An amusing trick is to press a finger down on a marble on a horizontal table top, in such a way that the marble is projected along the table with an initial linear speed $\mathit{v}$and an initial backward rotational speed $\mathit{\omega }$about a horizontal axis perpendicular to $\mathit{v}$. The coefficient of sliding friction between marble and top is constant. The marble has radius $\mathit{R}$. (a) if the marble slides to a complete stop, What was $\mathit{\omega }$in terms of $\mathit{v}$and$\mathit{R}$? (b) if the marble, skids to a stop, and then starts returning toward its initial position, with a final constant speed of $\mathbf{(}\mathbf{3}\mathbf{/}\mathbf{7}\mathbf{)}\mathit{v}\mathbf{,}$What was $\mathit{\omega }$in terms of $\mathit{v}$and$\mathit{R}$? Hint for part (b): when the marble rolls without slipping, the relationship between speed and angular speed is$\mathit{v}\mathbf{=}\mathit{\omega }\mathit{R}$.

The rate of change of a central angle corresponding to time of a rotating body is defined as angular speed.

The multiplication of mass and the square of a distance of the particle from the rotation axis are known as the moment of inertia.

The diagram represents a marble acquired translational velocity towards right and rotational speed in anti-clockwise (backward) direction about an axis through CM and normal to plane of the paper.

Net linear velocity of the bottom most point P of the marble in contact with floor is $v_{net}=v+R\omega$ towards right. As a result, the marble has sliding motion also towards right. Due to this, frictional force acts on the marble towards left.

This sliding friction is,

$f_k=\mu mg$

Here, $m$is mass of the marble.

The acceleration (retardation) due to sliding friction is,

$a=\frac{f_k}{m}$

Substitute $\mu mg$for $f_k$.

$\begin{array}{rcl}a& =& \frac{\mu mg}{m}\\ & =& \mu g\\ & & \end{array}$

$\begin{array}{rcl}a& =& \frac{\mu mg}{m}\\ & =& \mu g\\ & & \end{array}$

The sliding friction produces torque in clockwise direction. Hence, the torque acting on it is

localid="1668497242060" $\tau =f_{k}R$

Substitute $\mu mgR$for $f_k$.

localid="1668497256868" $\tau =\mu mgR$

The angular acceleration (retardation) is,

$\alpha =\frac{\tau }{I}$

The moment of inertia of the marble is,

$I=\frac{2}{5}m{R}^2$

Substitute all these values in the equation $\alpha =\frac{\tau }{I}$ and simplify.

$\begin{array}{rcl}\alpha & =& \frac{\mu mgR}{\frac{2}{5}m{R}^2}\\ & =& \frac{5\mu g}{2R}\\ & & \end{array}$

The marble slides to a complete stop. This indicates that both translation and rotational final velocities of the marble are zero.

For translation motion, the final velocity of the marble is as follows.

$v_f=v_i+at$

Here, initial and final velocities of the marble are.

$\begin{array}{rcl}d{u}_i& =& u\\ u_f& =& 0\end{array}$

The acceleration of the marble is,

$a=-\mu g$

Substitute all these values in the equation $v_f=v_i+at$ and simplify.

Use the kinematic equation for the rotatory motion, we have

${\omega }_f={\omega }_i+\alpha t$

Here, the initial and final angular velocities of the marble is,

$\begin{array}{rcl}{\omega }_i& =& \omega \\ {\omega }_f& =& 0\\ & & \end{array}$

The angular acceleration of the marble is,

$\alpha =\frac{-5\mu gt}{2R}$

Substitute all this data in the equation ${\omega }_f={\omega }_i+\alpha t$ and simplify.

$\begin{array}{rcl}0& =& \omega -\frac{5\mu gt}{2R}\\ \omega & =& \frac{5\mu gt}{2R}\\ & & \end{array}$

Substitute $\mu gt$for$v$.

$\omega =\frac{5v}{2R}$

Therefore, the value is$\omega =\frac{5v}{2R}$.

$a=-\mu g$

In this case, when translational velocity becomes zero, at time rotational velocity will not be zero.

For translational motion, the final velocity of the marble is as follows.

$v_f=v_i+a{t}_1$

Here, initial and final velocity of the marbles are,

$$\begin{gathered} u_i=u \\ {u}_f=0 \end{gathered}$$

The acceleration of the marble is,

$a=-\mu g$

Substitute all these value in the equation $v_f=v_i+at$and simplify.

$\begin{array}{rcl}0& =& v-\mu g{t}_1\\ v& =& \mu g{t}_1\\ & & \end{array}$

Substitute $\frac{-5v}{2R}$for $\alpha t_1$in the equation ${\omega }_1={\omega }_i+\alpha t_1$and simplify.

${\omega }_1=\omega -\frac{5v}{2R}$

At time $t_1$,At even though translation velocity is zero, due to rotational velocity, the marble will have sliding motion at P. Hence, friction still acting towards left. Due to this friction, the marble acquires translational velocity towards left. The torque due to this friction reduces further the angular velocity.

Here, initial and final velocities of the marble are,

$\begin{array}{rcl}{v}_i& =& 0\\ v_f& =& \frac{3v}{7}\\ & & \end{array}$

The acceleration of the marble is,

$a=\mu g$

Substitute all these values in the equation $v_f=v_i+a{t}_2$ and simplify.

$\begin{array}{rcl}\frac{3v}{7}& =& 0+\mu g{t}_2\\ \frac{3v}{7}& =& \mu g{t}_2\\ & & \end{array}$

The initial angular velocity of the marble is,

$\begin{array}{rcl}{\omega }_i& =& {\omega }_1\\ & =& \omega -\frac{5v}{2R}\\ & & \end{array}$

For rotational motion, we have

${\omega }_2={\omega }_i+\alpha t_2$

Substitute all these values in the above equation, we get

${\omega }_2=\omega -\frac{5v}{2R}-\frac{5\mu g}{2R}{t}_2$

Substitute $\frac{3v}{7}$ for$\mu g{t}_2$.

$\begin{array}{rcl}{\omega }_2& =& \omega -\frac{5v}{2R}-\frac{5v}{2R}·\frac{3v}{7}\\ & =& \omega -\frac{25v}{7R}\\ & & \\ & & \end{array}$

The translational speed attains constant value when the body acquires pure rolling. In pure rolling,

$\begin{array}{rcl}\frac{3v}{7}& =& R{\omega }_2\\ & =& R\left[\omega -\frac{25v}{7R}\right]\\ \frac{3v}{7}& =& R\omega -\frac{25v}{7}\\ R\omega & =& \frac{3v}{7}+\frac{25v}{7}\\ & =& \frac{28v}{7}\\ R\omega & =& 4v\\ \omega & =& \frac{4v}{R}\\ & & \end{array}$

In the case, when the marble acquires pure rolling $\omega$is $\frac{4v}{R}$.

$v_f=v_i+a{t}_1$