A bowler throws a bowling ball of radius $\mathit{R}\mathbf{}\mathbf{}\mathbf{=}\mathbf{11}\mathbf{}\mathbf{\text{cm}}$along a lane. The ball (in figure) slides on the lane with initial speed ${\mathit{v}}_{\mathbf{c}\mathbf{o}\mathbf{m}\mathbf{,}\mathbf{0}}\mathbf{}\mathbf{}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{\text{m/s}}$and initial angular speed ${\mathit{\omega }}_{\mathbf{0}}\mathbf{}\mathbf{=}\mathbf{0}$. The coefficient of kinetic friction between the ball and the lane is. The kinetic frictional force$\overrightarrow{f_k}$acting on the ball causes a linear acceleration of the ball while producing a torque that causes an angular acceleration of the ball. When speed ${\mathit{v}}_{\mathbf{c}\mathbf{o}\mathbf{m}}$ has decreased enough and angular speed v has increased enough, the ball stops sliding and then rolls smoothly.

(a) What then is ${\mathit{v}}_{\mathbf{c}\mathbf{o}\mathbf{m}}$in terms ofV? During the sliding (b) What is the ball’s linear acceleration (c) What is the angular acceleration? (d) How long does the ball slide? (e) How far does the ball slide? (f) What is the linear speed of the ball when smooth rolling begins?

$$\begin{gathered} R=11\text{cm} \\ {v}_{com,0}=8.5\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \\ {\omega }_0=0 \\ {\mu }_k=0.21 \end{gathered}$$

The problem is based on the equations and relations of kinematics as per transverse and angular relations of the motion in transverse and rotational motion respectively. When a body is revolving in a circular path, a torque is applied to it due to a tangential force. Here, the force acting tangentially is the frictional force acting on the body due to its motion.

Formulae:

The transverse velocity of the body,

$V_{com}=R\omega$

Where,R is the radius of the circle,is the angular velocity.

Frictional force,

$f_k={\mu }_{k}mg$

Where, m is the mass andis the acceleration due to gravity

Frictional force,${\alpha }_k=m{a}_{com}$

Where, m is the mass and$a_{com}$is the acceleration of center of mass.

The torque applied on a body due to its moment of inertia,

$\tau =I\times \alpha$

Where, m is the moment of inertia of the body,is the angular acceleration of the body.

The torque applied due to tangential force,

$\tau =F_f\times R$

Where,$F_f$is the tangential force and R is the radius of the circular path.

The final angular velocity according to analogy of rotational motion,

$\omega ={\omega }_0+\alpha t$

Where,ω₀ is the initial angular velocity and α is the angular acceleration and t is the time taken.

The horizontal distance travelled by the body according to third law of kinematics,

$\Delta x=v_{com}t+\frac{1}{2}a{t}^2$

where, $v_{com}$is the velocity of center of mass, a is the acceleration, t is the time taken.

We know the relation between linear and angular velocity$V_{com}=R\omega$

Taking clockwise motion as negative and anti-clockwise motion as positive,$V_{com}=-R\omega$

$V_{com}=\left(-0.11\text{m}\right)\omega$

We know that the frictional force acting on a body in accelerated motion is given as:

$$\begin{gathered} \text{frictionalforce}={\mu }_{k}mg \\ =m{a}_{com} \end{gathered}$$

Frictional force is points left, so we write

$\begin{array}{rcl}{a}_{com}& =& -{\mu }_{k}g\dots \dots \dots \left(\text{i}\right)\\ & =& -\left(0.21\right)\left(9.8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.\right)\\ & =& -2.06\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.\end{array}$

Negative value of acceleration shows that ball is decelerating.

By the definition of torque,

$\begin{array}{rcl}\tau & =& I\times \alpha \\ & =& F_f\times R\\ I\times \alpha & =& \left({\mu }_{k}mg\right)\times R\\ & & \end{array}$

Rearranging it for$\alpha$we get

$\alpha =\frac{\left({\mu }_{k}mg\right)\times R}{I}$

In addition, moment of inertial of ball is$I=\frac{2}{5}M{R}^2$
$\begin{array}{rcl}\alpha & =& \frac{\left({\mu }_{k}mg\right)\times R}{\frac{2}{5}M{R}^2}\\ & =& \frac{\left(-5{\mu }_{k}g\right)}{2R}\\ & =& \frac{\left({\displaystyle -5\times \left(0.21\right)\times \left(9.8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^2}$}\right.\right)}\right)}{{\displaystyle 2\left(0.11\text{m}\right)}}\\ & =& -47\raisebox{1ex}{$rad$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.\end{array}$

Negative value of angular acceleration suggests that it is in clockwise direction.

Velocity of ball decelerates and angular velocity increases from zero to

$\begin{array}{rcl}\omega & =& {\omega }_0+\alpha t\\ & =& 0+\alpha t\\ & =& \frac{\left(5{\mu }_{k}g\right)}{2R}t\\ & & \end{array}$

Also,$\omega =\frac{V_{com}}{R}$

So,
$\begin{array}{rcl}\frac{v_{com}}{R}& =& \frac{\left(5{\mu }_{k}g\right)}{2R}t\\ t& =& \frac{2v_{com}}{7\mu g}\\ & =& \frac{2\left(8.5 {\displaystyle \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}\right)}{7\left(0.21\right)\left(9.8{\displaystyle \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.}\right)}\\ & =& 1.2sec\end{array}$

Distance travelled by the ball during $t=1.2\text{s}$

$\begin{array}{rcl}∆x& =& V_{com}t-\frac{1}{2}\left(\mu g\right)t^2\\ & =& \left(8.5\text{m/s}\right)\left(1.2\text{s}\right)-\frac{1}{2}\left(0.21\right)\left(9.8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.\right){\left(1.2\text{s}\right)}^2\\ & =& 8.6m\end{array}$

We can find linear velocity of the ball using kinematic equation as,

$v=v_{com}+at$

From equation (i), we can write,

$\begin{array}{rcl}v& =& v_{com}-\mu gt\\ & =& \left(8.5\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right)-\left(0.21\right)\left(9.8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.\right)\left(1.2 \text{s}\right)\\ & =& 6.1\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\end{array}$