Chapter 9, Problem 107P

(a)

To determine

The speed of the ball as it begins rolling without slipping.

Answer to Problem 107P

  $v=\frac{11}{7}{v}_0$

Explanation of Solution

Given:

Radius of the ball is $R$

Initial speed of the ball is $v_0$

The coefficient of kinetic friction between the ball and billiard table is ${\mu }_k$

Forward spin of the ball just after its release is ${\omega }_0=\frac{3v_0}{R}$

Formula Used:

  

FIGURE: 1

Constant acceleration equation that relates the speed of the ball to the acceleration and time,

  $v=v_0+a\Delta t$  $\left(1\right)$

Where,

  $v$ is speed of the ball

  $v_0$ is speed of the ball just after impact

  $a$ is the acceleration ball

  $\Delta t$ is elapsed time

Referring to the force diagram shown in figure 1, applying Newton’s second law to the ball when it is rolling without slipping,

  $\Sigma F_x=f_k=ma$  $\left(2\right)$

  $\Sigma F_y=F_n-mg=0$  $\left(3\right)$

And

  $\Sigma {\tau }_0=-f_{k}R=I_0\alpha$  $\left(4\right)$

Where,

  $F_x$ is force in the X-direction

  $f_k$ is force of friction in X-direction

  $F_y$ is force in the Y-direction

  $f_n$ is force of friction in Y-direction

  $m$ is mass of the ball

  $a$ is the acceleration of the ball

  $g$ is acceleration due to gravity

  $R$ is radius of the ball

  ${\tau }_0$ is torque about the center of the ball

  $I_0$ is the moment of inertia with respect to an axis through the center of the ball

  $\alpha$ is angular acceleration of the ball

  $g$ is acceleration due to gravity, $g=9.8m/s^2$

Calculation:

  $f_k={\mu }_k{F}_n$  $\left(5\right)$

Where, ${\mu }_k$ is coefficient of kinetic friction

From equation $\left(3\right)$ , $F_n=mg$

Substituting this in equation $\left(5\right)$ ,

  $f_k={\mu }_{k}mg$

Now,substituting the expression for $f_k$ in equation $\left(2\right)$ ,

  ${\mu }_{k}mg=ma\Rightarrow a={\mu }_{k}g$  $\left(6\right)$

Substituting for $a$ from equation $\left(6\right)$ in equation $\left(1\right)$ ,

  $v=v_0+{\mu }_{k}g\Delta t$  $\left(7\right)$

From equation

  $\left(4\right)$ angular acceleration,

  $\alpha =\frac{-f_{k}R}{I_0}$

Moment of inertia with respect to an axis through the center of the ball is

  $I_0=\frac{2}{5}m{R}^2$

Substituting for $f_k$ and $I_0$ ,

  $\alpha =\frac{-{\mu }_{k}mgR}{\frac{2}{5}m{R}^2}=-\frac{5{\mu }_{k}g}{2R}$

Now let us write constant-acceleration equation that connects angular speed of the ball to the angular acceleration and time,

  $\omega ={\omega }_0+\alpha \Delta t={\omega }_0-\frac{5{\mu }_{k}g}{2R}\Delta t$  $\left(8\right)$

Imposing the condition for rolling the ball without slipping,

  $\begin{array}{l}v=R\omega =R\left({\omega }_0-\frac{5{\mu }_{k}g}{2R}\Delta t\right)\\ \end{array}$

Substituting for ${\omega }_0$ ,

  $\Rightarrow v=R\left(\frac{3v_0}{R}-\frac{5{\mu }_{k}g}{2R}\Delta t\right)$

  $\Rightarrow v=3v_0-\frac{5{\mu }_{k}g}{2}\Delta t$  $\left(9\right)$

Now equate the equations $\left(7\right)$ and $\left(9\right)$ ,

  $v_0+{\mu }_{k}g\Delta t=3v_0-\frac{5{\mu }_{k}g}{2}\Delta t$

  $\Rightarrow {\mu }_{k}g\Delta t+\frac{5{\mu }_{k}g}{2}\Delta t=3v_0-v_0$

  $\Rightarrow \frac{7{\mu }_{k}g}{2}\Delta t=2v_0\Rightarrow \Delta t=\frac{4v_0}{7{\mu }_{k}g}$

Substituting this $\Delta t$ in equation $\left(7\right)$ ,

  $v=v_0+{\mu }_{k}g\left(\frac{4}{7}\frac{v_0}{{\mu }_{k}g}\right)=\frac{11}{7}{v}_0$

Conclusion:

The speed of the ball as it begins rolling without slipping is $\frac{11}{7}{v}_0$ .

(b)

To determine

The time the ball moves before it begins to rolling without slipping .

Answer to Problem 107P

  $\Delta t=\frac{4v_0}{7{\mu }_{k}g}$

Explanation of Solution

Given:

Radius of the ball is $R$

Initial speed of the ball is $v_0$

The coefficient of kinetic friction between the ball and billiard table is ${\mu }_k$

Forward spin of the ball just after its release is ${\omega }_0=\frac{3v_0}{R}$

Formula Used:

  

FIGURE: 2

Constant acceleration equation that relates the speed of the ball to the acceleration and time,

  $v=v_0+a\Delta t$  $\left(10\right)$

Where,

  $v$ is speed of the ball

  $v_0$ is speed of the ball just after impact

  $a$ is the acceleration ball

  $\Delta t$ is elapsed time

Referring to the force diagram shown in figure 2, applying Newton’s second law to the ball when it is rolling without slipping,

  $\Sigma F_x=f_k=ma$  $\left(11\right)$

  $\Sigma F_y=F_n-mg=0$  $\left(12\right)$

And

  $\Sigma {\tau }_0=-f_{k}R=I_0\alpha$  $\left(13\right)$

Where,

  $F_x$ is force in the X-direction

  $f_k$ is force of friction in X-direction

  $F_y$ is force in the Y-direction

  $f_n$ is force of friction in Y-direction

  $m$ is mass of the ball

  $a$ is the acceleration of the ball

  $g$ is acceleration due to gravity

  $R$ is radius of the ball

  ${\tau }_0$ is torque about the center of the ball

  $I_0$ is the moment of inertia with respect to an axis through the center of the ball

  $\alpha$ is angular acceleration of the ball

  $g$ is acceleration due to gravity, $g=9.8{\text{m/s}}^{\text{2}}$

Calculation:

  $f_k={\mu }_k{F}_n$  $\left(14\right)$

Where, ${\mu }_k$ is coefficient of kinetic friction

From equation $\left(12\right)$ , $F_n=mg$

Substituting this in equation $\left(14\right)$ ,

  $f_k={\mu }_{k}mg$

Now,substituting the expression for $f_k$ in equation $\left(11\right)$ ,

  ${\mu }_{k}mg=ma\Rightarrow a={\mu }_{k}g$  $\left(15\right)$

Substituting for $a$ from equation $\left(15\right)$ in equation $\left(10\right)$ ,

  $v=v_0+{\mu }_{k}g\Delta t$  $\left(16\right)$

From equation $\left(13\right)$ angular acceleration,

  $\alpha =\frac{-f_{k}R}{I_0}$

Moment of inertia with respect to an axis through the center of the ball is

  $I_0=\frac{2}{5}m{R}^2$

Substituting for $f_k$ and $I_0$ ,

  $\alpha =\frac{-{\mu }_{k}mgR}{\frac{2}{5}m{R}^2}=-\frac{5{\mu }_{k}g}{2R}$

Now let us write constant-acceleration equation that connects angular speed of the ball to the angular acceleration and time,

  $\omega ={\omega }_0+\alpha \Delta t={\omega }_0-\frac{5{\mu }_{k}g}{2R}\Delta t$  $\left(17\right)$

Imposing the condition for rolling the ball without slipping,

  $\begin{array}{l}v=R\omega =R\left({\omega }_0-\frac{5{\mu }_{k}g}{2R}\Delta t\right)\\ \end{array}$

Substituting for ${\omega }_0$ ,

  $\Rightarrow v=R\left(\frac{3v_0}{R}-\frac{5{\mu }_{k}g}{2R}\Delta t\right)$

  $\Rightarrow v=3v_0-\frac{5{\mu }_{k}g}{2}\Delta t$  $\left(18\right)$

Now equate the equations $\left(16\right)$ and $\left(18\right)$ ,

  $v_0+{\mu }_{k}g\Delta t=3v_0-\frac{5{\mu }_{k}g}{2}\Delta t$

  $\Rightarrow {\mu }_{k}g\Delta t+\frac{5{\mu }_{k}g}{2}\Delta t=3v_0-v_0$

  $\Rightarrow \frac{7{\mu }_{k}g}{2}\Delta t=2v_0\Rightarrow \Delta t=\frac{4v_0}{7{\mu }_{k}g}$

Conclusion:

The time the ball moves before it begins to rolling without slipping $\Delta t=\frac{4v_0}{7{\mu }_{k}g}$ .

(c)

To determine

The distance slide down the lane by the ball before it begins rolling without slipping.

Answer to Problem 107P

  $\Delta x=\left(\frac{36v_0^2}{49{\mu }_{k}g}\right)$

Explanation of Solution

Given:

Radius of the ball is $R$

Initial speed of the ball is $v_0$

The coefficient of kinetic friction between the ball and billiard table is ${\mu }_k$

Forward spin of the ball just after its release is ${\omega }_0=\frac{3v_0}{R}$

Calculation:

Let $\Delta x$ be the distance slide down the lane by the ball before it begins rolling without slipping.

Now let us write expression that relates $\Delta x$ to the average speed of the ball $\left(v_{av}\right)$ and the time it moves before beginning to roll without slipping $\left(\Delta t\right)$ ,

  $\Delta x=v_{av}\Delta t$  $\left(19\right)$

Average speed of the ball is,

  $v_{av}=\frac{1}{2}\left(v_0+v\right)$

Substituting this average speed in equation $\left(19\right)$ ,

  $\Delta x=\frac{1}{2}\left(v_0+v\right)\Delta t$

Substituting for $v$ and $\Delta t$ from part $\left(a\right)$ and $\left(b\right)$ ,

  $\Delta x=\frac{1}{2}\left(v_0+\frac{11}{7}{v}_0\right)\left(\frac{4v_0}{7{\mu }_{k}g}\right)$

  $=\frac{1}{2}\left(\frac{18}{7}{v}_0\right)\left(\frac{4v_0}{7{\mu }_{k}g}\right)$

  $=\left(\frac{36v_0^2}{49{\mu }_{k}g}\right)$

Conclusion:

The distance slide down the lane by the ball before it begins rolling without slipping is $\frac{36v_0^2}{49{\mu }_{k}g}$ .