Chapter 16.1, Problem 16.72P

Solve Prob. 16.71, assuming that the bowler projects the ball with the same forward velocity but with a backspin of 18 rad/s.

To determine

(a)

Find time $t_1$.

Answer to Problem 16.72P

Time $t_{\text{1}}=1.863\text{ sec}$

Explanation of Solution

Given information:

Mass $m=8\text{ lb}$

Radius $r=4\text{ in}$

Initial velocity $v_0=15\text{ ft/s}$

Friction coefficient ${\mu }_k=0.1$

Angular velocity ${\omega }_0=18\text{ rad/s}$

Concept used:

Following formula is used:

1. Sum of horizontal forces, $\sum F_x=ma.$

2. Sum of moments about mass center, $\sum M_G=I\alpha .$

Calculation:

Friction force,

$\begin{array}{l}f={\mu }_{k}N\\ f={\mu }_{k}mg\end{array}$

Sum of horizontal forces,

$\begin{array}{l}\sum F_x=ma\\ f=ma\\ {\mu }_{k}mg=ma\\ a={\mu }_{k}g\leftarrow \end{array}$

Sum of moments about mass center,

$\begin{array}{l}\sum M_G=I\alpha \\ f\times r=m{k}^2\alpha \\ {\mu }_{k}mg\times r=m{k}^2\alpha \text{ for sphere }{k}^2=\frac{2}{5}{r}^2\\ \alpha =\frac{5{\mu }_{k}gr}{2r^2}{\text{rad/s}}^{\text{2}}\end{array}$

Velocity equation,

$\begin{array}{l}{v}_{}=v_0-at\to \\ v_{}=v_0-{\mu }_{k}gt\end{array}$

Angular velocity equation,

$\begin{array}{l}\omega ={\omega }_0-\alpha t\\ \omega ={\omega }_0-\frac{5{\mu }_{k}gr}{2r^2}t\end{array}$

From above both equation,

$\begin{array}{l}{\text{when t=t}}_{\text{1}}\text{, }\\ v=r\omega \\ v_0-{\mu }_{k}g{t}_{\text{1}}=r(-{\omega }_0+\frac{5{\mu }_{k}gr}{2r^2}{t}_{\text{1}})\end{array}$

$\begin{array}{l}{t}_{\text{1}}=\frac{2(v_0+r{\omega }_0)}{7{\mu }_{k}g}\\ t_{\text{1}}=\frac{2(15+\frac{4}{12}\times 18)}{7\times 0.1\times 32.2}\\ t_{\text{1}}=1.863\text{ sec}\end{array}$

Conclusion:

Thus we get,

Time $t_{\text{1}}=1.863\text{ sec}\text{.}$

To determine

(b)

Find speed of ball at that time.

Answer to Problem 16.72P

Speed $v_1=9.0\text{ ft/s}\text{.}$

Explanation of Solution

Given information:

Mass $m=8\text{ lb}$

Radius $r=4\text{ in}$

Initial velocity $v_0=15\text{ ft/s}$

Friction coefficient ${\mu }_k=0.1$

Angular velocity ${\omega }_0=18\text{ rad/s}$

Concept used:

Following formula is used:

1. Sum of horizontal forces, $\sum F_x=ma.$

2. Sum of moments about mass center, $\sum M_G=I\alpha .$

Calculation:

Friction force,

$\begin{array}{l}f={\mu }_{k}N\\ f={\mu }_{k}mg\end{array}$

Sum of horizontal forces,

$\begin{array}{l}\sum F_x=ma\\ f=ma\\ {\mu }_{k}mg=ma\\ a={\mu }_{k}g\leftarrow \end{array}$

Sum of moments about mass center,

$\begin{array}{l}\sum M_G=I\alpha \\ f\times r=m{k}^2\alpha \\ {\mu }_{k}mg\times r=m{k}^2\alpha \text{ for sphere }{k}^2=\frac{2}{5}{r}^2\\ \alpha =\frac{5{\mu }_{k}gr}{2r^2}{\text{rad/s}}^{\text{2}}\end{array}$

Velocity equation,

$\begin{array}{l}{v}_{}=v_0-at\to \\ v_{}=v_0-{\mu }_{k}gt\end{array}$

Angular velocity equation,

$\begin{array}{l}\omega ={\omega }_0-\alpha t\\ \omega ={\omega }_0-\frac{5{\mu }_{k}gr}{2r^2}t\end{array}$

From above both equation,

$\begin{array}{l}{\text{when t=t}}_{\text{1}}\text{, }\\ v=r\omega \\ v_0-{\mu }_{k}g{t}_{\text{1}}=r(-{\omega }_0+\frac{5{\mu }_{k}gr}{2r^2}{t}_{\text{1}})\end{array}$

$\begin{array}{l}{t}_{\text{1}}=\frac{2(v_0+r{\omega }_0)}{7{\mu }_{k}g}\\ t_{\text{1}}=\frac{2(15+\frac{4}{12}\times 18)}{7\times 0.1\times 32.2}\\ t_{\text{1}}=1.863\text{ sec}\end{array}$

Speed

$v_1=v_0-{\mu }_{k}gt=15-0.1\times 32.2\times 1.863=9.0\text{ ft/s}\text{.}$

Conclusion:

Thus we get,

Speed $v_1=9.0\text{ ft/s}\text{.}$

To determine

(c)

Find distance travelled by ball.

Answer to Problem 16.72P

Distance travelled $s_1=22.4\text{ ft}\text{.}$

Explanation of Solution

Given information:

Mass $m=8\text{ lb}$

Radius $r=4\text{ in}$

Initial velocity $v_0=15\text{ ft/s}$

Friction coefficient ${\mu }_k=0.1$

Angular velocity ${\omega }_0=18\text{ rad/s}$

Concept used:

Following formula is used:

1. Sum of horizontal forces, $\sum F_x=ma.$

2. Sum of moments about mass center, $\sum M_G=I\alpha .$

Calculation:

Friction force,

$\begin{array}{l}f={\mu }_{k}N\\ f={\mu }_{k}mg\end{array}$

Sum of horizontal forces,

$\begin{array}{l}\sum F_x=ma\\ f=ma\\ {\mu }_{k}mg=ma\\ a={\mu }_{k}g\leftarrow \end{array}$

Sum of moments about mass center,

$\begin{array}{l}\sum M_G=I\alpha \\ f\times r=m{k}^2\alpha \\ {\mu }_{k}mg\times r=m{k}^2\alpha \text{ for sphere }{k}^2=\frac{2}{5}{r}^2\\ \alpha =\frac{5{\mu }_{k}gr}{2r^2}{\text{rad/s}}^{\text{2}}\end{array}$

Velocity equation,

$\begin{array}{l}{v}_{}=v_0-at\to \\ v_{}=v_0-{\mu }_{k}gt\end{array}$

Angular velocity equation,

$\begin{array}{l}\omega ={\omega }_0-\alpha t\\ \omega ={\omega }_0-\frac{5{\mu }_{k}gr}{2r^2}t\end{array}$

From above both equation,

$\begin{array}{l}{\text{when t=t}}_{\text{1}}\text{, }\\ v=r\omega \\ v_0-{\mu }_{k}g{t}_{\text{1}}=r(-{\omega }_0+\frac{5{\mu }_{k}gr}{2r^2}{t}_{\text{1}})\end{array}$

$\begin{array}{l}{t}_{\text{1}}=\frac{2(v_0+r{\omega }_0)}{7{\mu }_{k}g}\\ t_{\text{1}}=\frac{2(15+\frac{4}{12}\times 18)}{7\times 0.1\times 32.2}\\ t_{\text{1}}=1.863\text{ sec}\end{array}$

Distance travelled,

$\begin{array}{l}{s}_1=v_0{t}_1-\frac{1}{2}{\mu }_{k}g{t}_1{}^2\\ s_1=15\times 1.863-\frac{1}{2}\times 0.1\times 32.2\times {1.863}^2\\ s_1=22.4\text{ ft}\end{array}$

Conclusion:

Thus we get,

Distance travelled $s_1=22.4\text{ ft}\text{.}$