories have been developed for centuries and
numerous situations have been studied in detail, but there are still many
unanswered questions, both in the framework of physics and mechanical
engineering. Among these are extensions to two-dimensions without any
restriction on the geometry or friction, replacing the particle with an
extended object, and adding an acceleration (translational or rotational) to
the substrate.

The two-dimensional motion of a particle sliding on a rough inclined plane has
been investigated in [5]. In [6], the motion of a particle on an arbitrary
surface but confined to a fixed vertical plane has been investigated. The
tangential motion at the contact of two solid objects has been studied in [7],
where it was shown that the friction force and the torque are inherently
coupled. The example studied there, is a disk sliding and spinning on a
horizontal flat surface. It could make the problem easier to study the motion
(on a plane) of symmetric extended objects; for example two-dimensional
symmetric objects such as a disk [8, 9, 10], or a hoop [12, 10]; or three-
dimensional symmetric objects such as a sphere, or a cylinder [13, 14]. Ref.
[15] is among the relatively old books in which some of such problems have
been discussed. The following and the references therein, provide some other
examples.

The dynamics of a steel disc spinning on a horizontal rough surface has been
investigated both experimentally and theoretically in [16]. In [17], the two-
dimensional motion of a generally non-circular non-uniform cylinder on a flat
horizontal surface has been addressed. In [5], the equation of motion for a
sphere on a rough inclined plane has been solved, using the solution of the
equation of motion for a point particle sliding on the same inclined plane
with a different friction coefficient. In [18], the motion of a point particle
sliding on a turntable has been studied. There are some other sources on the
motion of a sphere rolling on a turntable. Examples are [19, 20, 21, 22].

Some of these models have found applications as well. For example in [23], the
results of [5] have been used for the study and analysis of granular materials
including the study of rotation phenomenon using particle methods.

Motions influenced by dry friction, can also result in the stick-slip
phenomenon. There are two modes in the stick-slip phenomenon: the stick mode
occurs when there is no relative motion between the two objects in contact,
and the slip mode occurs when there is a relative motion. This phenomenon is
frequently seen all around us. Examples are squeaking doors, and earthquakes
taking place during periods of rapid slip. Stick-slip phenomena can also
produce sounds through induced vibrations. Examples are drawing chalk on a
blackboard, or moving a violin bow. Stick-slip phenomena are also important in
mechanical engineering and many studies have been devoted to it. Examples are
[25, 26, 24] and references therein. Typically, in the slip-stick phenomenon
for the friction between the involved surfaces the coefficient of static
friction is greater than the coefficient of kinetic friction. But in the
situation studied here this is not the case.

Here a situation is studied that there are two solids directly in contact,
without an intermediate fluid lubrication layer. Friction is taken to be
proportional to the normal force, with the same value for static and dynamic
coefficients. Dry friction imposes the resistance in the relative motion of
two objects in contact.

More specifically, here the two-dimensional motion of an object on an
accelerating rough horizontal plane is investigated. The plane’s acceleration
may be translational or rotational (produced by a turntable, for example).
This paper has essentially two parts. In the first part, sections 2 and 3, the
plane has a translation acceleration and the motion of a point particle or a
homogeneous sphere is studied. It is shown that the solution to the latter
problem can be expressed in terms of the solution to the former. Examples of
constant acceleration and periodic acceleration along a fixed line, and
specifically sinusoidal acceleration along a fixed line, are studied in more
detail. The motion on an inclined plane is equivalent to the motion on a
surface with constant acceleration. So the results obtained here for constant
acceleration are expected to be related to those of [5]. But this does not
apply to the results corresponding to a time-dependent acceleration. In the
case of sinusoidal acceleration along a fixed line, there may be stick-slip
motions, which are investigated in detail. Also a situation is investigated
where the friction is anisotropic, that is the friction coefficient depends on
the direction of the velocity. This leads to a much richer dynamic behavior.
The second part of the article, section 4, is on the motion of a point
particle on a rough turntable. Here the evolution equation is reduced to a set
of three first order coupled differential equations. The large-time behavior
of the system is studied in more detail, and some results are obtained about
the dependence of the large-time behavior on the initial condition;
specifically, which initial conditions result in final rest and which result
in perpetual motion. Finally, section 5 is devoted to the conclusion.

##  2 A point particle sliding on a rough accelerating surface

Consider a horizontal rough plane that is being pulled with the acceleration
$\bm{A}$. A particle of mass $m$ moves on this plane. There is friction
between the particle and the plane, with the coefficient $\mu$ which is
assumed to be the same for static and kinetic frictions. The equation of
motion of the particle, when it is not at rest, in the non-inertial frame of
the accelerating plane is

$\displaystyle\frac{\mathrm{d}\,\bm{v}}{\mathrm{d}\,t}$
$\displaystyle=-\bm{A}-\mu\,g\,\frac{\bm{v}}{v},$ (1) where ${\bm{v}}$ is the
particle’s velocity, $g$ is the gravity acceleration, and $|\bm{v}|$ is
denoted by $v$. Multiplying this equation by $(\bm{v}/v)$, one arrives at
$\displaystyle\frac{\mathrm{d}\,v}{\mathrm{d}\,t}$
$\displaystyle=-\bm{A}\cdot\frac{\bm{v}}{v}-\mu\,g.$ (2)

This time evolution has some general properties:

  * $\bullet$

If the maximum of $|\bm{A}|$ is less than $(\mu\,g)$, the right-hand side of
(2) is negative and the particle’s speed decreases with time so that it will
be at rest in a finite time.

  * $\bullet$

If for any time $t$ there is a larger time at which $|\bm{A}|$ is more than
$(\mu\,g)$, then the particle cannot remain at rest indefinitely.

  * $\bullet$

If the direction of $\bm{A}$ is fixed and $|\bm{A}|$ is always larger than
$(\mu\,g)$, then at sufficiently large times $v$ is increasing with time.

  * $\bullet$

If the direction of $\bm{A}$ is fixed, then $|\bm{v}_{\perp}|$ decreases with
time, where $\bm{v}_{\perp}$ is the part of $\bm{v}$ which is perpendicular to
$\bm{A}$. If $|\bm{A}|$ has an upper bound, then $\bm{v}_{\perp}$ tends to
zero at large times, so that the motion becomes essentially one dimensional.

###  2.1 Constant acceleration

A special case is when the acceleration is a constant vector. One can choose
the axes so that

$\displaystyle\bm{A}$ $\displaystyle=-A\,\hat{\bm{x}},$ (3) where $A$ is
positive. Denoting the angle of $\bm{v}$ with the $x$ axis by $\theta$,
$\displaystyle\bm{v}$
$\displaystyle=v\,(\hat{\bm{x}}\,\cos\theta+\hat{\bm{y}}\,\sin\theta),$ (4)
where $(x,y)$ are Cartesian coordinates, and defining $\lambda$ through
$\displaystyle\lambda$ $\displaystyle=\frac{\mu\,g}{A},$ (5) equation (1)
becomes $\displaystyle\frac{\mathrm{d}\,(v\,\cos\theta)}{\mathrm{d}\,t}$
$\displaystyle=A\,(1-\lambda\,\cos\theta).$ (6)
$\displaystyle\frac{\mathrm{d}\,(v\,\sin\theta)}{\mathrm{d}\,t}$
$\displaystyle=-A\,\lambda\,\sin\theta.$ (7) Eliminating $t$, one arrives at
$\displaystyle\frac{\mathrm{d}\,(v\,\sin\theta)}{\mathrm{d}\,(v\,\cos\theta)}$
$\displaystyle=-\frac{\lambda\,\sin\theta}{1-\lambda\,\cos\theta}.$ (8) So,
$\displaystyle\frac{\mathrm{d}\,v}{v}$
$\displaystyle=\frac{\lambda-\cos\theta}{\sin\theta}\,\mathrm{d}\,\theta,$ (9)
resulting in $\displaystyle
v\,(\sin\theta)\,\left(\tan\frac{\theta}{2}\right)^{-\lambda}$
$\displaystyle=\mbox{constant},$ (10)

or,

$v=v_{0}\,\frac{\sin\theta_{0}}{\sin\theta}\,\left(\frac{\displaystyle{\tan\frac{\theta}{2}}}{\displaystyle{\tan\frac{\theta_{0}}{2}}}\right)^{\lambda}.$
(11)

Putting this in equation (7),

$A\,\mathrm{d}\,t=-\frac{v_{0}\,\sin\theta_{0}}{\displaystyle{\left(\tan\frac{\theta_{0}}{2}\right)^{\lambda}}}\,\frac{\mathrm{d}\,\displaystyle{\left[\left(\tan\frac{\theta}{2}\right)^{\lambda}\right]}}{\lambda\,\sin\theta},$
(12)

resulting in

$t=\frac{v_{0}\,(\lambda+\cos\theta_{0})}{A\,(1-\lambda^{2})}\,\left[\frac{(\lambda+\cos\theta)\,\sin\theta_{0}}{(\lambda+\cos\theta_{0})\,\sin\theta}\,\left(\frac{\displaystyle{\tan\frac{\theta}{2}}}{\displaystyle{\tan\frac{\theta_{0}}{2}}}\right)^{\lambda}-1\right].$
(13)

Using (11) and (13), one arrives at

$v=\frac{(1-\lambda^{2})\,A\,t+v_{0}\,(\lambda+\cos\theta_{0})}{(\lambda+\cos\theta)}.$
(14)

Depending on the value of $\lambda$, different behaviors occur:

  * •

$\lambda=0$.
The surface is friction-less. The $y$-component of the particle’s velocity is
constant, and the $x$-component of the particle’s velocity grows linearly with
time:

$\displaystyle v_{x}$ $\displaystyle=v_{0\,x}+A\,t.$ (15) $\displaystyle
v_{y}$ $\displaystyle=v_{0\,y}.$ (16)

The particle’s trajectory is generally a parabola.

  * •

$\lambda\leq 1$.
The friction is not large enough to keep the particle at rest. At large time,
$\theta$ tends to zero, and the particle slides along the plane’s
acceleration. For the special case $\lambda=1$,

$\displaystyle v\,(1+\cos\theta)$ $\displaystyle=v_{0}\,\,(1+\cos\theta_{0}).$
(17) At large times the particle’s velocity tends to a constant:
$\displaystyle v_{x}(\infty)$
$\displaystyle=\frac{v_{0}\,\,(1+\cos\theta_{0})}{2}.$ (18) $\displaystyle
v_{y}(\infty)$ $\displaystyle=0.$ (19)

  * •

$\lambda>1$.
The friction is large enough to prevent the particle from sliding. The
particle eventually stops at the time T:

$\displaystyle T=\frac{v_{0}\,(\lambda+\cos\theta_{0})}{A\,(\lambda^{2}-1)}.$
(20)

And as this time is approached, $\theta$ tends to zero, so the direction of
the particle’s velocity approaches the $x$ axis.

These can be visualized in figure (1), which is the phase portrait of
$(v_{x},v_{y})$ for different values of $\lambda$. For $\lambda\leq 1$, the
friction isn’t large enough to prevent the particle from sliding: at large
times, the particle slides along $x$ direction (the plane’s acceleration). For
$\lambda>1$, the friction is large enough and the particle eventually comes to
rest, after a finite time. At this point, the direction of the velocity tends
to the $x$ axis.

(a)(b)

Figure 1: The velocity phase portrait for different values of $\lambda$. For
$\lambda\leq 1$, the friction is not large enough to prevent the particle from
sliding. At large times, the particle slides along the plane’s acceleration.
For $\lambda>1$, the friction is large enough and the particle eventually
comes to rest, in a finite time.

These results are consistent with those of ref. [5].

###  2.2 Sinusoidal linear acceleration

Consider a case where the acceleration is sinusoidal and along a fixed line.
The axes are chosen so that

$\displaystyle\bm{A}$ $\displaystyle=-A_{0}\,\sin(\varpi\,t)\,\hat{\bm{x}}.$
(21) At large times, the $y$-component of the velocity vanishes. If $A_{0}$
(which is taken to be positive) is smaller than $(\mu\,g)$, $v_{x}$ vanishes
at large times as well. Here the case is studied that $A_{0}$ is larger than
$(\mu\,g)$. So $v_{x}$ doesn’t tend to zero at large times. The dimensionless
parameters $\lambda_{0}$, $\phi$, and $\mathfrak{w}$ are defined through
$\displaystyle\lambda_{0}$ $\displaystyle=\frac{\mu\,g}{A_{0}}.$ (22)
$\displaystyle\phi$ $\displaystyle=\varpi\,t.$ (23)
$\displaystyle\mathfrak{w}$ $\displaystyle=\frac{\varpi\,v_{x}}{A_{0}}.$ (24)
The assumption that $A_{0}$ is larger than $(\mu\,g)$ means that
$\displaystyle\lambda_{0}$ $\displaystyle<1.$ (25)

At large times, the $v_{y}$ vanishes. If initially $v_{y}$ is much larger than
the $v_{x}$, then initially the friction is approximately along the $y$ axis,
and is constant. That produces a constant acceleration along the $y$ axis
which decreases $v_{y}$ linearly with time. But at some time $v_{y}$ becomes
negligible compared to $v_{x}$. After this, the friction is nearly along the
$x$ axis, with a small $y$-component which is proportional to $v_{y}$. This
results in a slower decrease of $v_{y}$, which is exponential in time.
Regarding $v_{x}$, its large time behavior doesn’t depend on the initial value
of $v_{y}$. But its transient behavior does. Figure 2 shows this behavior for
several initial values of $v_{y}$.

Figure 2: At large times, the velocity becomes parallel to the vector $\bm{A}$
(here parallel to the $x$ axis). The large time behavior of $v_{x}$ doesn’t
depend on the initial value of $v_{y}$. But the transient behavior of $v_{x}$
does depend on the initial value of $v_{y}$. The figure shows the time
dependence of $v_{x}$ for a fixed initial value of $v_{x}$ and several initial
values of $v_{y}$.

The large-time equation for $v_{x}$, when it is not zero, is rewritten as

$\displaystyle\frac{\mathrm{d}\,\mathfrak{w}}{\mathrm{d}\,\phi}$
$\displaystyle=\sin\phi-\lambda_{0}\,\frac{\mathfrak{w}}{|\mathfrak{w}|}.$
(26) $|\sin\phi|$ becomes equal to $\lambda_{0}$ at four points (in each
period). For $\phi$ in $[0,2\,\pi]$, these points are denoted by $\phi_{1}$,
$\phi_{2}$, $\phi_{3}$, and $\phi_{4}$: $\displaystyle\phi_{1}$
$\displaystyle=\sin^{-1}\lambda_{0}.$ (27) $\displaystyle\phi_{2}$
$\displaystyle=\pi-\sin^{-1}\lambda_{0}.$ (28) $\displaystyle\phi_{3}$
$\displaystyle=\pi+\sin^{-1}\lambda_{0}.$ (29) $\displaystyle\phi_{4}$
$\displaystyle=2\,\pi-\sin^{-1}\lambda_{0}.$ (30)

If there are time intervals in which $\mathfrak{w}$ is zero, then those time
intervals would end at $\phi_{1}$ or $\phi_{3}$ (in the first period).
Assuming that $\mathfrak{w}$ is zero at $\phi_{1}$, one arrives at the
following expression for $\mathfrak{w}$, for the interval in which
$\mathfrak{w}$ is positive.

$\mathfrak{w}=-\lambda_{0}\,(\phi-\phi_{1})+\cos\phi_{1}-\cos\phi.$ (31)

This is valid for $\phi$ in $[\phi_{1},\tilde{\phi}_{2}]$, where
$\tilde{\phi}_{2}$ is the smallest value larger than $\phi_{1}$ which
satisfies

$-\lambda_{0}\,(\tilde{\phi}_{2}-\phi_{1})+\cos\phi_{1}-\cos\tilde{\phi}_{2}=0.$
(32)

It is seen that $\tilde{\phi}_{2}$ is larger than $\phi_{2}$ but smaller than
$\phi_{4}$. If $\tilde{\phi}_{2}$ is smaller than $\phi_{3}$, then there is an
interval $[\tilde{\phi}_{2},\phi_{3}]$ in which $\mathfrak{w}$ is zero. Then
$\mathfrak{w}$ is of the following form.

$\mathfrak{w}=\begin{cases}-\lambda_{0}\,(\phi-\phi_{1})+\cos\phi_{1}-\cos\phi,&\phi_{1}<\phi<\tilde{\phi}_{2}\\\
0,&\tilde{\phi}_{2}<\phi<\phi_{3}\\\
\lambda_{0}\,(\phi-\phi_{3})+\cos\phi_{3}-\cos\phi,&\phi_{3}<\phi<\tilde{\phi}_{2}+\pi\\\
0,&\tilde{\phi}_{2}+\pi<\phi<\phi_{1}+2\,\pi\end{cases}.$ (33)

And of course $\mathfrak{w}$ continues periodically in $\phi$, with the period
$(2\,\pi)$. For this behavior to happen, $\tilde{\phi}_{2}$ should be smaller
than $\phi_{3}$, or

$\displaystyle-\lambda_{0}\,(\phi_{3}-\phi_{1})+\cos\phi_{1}-\cos\phi_{3}$
$\displaystyle<0.$ (34) Expressing everything in terms of $\lambda_{0}$, one
arrives at the following condition for $\lambda_{0}$:
$\displaystyle-\pi\,\lambda_{0}+2\,\sqrt{1-\lambda_{0}^{2}}$
$\displaystyle<0,$ (35)

or,

$\displaystyle\lambda_{0}$ $\displaystyle>\lambda_{\mathrm{c}}.$ (36) Where,
$\displaystyle\lambda_{\mathrm{c}}$
$\displaystyle=\frac{2}{\sqrt{4+\pi^{2}}}.$ (37)

If the condition (36) is not satisfied, then the interval
$[\phi_{1},\tilde{\phi}_{2}]$ in which the frictional acceleration is
$(-\mu\,g)$ is more than $\pi$, which is half the period. So whatever the
friction be in the interval $[\tilde{\phi}_{2},\phi+2\,\pi]$, the integral of
the frictional acceleration from $\phi_{1}$ to $(\phi_{1}+2\,\pi)$ is
negative, and as the integral of $A$ in that interval is zero,
$[\mathfrak{w}(\phi_{1}+2\,\pi)]$ would be negative. This means that if (36)
is not satisfied, then there is no periodic solution for $\mathfrak{w}$ which
vanishes on some interval. ($\mathfrak{w}$ still does vanish at some points,
but not on any interval.) The friction is simply too small to do so. In that
case, the large-time behavior of $\mathfrak{w}$ is still periodic, but with
the following form.

$\mathfrak{w}=\begin{cases}-\lambda_{0}\,(\phi-\tilde{\phi}_{1})+\cos\tilde{\phi}_{1}-\cos\phi,&\tilde{\phi}_{1}<\phi<\tilde{\phi}_{1}+\pi\\\
\lambda_{0}\,(\phi-\tilde{\phi}_{1}-\pi)-\cos\tilde{\phi}_{1}-\cos\phi,&\tilde{\phi}_{1}+\pi<\phi<\tilde{\phi}_{1}+2\,\pi\end{cases},$
(38)

where

$\displaystyle\tilde{\phi}_{1}$
$\displaystyle=\cos^{-1}\frac{\pi\,\lambda_{0}}{2}.$ (39) And it is seen that
when this behavior occurs, for which $\displaystyle\lambda_{0}$
$\displaystyle<\lambda_{\mathrm{c}},$ (40) then
$\displaystyle\tilde{\phi}_{1}$ $\displaystyle>\phi_{1}.$ (41)

To summarize, the qualitative large-time behavior of $\mathfrak{w}$ (hence
$v_{x}$) depends on $\lambda_{0}$:

  * $\bullet$

$\lambda_{0}<\lambda_{\mathrm{c}}$.
In this small-friction situation, at each period $v_{x}$ vanishes at exactly
two points.

  * $\bullet$

$\lambda_{\mathrm{c}}<\lambda_{0}<1$.
In this large-friction situation, each period contains two time intervals when
$v_{x}$ is zero.

  * $\bullet$

$1<\lambda_{0}$.
In this very-large-friction situation, $v_{x}$ is identically zero.

Figures 3 and 4 show examples of large-time behavior of $v_{x}$ versus time,
respectively.

$\phi$$\mathfrak{w}$$4\,\pi$$0.8$

Figure 3:  The large-time behavior of $\mathfrak{w}$ (dimensionless $v_{x}$ or
$u_{x}$) versus $\phi$ (dimensionless $t$), for the small-friction case
$\lambda_{0}=\displaystyle{\frac{1}{\pi}}$.

$\phi$$\mathfrak{w}$$4\,\pi$$0.3$

Figure 4:  The large-time behavior of $\mathfrak{w}$ (dimensionless $v_{x}$ or
$u_{x}$) versus $\phi$ (dimensionless $t$), for the large-friction case
$\lambda_{0}=\displaystyle{\frac{1}{\sqrt{2}}}$.

###  2.3 More general periodic linear acceleration

The arguments of subsection 2.2 can qualitatively be repeated for the case the
acceleration is periodic. Consider a function $h$ substituting the sine in the
right hand side of (26). It is assumed that $h$ is periodic with the period
$(2\,\pi)$, and the maximum of $|h|$ is $1$. The aim is to study the large-
time behavior of $\mathfrak{w}$. If $\lambda_{0}$ is larger than $1$ (very
large friction), the particle will always be at rest. For a generic $h$, it is
expected that when $\lambda_{0}$ is slightly less than $1$ (large friction)
the particle is at rest for a big fraction of time, and when $\lambda_{0}$ is
very small (small friction) the velocity doesn’t vanish on some interval. So
there should be a $\lambda_{\mathrm{c}}$ determining the boundary of large and
small friction. The actual situation could be more complicated. There could be
more that one value of $\lambda$ at which such change-of-behaviors occur:
There could be cases when the number of zero-velocity intervals in each period
changes.

As a simple no-so-generic example, consider

$\displaystyle h(\phi)$ $\displaystyle=\begin{cases}1,&0<\phi<\eta\,\pi\\\
0,&\eta\,\pi<\phi<\pi\end{cases}.$ (42) $\displaystyle h(\pi+\phi)$
$\displaystyle=-h(\phi).$ (43)

Where $\eta$ is a constant between $0$ and $1$. It is seen that if
$\lambda_{0}$ is less than $1$ but not very much less than $1$, then

$\displaystyle\mathfrak{w}$
$\displaystyle=\begin{cases}(1-\lambda_{0})\,\phi,&0<\phi<\eta\,\pi\\\
(1-\lambda_{0})\,\eta\,\pi-\lambda_{0}\,(\phi-\eta\,\pi),&\eta\,\pi<\phi<(\eta\,\pi/\lambda_{0})\\\
0,&(\eta\,\pi/\lambda_{0})<\phi<\pi\end{cases}.$ (44)
$\displaystyle\mathfrak{w}(\pi+\phi)$ $\displaystyle=-\mathfrak{w}(\phi).$
(45)

The condition that $\lambda_{0}$ is not very much less than $1$, is that
$\mathfrak{w}$ vanishes before the friction becomes nonzero again. That is,
$\lambda_{0}$ being large than $\lambda_{\mathrm{c}}$, where

$\displaystyle\lambda_{\mathrm{c}}$ $\displaystyle=\eta.$ (46)

This is qualitatively similar to what was found in subsection 2.2. It is seen
that for $\eta=1$ (the acceleration never vanishes), there is no large-
friction phase. And for $\eta=0$ (no acceleration), there is no small-friction
phase.

###  2.4 Anisotropic friction coefficient

Up to now, it has been assumed that the friction coefficient is isotropic.
That is, $\mu$ is a constant, specifically independent of the direction of
$\bm{v}$. Here a situation is studied in which this is not the case. The case
of linear sinusoidal acceleration is reexamined. Again, at large times only
the component of the velocity along the acceleration could be nonzero. Then,
for the large-time behavior equations similar to (21) to (24) are used, except
that (22) is substituted with

$\displaystyle\lambda_{+}$ $\displaystyle=\frac{\mu_{+}\,g}{A_{0}}.$ (47)
$\displaystyle\lambda_{-}$ $\displaystyle=\frac{\mu_{-}\,g}{A_{0}}.$ (48)
where $\mu_{+}$ ($\mu_{-}$) corresponds to positive (negative) $v_{x}$.
$\lambda_{+}$ and $\lambda_{-}$ are reparametrized through
$\displaystyle\lambda_{+}$ $\displaystyle=\lambda_{0}.$ (49)
$\displaystyle\lambda_{-}$ $\displaystyle=q\,\lambda_{0}.$ (50) Then the
equation (26) becomes
$\displaystyle\frac{\mathrm{d}\,\mathfrak{w}}{\mathrm{d}\,\phi}$
$\displaystyle=\sin\phi-\lambda_{0},$ $\displaystyle\quad\mathfrak{w}$
$\displaystyle>0.$ (51)
$\displaystyle\frac{\mathrm{d}\,\mathfrak{w}}{\mathrm{d}\,\phi}$
$\displaystyle=\sin\phi+q\,\lambda_{0},$ $\displaystyle\quad\mathfrak{w}$
$\displaystyle<0.$ (52)

Without loss of generality, one could consider $q$ to be not less than $1$.
Then, arguments similar to those presented for the isotropic case result in
the following regions for the qualitative large-time behavior of
$\mathfrak{w}$, in terms of $(q,\lambda_{0})$. The change of behavior in this
parameter space occurs on the following curves:

A $\displaystyle
K=\frac{\pi\,q}{q+1},\quad\frac{\sqrt{1-\lambda_{0}^{2}}}{\lambda_{0}}$
$\displaystyle=\frac{K}{\sin^{2}K}-\cot K.$ (53) B
$\displaystyle\quad\sqrt{1-\lambda_{0}^{2}}+\sqrt{1-(q\,\lambda_{0})^{2}}$
$\displaystyle=\lambda_{0}\,[\pi+\sin^{-1}(q\,\lambda_{0})-\sin^{-1}(\lambda_{0})].$
(54) C $\displaystyle\quad\lambda_{0}$ $\displaystyle=\frac{1}{q}.$ (55) D
$\displaystyle\quad\lambda_{0}$ $\displaystyle=1.$ (56)

The parameter space is divided into the following regions

  * I

Below A
Each period contains an interval of positive speed and an interval of negative
speed, with no rest intervals.

  * II

Above A and below B and C
Each period contains an interval of positive speed, immediately followed by an
interval of negative speed, and then an interval of rest.

  * III

Between B and C
Each period contains an interval of positive speed, an interval of negative
speed, and two intervals of rest between these.

  * IV

Between C and D
Each period contains an interval of positive speed and an interval of rest.

  * V

Above D
The particle is at rest.

The curves A and B have an intersection at the point $X$, and the curves B and
C have an intersection at the point $Y$. Actually they are tangent to each
other at $Y$:

$\displaystyle X$ $\displaystyle=(q=1,\lambda_{0}=\lambda_{\mathrm{c}}).$ (57)
$\displaystyle Y$ $\displaystyle=(q=4.603,\lambda_{0}=0.217)$ (58)

The regions of the parameter space are illustrated in the figure 5. The large-
time behavior of the velocity in each region is illustrated in figure 6.

$q$$\lambda_{0}$$1$$0.7$$0.3$$0$$1$$3$$5$DCBA$X$$\scriptstyle{\bullet}$$Y$$\scriptstyle{\bullet}$IIIIIIIVV

Figure 5:  The regions in the parameter space of the anisotropic friction,
corresponding to the large-time behavior of the velocity with linear
sinusoidal acceleration. The curve B is the dotted curve.

$\mathfrak{w}$$1$$\phi$$4\,\pi$I

$\mathfrak{w}$$1$$\phi$$4\,\pi$II

$\mathfrak{w}$$1$$\phi$$4\,\pi$III

$\mathfrak{w}$$1$$\phi$$4\,\pi$IV

$\mathfrak{w}$$1$$\phi$$4\,\pi$V

Figure 6:  The large-time behavior of the velocity in each region in the
parameter space of the anisotropic friction, corresponding to the large-time
behavior of the velocity with linear sinusoidal acceleration. Each plot
contains two periods.

##  3 The motion of a sphere on a rough accelerating horizontal surface

In this section, the motion of a homogeneous sphere on a rough accelerating
horizontal plane is investigated. The sphere is of radius $R$ and mass $m$,
and it can roll and slide on the plane. The equation of motion for the sphere
(in the accelerated frame) are

$\displaystyle m\,\frac{\mathrm{d}^{2}\,\bm{r}}{\mathrm{d}\,t^{2}}$
$\displaystyle=-m\,\bm{A}+\bm{F},$ (59) $\displaystyle
I\,\frac{\mathrm{d}\,\bm{\omega}}{\mathrm{d}\,t}$
$\displaystyle=\bm{R}\times\bm{F},$ (60) where $\bm{r}$ is the two-dimensional
position of the center of the sphere, $\bm{\omega}$ is the angular velocity of
the sphere, $I$ is the moment of inertia of the sphere, $\bm{F}$ is the
friction force, and $\displaystyle\bm{R}$ $\displaystyle=-R\,\hat{\bm{z}}.$
(61) The $z$ axis is normal to the plane and upward. Defining $\kappa$ through
$\displaystyle\kappa$ $\displaystyle=\frac{I}{m\,R^{2}},$ (62) the equations
of motion for the case the sphere slids become
$\displaystyle\frac{\mathrm{d}^{2}\,\bm{r}}{\mathrm{d}\,t^{2}}$
$\displaystyle=-\bm{A}-\mu\,g\,\left|\frac{\mathrm{d}\,\bm{r}}{\mathrm{d}\,t}+\bm{\omega}\times\bm{R}\right|^{-1}\,\left(\frac{\mathrm{d}\,\bm{r}}{\mathrm{d}\,t}+\bm{\omega}\times\bm{R}\right).$
(63)
$\displaystyle\kappa\,R^{2}\,\frac{\mathrm{d}\,\bm{\omega}}{\mathrm{d}\,t}$
$\displaystyle=-\mu\,g\,\bm{R}\times\left\\{\left|\frac{\mathrm{d}\,\bm{r}}{\mathrm{d}\,t}+\bm{\omega}\times\bm{R}\right|^{-1}\,\left(\frac{\mathrm{d}\,\bm{r}}{\mathrm{d}\,t}+\bm{\omega}\times\bm{R}\right)\right\\}.$
(64)

So,

$\displaystyle\frac{\mathrm{d}\,(\bm{\omega}\times\bm{R})}{\mathrm{d}\,t}$
$\displaystyle=-\frac{\mu\,g}{\kappa}\,\left|\frac{\mathrm{d}\,\bm{r}}{\mathrm{d}\,t}+\bm{\omega}\times\bm{R}\right|^{-1}\,\left(\frac{\mathrm{d}\,\bm{r}}{\mathrm{d}\,t}+\bm{\omega}\times\bm{R}\right).$
(65)
$\displaystyle\frac{\mathrm{d}}{\mathrm{d}\,t}\,\left(\frac{\mathrm{d}\,\bm{r}}{\mathrm{d}\,t}+\bm{\omega}\times\bm{R}\right)$
$\displaystyle=-\bm{A}-b\,\mu\,g\,\left|\frac{\mathrm{d}\,\bm{r}}{\mathrm{d}\,t}+\bm{\omega}\times\bm{R}\right|^{-1}\,\left(\frac{\mathrm{d}\,\bm{r}}{\mathrm{d}\,t}+\bm{\omega}\times\bm{R}\right),$
(66)

where,

$\displaystyle b$ $\displaystyle=1+\frac{1}{\kappa}.$ (67)

Using $R$ and $(\mu\,g)$, one can make the quantities dimensionless:

$\displaystyle\bm{\rho}$ $\displaystyle=\frac{\bm{r}}{R}.$ (68)
$\displaystyle\tau$ $\displaystyle=\sqrt{\frac{\mu\,g}{R}}\,t.$ (69)
$\displaystyle\bm{\zeta}$
$\displaystyle=\sqrt{\frac{R}{\mu\,g}}\,\bm{\omega}.$ (70)
$\displaystyle\bm{u}$
$\displaystyle=\frac{1}{\sqrt{\mu\,g\,R}}\,\left(\frac{\mathrm{d}\,\bm{r}}{\mathrm{d}\,t}+\bm{\omega}\times\bm{R}\right).$
(71) $\displaystyle\bm{\gamma}$ $\displaystyle=\frac{\bm{A}}{\mu\,g}.$ (72)

Denoting differentiation with respect to $\tau$ by dot, and denoting
$|\bm{u}|$ by $u$, the equations of motion become

$\displaystyle\ddot{\bm{\rho}}$ $\displaystyle=-\bm{\gamma}-\frac{\bm{u}}{u}.$
(73) $\displaystyle\dot{\bm{u}}$
$\displaystyle=-\bm{\gamma}-b\,\frac{\bm{u}}{u}.$ (74) Of course one also has
$\displaystyle\bm{\zeta}$
$\displaystyle=\hat{\bm{z}}\,\zeta_{3}+\hat{\bm{z}}\times(\bm{u}-\dot{\bm{\rho}}).$
(75)

And it is seen that $\zeta_{3}$ is a constant and does not enter the evolution
of other parameters.

Defining $\bm{\nu}$ as

$\displaystyle\dot{\bm{\nu}}$ $\displaystyle=\bm{\gamma},$ (76) one arrives at
$\displaystyle(\dot{\bm{\rho}}+\bm{\nu})^{\dot{}}$
$\displaystyle=-\frac{\bm{u}}{u}.$ (77)
$\displaystyle(\bm{u}+\bm{\nu})^{\dot{}}$
$\displaystyle=-b\,\frac{\bm{u}}{u}.$ (78) So, $\displaystyle\dot{\bm{\rho}}$
$\displaystyle=-\left(1-\frac{1}{b}\right)\,\bm{\nu}+\frac{\bm{u}}{b}+\bm{c},$
(79)

where $\bm{c}$ is a constant vector. So the problem of finding the velocity
and the angular velocity of the sphere as a function of time, is reduced to
finding $\bm{u}$ as a function of time. In other words to investigate the
motion of a sphere on a rough accelerating surface with the friction
coefficient $\mu$ is equivalent to study the motion of a particle on a rough
surface with the friction coefficient $(b\,\mu)$ and the same acceleration.

###  3.1 Constant acceleration

Consider a special choice that $\bm{A}$ is a constant vector. The axes are
chosen like (3). As noted before, the problem of finding $\bm{u}$ is the same
as the problem of finding $\bm{v}$ for the motion of a particle (without
rotation), but with $\mu$ replaced by $(b\,\mu)$. So one can use the results
for $\bm{v}$ to obtain $\bm{u}$. Denoting the angle of $\bm{u}$ with the $x$
axis by $\theta$, and defining $\lambda$ through

$\displaystyle\lambda$ $\displaystyle=\frac{b\,\mu\,g}{A},$ (80) one arrives
at $\displaystyle u$
$\displaystyle=u_{0}\,\frac{\sin\theta_{0}}{\sin\theta}\,\left(\frac{\displaystyle{\tan\frac{\theta}{2}}}{\displaystyle{\tan\frac{\theta_{0}}{2}}}\right)^{\lambda}.$
(81) $\displaystyle\tau$
$\displaystyle=\frac{u_{0}\,(\lambda+\cos\theta_{0})}{1-\lambda^{2}}\,\left[\frac{(\lambda+\cos\theta)\,\sin\theta_{0}}{(\lambda+\cos\theta_{0})\,\sin\theta}\,\left(\frac{\displaystyle{\tan\frac{\theta}{2}}}{\displaystyle{\tan\frac{\theta_{0}}{2}}}\right)^{\lambda}-1\right].$
(82) $\displaystyle u$
$\displaystyle=\frac{(1-\lambda^{2})\,\tau+u_{0}\,(\lambda+\cos\theta_{0})}{(\lambda+\cos\theta)}.$
(83)

###  3.2 Sinusoidal linear acceleration

The (dimensionless) time evolution of $\bm{u}$ is the same as the
(dimensionless) time evolution of $\bm{v}$ as discussed in subsection 2.2,
except that here $\mu$ should be substituted with $(b\,\mu)$, so that here

$\lambda_{0}=\frac{b\,\mu\,g}{A_{0}}.$ (84)

Also, here a vanishing $\bm{u}$ means that the sphere rolls without slipping.
So the qualitative large-time behavior can be summarized as:

  * $\bullet$

$\lambda_{0}<\lambda_{\mathrm{c}}$.
In this small-friction situation, at each period $u_{x}$ vanishes at exactly
two points. At these points rolling occurs.

  * $\bullet$

$\lambda_{\mathrm{c}}<\lambda_{0}<1$.
In this large-friction situation, each period contains two time intervals when
$u_{x}$ is zero. At these intervals rolling occurs.

  * $\bullet$

$1<\lambda_{0}$.
In this very-large-friction situation, $u_{x}$ is identically zero. This means
that in this case, eventually the motion of the sphere will be rolling without
slipping.

Figures 3 and 4 show examples of large-time behavior of the dimensionless
$u_{x}$ versus time, respectively.

###  3.3 More general periodic linear acceleration

Similar to the subsection 3.2, one can bring the arguments of subsection 2.3
here, with $\lambda_{0}$ now defined like (84). Also, for the example of $h$
defined as (42) and (43), the boundary of small and large friction is
$\lambda_{\mathrm{c}}$, defined in (46).

###  3.4 Anisotropic friction coefficient

Again, the (dimensionless) time evolution of $\bm{u}$ is the same as the
(dimensionless) time evolution of $\bm{v}$ as discussed in subsection 2.4,
except that here $\mu$ should be substituted with $(b\,\mu)$, so that here

$\lambda_{0}=\frac{b\,\mu_{+}\,g}{A_{0}}.$ (85)

##  4 A point particle sliding on a rough turntable

In this section the motion of a point particle on a turntable of infinite
extent is investigated. It is assumed that $\bm{\Omega}$, the angular
frequency of the rotation of the table, is perpendicular to the plane of the
table and is constant. The equation of motion in the rotating frame is

$\displaystyle\frac{\mathrm{d}^{2}\,\bm{r}}{\mathrm{d}\,t^{2}}$
$\displaystyle=\Omega^{2}\,\bm{r}+2\,\Omega\,\frac{\mathrm{d}\,\bm{r}}{\mathrm{d}\,t}\times\hat{\bm{z}}-\mu\,g\,\left|\frac{\mathrm{d}\,\bm{r}}{\mathrm{d}\,t}\right|^{-1}\,\frac{\mathrm{d}\,\bm{r}}{\mathrm{d}\,t}.$
(86)

From now on, it is assumed that $\Omega$ is positive. This is no loss of
generality, as any solution to the above equation with the sign of $\Omega$
changed, is the mirror-reflected of a solution of the original problem. Using
$\Omega$ and $(\mu\,g)$, two dimensionless quantities $\bm{\rho}$ and $\tau$
are defined:

$\displaystyle\bm{\rho}$ $\displaystyle=\frac{\Omega^{2}}{\mu\,g}\,\bm{r}.$
(87) $\displaystyle\tau$ $\displaystyle=\Omega\,t.$ (88) The equation of
motion becomes $\displaystyle\ddot{\bm{\rho}}$
$\displaystyle=\bm{\rho}+2\,\dot{\bm{\rho}}\times\hat{\bm{z}}-\frac{\dot{\bm{\rho}}}{|\dot{\bm{\rho}}|},$
(89)

where dot means differentiation with respect to $\tau$. One has

$\displaystyle\dot{\bm{\rho}}\cdot\ddot{\bm{\rho}}$
$\displaystyle=\bm{\rho}\cdot\dot{\bm{\rho}}-|\dot{\bm{\rho}}|.$ (90)
$\displaystyle\hat{\bm{z}}\cdot\bm{\rho}\times\ddot{\bm{\rho}}$
$\displaystyle=-2\,\bm{\rho}\cdot\dot{\bm{\rho}}-\frac{\hat{\bm{z}}\cdot\bm{\rho}\times\dot{\bm{\rho}}}{|\dot{\bm{\rho}}|}.$
(91)
$\displaystyle\bm{\rho}\cdot\ddot{\bm{\rho}}+\dot{\bm{\rho}}\cdot\dot{\bm{\rho}}$
$\displaystyle=\bm{\rho}\cdot\bm{\rho}+2\,\hat{\bm{z}}\cdot\bm{\rho}\times\dot{\bm{\rho}}-\frac{\bm{\rho}\cdot\dot{\bm{\rho}}}{|\dot{\bm{\rho}}|}+\dot{\bm{\rho}}\cdot\dot{\bm{\rho}}.$
(92)

The length of $\dot{\bm{\rho}}$ is denoted by $p$:

$\displaystyle p$ $\displaystyle=|\dot{\bm{\rho}}|,$ (93) and $\psi$ is
defined as the angle of the position vector with respect to the velocity
vector, counterclockwise. Further, $\xi$ and $\ell$ are defined as
$\displaystyle\xi$ $\displaystyle=\rho\,\cos\psi.$ (94) $\displaystyle\ell$
$\displaystyle=-\rho\,\sin\psi.$ (95) So, $\displaystyle p\,\ell$
$\displaystyle=\hat{\bm{z}}\cdot\bm{\rho}\times\dot{\bm{\rho}},$ (96)
$\displaystyle p\,\xi$ $\displaystyle=\bm{\rho}\cdot\dot{\bm{\rho}},$ (97)
Then, using $\displaystyle\bm{\rho}\cdot\bm{\rho}$
$\displaystyle=\ell^{2}+\xi^{2},$ (98) one arrives at $\displaystyle\dot{p}$
$\displaystyle=\xi-1.$ (99) $\displaystyle\dot{\ell}$
$\displaystyle=-\frac{\xi\,(2\,p+\ell)}{p}.$ (100) $\displaystyle\dot{\xi}$
$\displaystyle=\frac{(p+\ell)^{2}}{p}.$ (101)

Equation (99) is the projection of Newton’s equation along the particle’s
velocity (or trajectory). Equation (100) is the projection of Newton’s law
along the azimuthal direction (the equation of change for the angular
momentum), combined with the evolution equation for $u$. Equation (101) is the
projection of Newton’s law along the radial direction, combined with the
evolution equation for $u$. Equations (99), (100), and (101) are three coupled
differential equations, governing the evolution of $p$, $\ell$, $\xi$. The
system contains a constant of motion. One has

$\displaystyle\rho\,\dot{\rho}$
$\displaystyle=\ell\,\dot{\ell}+\xi\,\dot{\xi},$ $\displaystyle=\xi\,p,$
$\displaystyle=p\,(\dot{p}+1).$ (102) So,
$\displaystyle\left(\frac{p^{2}-\rho^{2}}{2}\right)^{\dot{}}$
$\displaystyle=-p,$ (103) or $\displaystyle\frac{p^{2}-\rho^{2}}{2}+s$
$\displaystyle=\mbox{constant},$ (104) where $s$ is the arc-length parameter:
$\displaystyle\dot{s}$ $\displaystyle=p.$ (105)

Re-dimensionalizing (104), one arrives at an equation proportional to

$\frac{m}{2}\,\left(\frac{\mathrm{d}\,\bm{r}}{\mathrm{d}\,t}\right)\cdot\left(\frac{\mathrm{d}\,\bm{r}}{\mathrm{d}\,t}\right)-\frac{m\,\Omega^{2}\,r^{2}}{2}+m\,g\,s=\mbox{constant}.$
(106)

The first term is the kinetic energy of the particle in the non-inertial frame
of the turntable, the second term is the potential energy associated to the
centrifugal force, and $(-m\,g\,s)$ is the work done by the friction. It is
noted that the Coriolis force does no work. So the above equation is the work-
energy theorem in the non-inertial frame.

Expressing $\ell$ and $\xi$ in terms of $\rho$ and $\psi$, the evolution
equations become

$\displaystyle\dot{p}$ $\displaystyle=\rho\,\cos\psi-1.$ (107)
$\displaystyle\dot{\rho}$ $\displaystyle=p\,\cos\psi.$ (108)
$\displaystyle\dot{\psi}$
$\displaystyle=2-\left(\frac{p}{\rho}+\frac{\rho}{p}\right)\,\sin\psi.$ (109)
One can also express $\rho$ and $p$ in terms of hyperbolic parameters. There
are two cases. Either, $\displaystyle p^{2}-\rho^{2}$ $\displaystyle<0.$ (110)
$\displaystyle p$ $\displaystyle=A\,\sinh\alpha.$ (111) $\displaystyle\rho$
$\displaystyle=A\,\cosh\alpha.$ (112) Then, $\displaystyle\dot{A}$
$\displaystyle=\sinh\alpha.$ (113) $\displaystyle\dot{\alpha}$
$\displaystyle=\cos\psi-\frac{\cosh\alpha}{A}.$ (114)
$\displaystyle\dot{\psi}$ $\displaystyle=2\,[1-(\sin\psi)\,\coth(2\,\alpha)].$
(115) Or, $\displaystyle p^{2}-\rho^{2}$ $\displaystyle>0.$ (116)
$\displaystyle p$ $\displaystyle=B\,\cosh\beta.$ (117) $\displaystyle\rho$
$\displaystyle=B\,\sinh\beta.$ (118) Then, $\displaystyle\dot{B}$
$\displaystyle=-\cosh\beta.$ (119) $\displaystyle\dot{\beta}$
$\displaystyle=\cos\psi+\frac{\sinh\beta}{B}.$ (120) $\displaystyle\dot{\psi}$
$\displaystyle=2\,[1-(\sin\psi)\,\coth(2\,\beta)].$ (121)

The pair $(B,\beta)$ can be related to the pair $(A,\alpha)$ through

$\displaystyle A$ $\displaystyle=-\mathrm{i}\,B.$ (122) $\displaystyle\alpha$
$\displaystyle=\frac{\mathrm{i}\,\pi}{2}+\beta.$ (123)

Consider the first case (real $\alpha$). The evolution of $\psi$ has two
quasi-fixed points:

$\displaystyle\psi_{1}$ $\displaystyle=\sin^{-1}[\tanh(2\,\alpha)].$ (124)
$\displaystyle\psi_{2}$ $\displaystyle=\pi-\sin^{-1}[\tanh(2\,\alpha)].$ (125)

These are not actual fixed points, as $\alpha$ is not constant. However,
$\psi_{1}$ is attractive while $\psi_{2}$ is repulsive. It is seen that $A$
increases indefinitely, unless $\alpha$ tends to zero. But if $\alpha$ is near
zero, then $\psi$ changes rapidly towards its attractive quasi-fixed point
$\psi_{1}$, which is near zero, if $\alpha$ is near zero. Then the evolution
of $\alpha$ becomes

$\displaystyle\dot{\alpha}$ $\displaystyle\approx 1-\frac{1}{A}.$ (126) If
$\displaystyle A$ $\displaystyle>1,$ (127)

then $\alpha$ will increase and ceases to be near zero. So the particle’s
speed will never vanish, and $A$ will further increase. $A$ never decreases.
So $A$ remains bigger than $1$, if its initial value is bigger than $1$, and
in that case $A$ increases indefinitely, and the particle’s speed never
vanishes.

Assuming that (127) holds, initially, one can find the asymptotic behavior of
the variables at large times. Approximating $\psi$ by its attractive quasi-
fixed point, one arrives at

$\dot{\alpha}\approx\frac{1}{\cosh(2\,\alpha)}-\frac{\cosh\alpha}{A}.$ (128)

$\alpha$ cannot remain finite, because in this case the right-hand side will
eventually become positive (as $A$ increases). So both $A$ and $\alpha$ should
increase indefinitely. A further approximation is to take $\alpha$ to be the
quasi-fixed point of its evolution:

$(\cosh\alpha)\,\cosh(2\,\alpha)\approx A.$ (129)

For large values of $A$, and hence $\alpha$, this becomes

$\alpha\approx\frac{1}{3}\,\ln(4\,A).$ (130)

So,

$\dot{A}\approx\frac{(4\,A)^{1/3}}{2},$ (131)

leading to

$\displaystyle A$ $\displaystyle\sim\frac{2\,\tau^{3/2}}{\sqrt{27}}.$ (132)
$\displaystyle\alpha$ $\displaystyle\sim\frac{1}{2}\,\ln\frac{4\,\tau}{3}.$
(133) $\displaystyle\psi$
$\displaystyle\sim\frac{\pi}{2}-\left(\frac{3}{\tau}\right)^{1/2}.$ (134)
$\displaystyle p$ $\displaystyle\sim\frac{2\,\tau^{2}}{9}.$ (135)
$\displaystyle\rho$ $\displaystyle\sim\frac{2\,\tau^{2}}{9}.$ (136)
$\displaystyle s$ $\displaystyle\sim\frac{2\,\tau^{3}}{27}.$ (137)

As a check, it is seen that

$\frac{A^{2}}{2}\sim s.$ (138)

Now consider cases when the particle eventually comes to rest. The condition
that the particle remain at rest, at $\rho=\sigma$, is

$\sigma\leq 1.$ (139)

The reason is that at $\rho>1$, the maximum of the static frictional force is
less than the centrifugal force. When the particle nears its rest, $p$ tends
to zero, so $\alpha$ tends to zero. Then from the evolution equation for
$\psi$ it is seen that $\psi$ tends to zero as well. So near the rest and
before that,

$\displaystyle\dot{\alpha}$ $\displaystyle=1-\frac{1}{\sigma}+\cdots,$ (140)
which results in $\displaystyle\alpha$
$\displaystyle=\frac{(\sigma-1)\,\tau}{\sigma}+\cdots.$ (141)

Putting this in the evolution equation for $\psi$, and noting that $\alpha$
and $\psi$ are both small, one arrives at

$\displaystyle\frac{\sigma-1}{\sigma}\,\frac{\mathrm{d}\,\psi}{\mathrm{d}\,\alpha}$
$\displaystyle\approx 2\,\left(1-\frac{\psi}{2\,\alpha}\right),$ (142) which
results in $\displaystyle\psi$
$\displaystyle=2\,\alpha\,\left[\frac{\sigma}{2\,\sigma-1}+c\,\alpha^{(2\,\sigma-1)/(1-\sigma)}\right]+\cdots,$
(143) where $c$ is an integration constant. Similarly, the evolution equation
for $A$ becomes
$\displaystyle\frac{\sigma-1}{\sigma}\,\frac{\mathrm{d}\,A}{\mathrm{d}\,\alpha}$
$\displaystyle\approx\alpha,$ (144) resulting in $\displaystyle A$
$\displaystyle=\sigma+\frac{\sigma}{\sigma-1}\,\frac{\alpha^{2}}{2}+\cdots.$
(145)

For each value of $\sigma$, equations (143) and (145) represent a surface
$\mathbb{S}_{\sigma}$ in the $3$-dimensional parameter space
$(A,\alpha,\psi)$. The envelope $\mathbb{S}$ of these surfaces is the boundary
between two regions of the parameter space: the region $\mathbb{V}_{0}$
corresponding to the initial values which result in an eventual rest, and the
region $\mathbb{V}_{1}$ corresponding to the initial values which result in
unbounded motions. To obtain the equation of $\mathbb{S}$, one notices that
the equation of $\mathbb{S}_{\sigma}$ is just (145), with $\psi$ being free.
Hence the parametric equation for the envelope consists of (145) and its
derivative with respect to the parameter $\sigma$:

$\displaystyle 0$
$\displaystyle=1-\frac{1}{(\sigma-1)^{2}}\,\frac{\alpha^{2}}{2}+\cdots.$ (146)
Eliminating $\sigma$ between this and (145), the equation for $\mathbb{S}$ is
determined as $\displaystyle A$
$\displaystyle=\left(1-\frac{\alpha}{\sqrt{2}}\right)^{2}+\cdots.$ (147)

The particle eventually comes to rest, if initially the value of $A$ is less
that the right-hand side. Otherwise its motion would be unbounded.

##  5 Conclusion

Even though friction-based problems in classical mechanics have been studied
extensively, there is still a substantial number of new researchs, with
compelling results. The problems stadied here were the two-dimensional motion
of a particle and a homogeneous sphere on a moving rough horizontal plane. It
was shown that the problem of the motion of a homogeneous sphere on a moving
plane with translational acceleration is reduced to that of the motion of a
point particle. Some examples were studied in more detail: constant
acceleration, periodic (and specifically sinusoidal) acceleration along a
fixed line, and a situation where the friction is anisotropic, which leads to
a much richer dynamical behavior. The motion of a point particle on a rough
turntable was also studied. It was shown that the evolution equation is
reduced to a set of three coupled first order differential equations. The
large-time behavior of the system was studied in more detail, and some results
were obtained about the dependence of the large-time behavior on the initial
condition; specifically, which initial conditions result in final rest and
which result in perpetual motion.
Acknowledgment: The work of M. Khorrami and A. Aghamohammadi was supported by
the research council of the Alzahra University.

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