Chapter 17.3, Problem 17.133P

In a game of pool, ball A is rolling without slipping with a velocity ${\bar{v}}_0$ as it hits obliquely ball B, which is at rest. Denoting by r the radius of each ball and by µ_k the coefficient of kinetic friction between a ball and the table, and assuming perfectly elastic impact, determine (a) the linear and angular velocity of each ball immediately after the impact, (b) the velocity of ball B after it has started rolling uniformly.

Fig. P17.133

To determine

The linear and angular velocity of each ball immediately after the impact.

Answer to Problem 17.133P

The linear velocity of ball A immediately after the impact is $\underset{\_}{\left(v_0\sin \theta \right)j}$.

The angular velocity of ball A immediately after the impact is $\underset{\_}{\frac{v_0}{r}\left(-\sin \theta i+\cos \theta j\right)}$.

The linear velocity of ball B immediately after the impact is $\underset{\_}{\left(v_0\cos \theta \right)i}$.

The angular velocity of ball B immediately after the impact is $\underset{\_}{0}$.

Explanation of Solution

Given information:

The mass of the each ball is m.

The radius of the each ball is r.

The velocity of the ball A before the impact is $v_0$.

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

Calculation:

Write the equation of moment of inertia $\left(I_{AB}\right)$ of each ball using the equation:

$\overline{I}=\frac{2}{5}m{r}^2$

Analyze the impact of ball A.

Here, G is the mass center of ball A.

Consider the conservation of momentum principle.

${\text{Syst Momenta}}_{\text{1}}+{\text{ Syst Ext Imp}}_{\text{1y2 }}={\text{ Syst Momenta}}_{\text{2}}$

Sketch the impulse and momentum diagram of the ball A as shown in Figure 1.

Here, ${\omega }_0$ is the angular velocity of ball A before the impact, $v_0$ is the velocity of ball A before the impact, $P$ is the impulsive reaction, $dt$ is the time interval, ${\omega }_A$ is the angular velocity of ball A immediately after the impact, and $v_A$ is the velocity of ball A immediately after the impact.

Refer Figure (1).

Consider the kinematics in position 1.

Write the equation of angular velocity $\left({\omega }_0\right)$ of ball A before the impact using kinematics.

${\omega }_0=\frac{v_0}{r}$

Consider the horizontal components of forces.

$m{v}_0\cos \theta -{\displaystyle \int Pdt}=m{\left(v_A\right)}_x$ (1)

Consider the vertical components of forces.

$m{v}_0\sin \theta +0=m{\left(v_A\right)}_y$ (2)

Take moments about y axis.

$\overline{I}{\omega }_0\cos \theta +0=\overline{I}{\omega }_A\cos \beta$ (3)

Take moments about x axis.

$-\overline{I}{\omega }_0\sin \theta +0=-\overline{I}{\omega }_A\sin \beta$ (4)

Analyze the impact of ball B.

Here, G is the mass center of ball B.

Consider the conservation of momentum principle.

${\text{Syst Momenta}}_{\text{1}}+{\text{ Syst Ext Imp}}_{\text{1y2 }}={\text{ Syst Momenta}}_{\text{2}}$

Sketch the impulse and momentum diagram of the ball B as shown in Figure (2).

Here, $P$ is the impulsive reaction, $dt$ is the time interval, ${\omega }_B$ is the angular velocity of ball B immediately after the impact, and $v_B$ is the velocity of ball B immediately after the impact.

Refer Figure (2),

Consider the horizontal components of forces.

${\displaystyle \int Pdt}=m{\left(v_B\right)}_x$ (5)

Consider the vertical components of forces.

$0+0=m{\left(v_B\right)}_y$ (6)

Take moments about y axis.

$0+0=\overline{I}{\omega }_B\cos \gamma$ (7)

Take moments about x axis.

$0+0=\overline{I}{\omega }_B\sin \gamma$ (8)

Add Equations (1) and (5).

$\begin{array}{l}m{v}_0\cos \theta -{\displaystyle \int Pdt}+{\displaystyle \int Pdt}=m{\left(v_A\right)}_x+m{\left(v_B\right)}_x\\ m{v}_0\cos \theta =m{\left(v_A\right)}_x+m{\left(v_B\right)}_x\\ {\left(v_B\right)}_x+{\left(v_A\right)}_x=v_0\cos \theta \end{array}$ (9)

The impact is perfectly plastic. Therefore ,the coefficient of restitution (e) is 1.

Consider the condition of impact equation.

${\left(v_B\right)}_x-{\left(v_A\right)}_x=e{v}_0\cos \theta$

Substitute 1 for e.

$\begin{array}{l}{\left(v_B\right)}_x-{\left(v_A\right)}_x=\left(1\right)v_0\cos \theta \\ {\left(v_B\right)}_x-{\left(v_A\right)}_x=v_0\cos \theta \end{array}$ (10)

Find the horizontal components of linear velocity ${\left(v_B\right)}_x$ of ball B immediately after the impact.

Solve Equations (9) and (10) simultaneously.

Add Equations (9) and (10).

$\begin{array}{l}{\left(v_B\right)}_x-{\left(v_A\right)}_x+{\left(v_B\right)}_x+{\left(v_A\right)}_x=v_0\cos \theta +v_0\cos \theta \\ 2{\left(v_B\right)}_x=2v_0\cos \theta \\ {\left(v_B\right)}_x=v_0\cos \theta \left(\text{or}\right)\end{array}$

Find the horizontal components linear velocity ${\left(v_A\right)}_x$ of ball A immediately after the impact using equation (10).

${\left(v_B\right)}_x-{\left(v_A\right)}_x=v_0\cos \theta$

Substitute $v_0\cos \theta$ for ${\left(v_B\right)}_x$.

$\begin{array}{l}{v}_0\cos \theta -{\left(v_A\right)}_x=v_0\cos \theta \\ {\left(v_A\right)}_x=v_0\cos \theta -v_0\cos \theta \\ {\left(v_A\right)}_x=0\end{array}$

Find the vertical components of linear velocity ${\left(v_B\right)}_y$ of ball B immediately after the impact using Equation (6).

$\begin{array}{l}0+0=m{\left(v_B\right)}_y\\ {\left(v_B\right)}_y=0\end{array}$

Find the vertical components of linear velocity ${\left(v_A\right)}_y$ of ball A immediately after the impact using Equation (2).

$\begin{array}{l}m{v}_0\sin \theta +0=m{\left(v_A\right)}_y\\ {\left(v_A\right)}_y=v_0\sin \theta \end{array}$

Find the linear velocity $\left(v_A\right)$ of ball A immediately after the impact.

$v_A={\left(v_A\right)}_{x}i+{\left(v_A\right)}_{y}j$

Substitute 0 for ${\left(v_A\right)}_x$ and $v_0\sin \theta$ for ${\left(v_A\right)}_y$.

$\begin{array}{c}{v}_A=0+v_0\sin \theta j\\ =v_0\sin \theta j\end{array}$

Thus, the linear velocity of ball A immediately after the impact is $\underset{\_}{0}$.

Find the linear velocity $\left(v_B\right)$ of ball B immediately after the impact.

$v_B={\left(v_B\right)}_{x}i+{\left(v_B\right)}_{y}j$

Substitute $v_0\cos \theta$ for ${\left(v_B\right)}_x$ and 0 for ${\left(v_B\right)}_y$.

$\begin{array}{c}{v}_B={\left(v_B\right)}_{x}i+{\left(v_B\right)}_{y}j\\ =v_0\cos \theta +0\\ =v_0\cos \theta \end{array}$

Thus, the linear velocity of ball B immediately after the impact is $\underset{\_}{v_0\cos \theta i}$.

Find the initial angular velocity of ball A using kinematics.

${\omega }_0=\frac{v_0}{r}$

Find the angular velocity $\left({\omega }_A\right)$ of ball A immediately after the impact.

Solve Equations (3) and (4) simultaneously.

Add Equations (3) and (4).

$\begin{array}{l}\overline{I}{\omega }_0\cos \theta +0-\overline{I}{\omega }_0\sin \theta +0=\overline{I}{\omega }_A\cos \beta -\overline{I}{\omega }_A\sin \beta \\ \overline{I}{\omega }_0\left(\cos \theta -\sin \theta \right)=\overline{I}{\omega }_A\left(\cos \beta -\sin \beta \right)\\ {\omega }_0\left(\cos \theta -\sin \theta \right)={\omega }_A\left(\cos \beta -\sin \beta \right)\end{array}$

Substitute $\frac{v_0}{r}$ for ${\omega }_0$ and 0 for $\beta$.

$\begin{array}{l}\frac{v_0}{r}\left(\cos \theta -\sin \theta \right)={\omega }_A\left(\cos 0-\sin 0\right)\\ {\omega }_A=\frac{v_0}{r}\left(\cos \theta -\sin \theta \right)\\ {\omega }_A=\frac{v_0}{r}\left(-\sin \theta i+\cos \theta j\right)\end{array}$

Thus, the angular velocity of ball A immediately after the impact is $\underset{\_}{\frac{v_0}{r}\left(-\sin \theta i+\cos \theta j\right)}$.

Find the angular velocity $\left({\omega }_B\right)$ of ball B immediately after the impact.

From Equations (7) and (8).

${\omega }_B=0$

Thus, the angular velocity of ball B immediately after the impact is $\underset{\_}{0}$.

To determine

Find the velocity of ball B after it has started rolling uniformly.

Answer to Problem 17.133P

The velocity of ball B after it has started rolling uniformly is $\underset{\_}{\frac{5}{7}\left(v_0\cos \theta \right)i}$.

Explanation of Solution

Calculation:

Consider the motion after impact of ball B.

Consider C is the mass center of ball A.

Consider the conservation of momentum principle.

${\text{Syst Momenta}}_{\text{1}}+{\text{ Syst Ext Imp}}_{\text{1y2 }}={\text{ Syst Momenta}}_{\text{2}}$

Sketch the motion of impulse and momentum diagram of the ball A after the impact as shown in Figure (3).

Here, ${\omega }_A$ is the angular velocity of ball A immediately after the impact, and $v_A$ is the velocity of ball A immediately after the impact, N is the normal force on ball A, F is the friction force between ball and floor, t is the rolling time, ${{\omega }^{\prime }}_A$ is the angular velocity of ball A after the impact, and ${v^{\prime }}_A$ is the velocity of ball A after the impact.

Consider the condition of rolling without slipping in kinematics.

${v^{\prime }}_A={{\omega }^{\prime }}_{A}r$

Refer Figure (3).

Take moments about C:

$\overline{I}{\omega }_A+0=\overline{I}{{\omega }^{\prime }}_A+m{v^{\prime }}_{A}r$

Substitute $\frac{2}{5}m{r}^2$ for $\overline{I}$, $\frac{v_1}{r}$ for ${\omega }_A$, and $r{{\omega }^{\prime }}_A$ for ${v^{\prime }}_A$.

$\begin{array}{l}\left(\frac{2}{5}m{r}^2\right)\left(\frac{v_1}{r}\right)+0=\left(\frac{2}{5}m{r}^2\right){{\omega }^{\prime }}_A+m\left(r{{\omega }^{\prime }}_A\right)r\\ \frac{2}{5}mr{v}_1=\left(\frac{2}{5}m{r}^2+m{r}^2\right){{\omega }^{\prime }}_A\\ \frac{2}{5}mr{v}_1=\frac{7}{5}m{r}^2{{\omega }^{\prime }}_A\\ 2v_1=7r{{\omega }^{\prime }}_A\\ {{\omega }^{\prime }}_A=\frac{2}{7}\frac{v_1}{r}\end{array}$

Find the velocity of sphere A after it has started rolling uniformly using the equation:

${v^{\prime }}_A={{\omega }^{\prime }}_{A}r$

Substitute $\frac{2}{7}\frac{v_1}{r}$ for ${{\omega }^{\prime }}_A$.

$\begin{array}{c}{v^{\prime }}_A=\left(\frac{2}{7}\frac{v_1}{r}\right)r\\ =\frac{2}{7}{v}_1\end{array}$

Thus, the velocity of ball A after it has started rolling uniformly is $\underset{\_}{\frac{2}{7}{v}_1}$.

Consider the motion after impact of ball A.

Consider C is the mass center of ball A.

Consider the conservation of momentum principle.

${\text{Syst Momenta}}_{\text{1}}+{\text{ Syst Ext Imp}}_{\text{1y2 }}={\text{ Syst Momenta}}_{\text{2}}$

Sketch the motion of impulse and momentum diagram of the ball B after the impact as shown in Figure (3).

Here, N is the normal force on ball A, F is the friction force between ball and floor, t is the rolling time, ${{\omega }^{\prime }}_B$ is the angular velocity of ball B after the impact, and ${v^{\prime }}_B$ is the velocity of ball B after the impact.

Consider the condition of rolling without slipping in kinematics.

${v^{\prime }}_B={{\omega }^{\prime }}_{B}r$

Refer Figure (3),

Take moments about C:

$m{v}_{B}r+0=\overline{I}{{\omega }^{\prime }}_B+m{v^{\prime }}_{B}r$

Substitute $v_0\cos \theta i$ for $v_B$, $\frac{2}{5}m{r}^2$ for $\overline{I}$, and $r{{\omega }^{\prime }}_B$ for ${v^{\prime }}_B$.

$\begin{array}{l}m\left(v_0\cos \theta i\right)r+0=\left(\frac{2}{5}m{r}^2\right){{\omega }^{\prime }}_B+m\left(r{{\omega }^{\prime }}_B\right)r\\ m{v}_{1}r=\left(\frac{2}{5}+1\right)m{r}^2{{\omega }^{\prime }}_B\\ v_0\cos \theta i=\frac{7}{5}{{\omega }^{\prime }}_{B}r\\ {{\omega }^{\prime }}_B=\frac{5}{7}\frac{v_0\cos \theta i}{r}\end{array}$

Find the velocity of sphere B after it has started rolling uniformly using the equation:

${v^{\prime }}_B={{\omega }^{\prime }}_{B}r$

Substitute $\frac{5}{7}\frac{v_0\cos \theta i}{r}$ for ${{\omega }^{\prime }}_B$.

$\begin{array}{c}{v^{\prime }}_B=\left(\frac{5}{7}\frac{v_0\cos \theta i}{r}\right)r\\ =\frac{5}{7}\frac{v_0\cos \theta i}{r}\end{array}$

Thus, the velocity of sphere B after it has started rolling uniformly is $\underset{\_}{\frac{5}{7}\frac{v_0\cos \theta i}{r}}$.