91Ó°ÊÓ

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

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,

$V_{com}=R\mathrm{Ó\neg }$

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

Frictional force,

$f_k={\mathrm{μ}}_{k}mg$

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

Frictional force,${\mathrm{Î\pm }}_k=m{a}_{com}$

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,

$\mathrm{Ï„}=I×\mathrm{Î\pm }$

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

The torque applied due to tangential force,

$\mathrm{τ}=F_f×R$

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,

$\mathrm{Ó\neg }={\mathrm{Ó\neg }}_0+\mathrm{Î\pm }t$

Where,Ó¬₀ 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,

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

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$V_{com}=R\mathrm{Ó\neg }$

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

$V_{com}=\left(-0.11\text{m}\right)\mathrm{Ó\neg }$

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

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

Frictional force is points left, so we write

$\begin{array}{rcl}{a}_{com}& =& -{\mathrm{μ}}_{k}g………\left(\text{i}\right)\\ & =& -\left(0.21\right)\left(9.8\raisebox{1ex}{$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,

$\begin{array}{rcl}\mathrm{Ï„}& =& I×\mathrm{Î\pm }\\ & =& F_f×R\\ I×\mathrm{Î\pm }& =& \left({\mathrm{μ}}_{k}mg\right)×R\\ & & \end{array}$

Rearranging it for$\mathrm{Î\pm }$we get

$\mathrm{Î\pm }=\frac{\left({\mathrm{μ}}_{k}mg\right)×R}{I}$

In addition, moment of inertial of ball is$I=\frac{2}{5}M{R}^2$
$\begin{array}{rcl}\mathrm{Î\pm }& =& \frac{\left({\mathrm{μ}}_{k}mg\right)×R}{\frac{2}{5}M{R}^2}\\ & =& \frac{\left(-5{\mathrm{μ}}_{k}g\right)}{2R}\\ & =& \frac{\left({\displaystyle -5×\left(0.21\right)×\left(9.8\raisebox{1ex}{$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

$\begin{array}{rcl}\mathrm{Ó\neg }& =& {\mathrm{Ó\neg }}_0+\mathrm{Î\pm }t\\ & =& 0+\mathrm{Î\pm }t\\ & =& \frac{\left(5{\mathrm{μ}}_{k}g\right)}{2R}t\\ & & \end{array}$

Also,$\mathrm{Ó\neg }=\frac{V_{com}}{R}$

So,
$\begin{array}{rcl}\frac{v_{com}}{R}& =& \frac{\left(5{\mathrm{μ}}_{k}g\right)}{2R}t\\ t& =& \frac{2v_{com}}{7\mathrm{μ}g}\\ & =& \frac{2\left(8.5  {\displaystyle \raisebox{1ex}{$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 $t=1.2\text{s}$

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

$\begin{array}{rcl}v& =& v_{com}-\mathrm{μ}gt\\ & =& \left(8.5\raisebox{1ex}{$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$}\right.\end{array}$