Chapter 9, Problem 106P

(a)

To determine

The angular speed of the ball just after the blow.

Answer to Problem 106P

  ${\omega }_0=-\frac{5v_0}{6R}$

Explanation of Solution

Given: The horizontal force applied on a billiard ball at a distance $h=\frac{2R}{3}$ , below the centerline

Speed of the ball just after the blow is $v_0$

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

Formula Used:

Newton’s second law of motion in rotational form

  $\sum \tau =r\times F=I_{cm}\alpha$

Calculation:

  

FIGURE: 1

Applying Newton’s second law in rotational form to ball,

  $\sum {\tau }_0=F_{av}\left(h-R\right)=I_{cm}\alpha =I_{cm}\frac{{\omega }_0}{\Delta t}$

  $\Rightarrow F_{av}\left(h-R\right)=I_{cm}\frac{{\omega }_0}{\Delta t}$

  ${\omega }_0=\frac{F_{av}\left(h-R\right)\Delta t}{I_{cm}}$  $\left(1\right)$

Where,

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

  $F_{av}$ is the average force on the ball

  $h$ is the below the centerline at which the force is applied on the ball

  $R$ is radius of the ball

  $I_{cm}$ is the moment of inertia with respect to an axis through the center of mass of the ball

  $\alpha$ is angular acceleration of the ball

  ${\omega }_0$ is angular speed of the ball after impact

  $\Delta t$ is elapsed time

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

  $I_{cm}=\frac{2}{5}m{r}^2$

Substituting this in equation $\left(1\right)$ ,

  ${\omega }_0=\frac{F_{av}\left(h-r\right)\Delta t}{\frac{2}{5}m{r}^2}$  $\left(2\right)$

Applying impulse-momentum theorem to the ball,

  $I=F_{av}\Delta t=\Delta p=m{v}_0$

  $\Rightarrow m{v}_0=F_{av}\Delta t$

  $v_0=\frac{F_{av}\Delta t}{m}$  $\left(3\right)$

Where, $m$ is mass of the ball

  $\Delta p$ is momentum, and

  $v_0$ is speed of the ball just after the impact

From equation $\left(3\right)$ , $v_0=\frac{F_{av}\Delta t}{m}\Rightarrow \Delta t=\frac{m{v}_0}{F_{av}}$

Substituting the expression for $\Delta t$ in equation $\left(3\right)$ ,

  $\begin{array}{l}{\omega }_0=\frac{F_{av}\left(h-R\right)\frac{m{v}_0}{F_{av}}}{\frac{2}{5}m{R}^2}=\frac{5v_0\left(h-R\right)}{2R^2}\\ \end{array}$

Substituting $h=\frac{2R}{3}$ in the above equation,

  ${\omega }_0=\frac{5v_0\left(\frac{2R}{3}-R\right)}{2R^2}=\frac{5v_0\left(\frac{2R-3R}{3}\right)}{2R^2}=\frac{5v_0\left(\frac{-R}{3}\right)}{2R^2}=-\frac{5v_0}{6R}$

Conclusion:

The angular speed of the ball just after the blow is $-\frac{5v_0}{6R}$ .

(b)

To determine

The speed of the ball once it begins to roll without slipping.

Answer to Problem 106P

  $v=\frac{5}{21}{v}_0$

Explanation of Solution

Given: Speed of the ball just after the blow is $v_0$

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

Formula Used:

  

FIGURE: 2

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

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

  $\Sigma F_y=F_n-mg=0$  $\left(5\right)$

And

  $\Sigma F_x=-f_k=ma$  $\left(6\right)$

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

  $I_{cm}$ is the moment of inertia with respect to an axis through the center of mass of the ball

  $\alpha$ is angular acceleration of the ball

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

But, $f_k={\mu }_k{F}_n$  $\left(7\right)$

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

Calculations:

From equation $\left(5\right)$ , $F_n=mg$

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

  $f_k={\mu }_{k}mg$

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

  $\alpha =\frac{f_{k}R}{I_{cm}}$

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

  $I_{cm}=\frac{2}{5}m{R}^2$

Substituting for $f_k$ and $I_{cm}$ ,

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

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

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

Now,substituting the expression for $f_k$ in equation $\left(6\right)$ ,

  $-{\mu }_{k}mg=ma\Rightarrow a=-{\mu }_{k}g$  $\left(9\right)$

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

Substituting for $a$ from equation $\left(9\right)$ in equation $\left(10\right)$ ,

  $v=v_0-{\mu }_{k}g\Delta t$  $\left(11\right)$

Imposing the condition for rolling the ball without slipping,

  $\begin{array}{l}R\left({\omega }_0+\frac{5{\mu }_{k}g}{2R}\Delta t\right)=v_0-{\mu }_{k}g\Delta t\\ \Rightarrow \Delta t=\frac{16}{21}\frac{v_0}{{\mu }_{k}g}\end{array}$

Substituting this $\Delta t$ in equation $\left(11\right)$ ,

  $v=v_0-{\mu }_{k}g\left(\frac{16}{21}\frac{v_0}{{\mu }_{k}g}\right)=\frac{5}{21}{v}_0$

Conclusion:

The speed of the ball once it begins to roll without slipping is $\frac{5}{21}{v}_0$ .

(c)

To determine

The kinetic energy of the ball just after the hit.

Answer to Problem 106P

  $K_i=\frac{23m{v}_0^2}{36}$

Explanation of Solution

Given: The horizontal force applied on a billiard ball at a distance $h=\frac{2R}{3}$ , below the centerline.

Speed of the ball just after the blow is $v_0$ .

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

Formula Used:

Initial kinetic energy of the ball can be written as,

  $K_i=K_{trans}+K_{rot}$  $\left(12\right)$

Where, $K_{trans}$ is kinetic energy due to translational motion of the ball, $K_{trans}=\frac{1}{2}m{v}_0^2$

  $K_{rot}$ is kinetic energy due to rotational energy of the ball, $K_{rot}=\frac{1}{2}I{\omega }_0^2$

Substituting the expressions for $K_{trans}$ and $K_{rot}$ in equation $\left(12\right)$ ,

  $K_i=\frac{1}{2}m{v}_0^2+\frac{1}{2}I{\omega }_0^2$  $\left(13\right)$

Where, $m$ is mass of the ball,

  $v_0$ is speed of the ball just after impact

  $I$ is moment of inertia of the ball

  ${\omega }_0$ is angular speed of the ball

Moment of inertia, $I=\frac{2}{5}m{R}^2$ , and Angular speed, ${\omega }_0=-\frac{5v_0}{6R}$

Substituting these in equation $\left(13\right)$ ,

  $K_i=\frac{1}{2}m{v}_0^2+\frac{1}{2}\left(\frac{2}{5}m{R}^2\right){\left(-\frac{5v_0}{6R}\right)}^2$

  $=\frac{1}{2}m{v}_0^2+\frac{1}{5}m{R}^2\times \frac{25}{36}\frac{v_0^2}{R^2}$

  $=\frac{1}{2}m{v}_0^2+\frac{5}{36}m{v}_0^2$

  $=\frac{18m{v}_0^2+5m{v}_0^2}{36}=\frac{23m{v}_0^2}{36}$

Conclusion:

The kinetic energy of the ball just after the hit is $\frac{23m{v}_0^2}{36}$ .