Question

A tennis ball is a hollow sphere with a thin wall. It is set rolling without slipping at $\mathbf{4}\mathbf{.}\mathbf{03}\mathbf{m}\mathbf{/}\mathbf{s}$ on a horizontal section of a track as shown in Figure P10.64. It rolls around the inside of a vertical circular loop of radius $\mathbf{r}\mathbf{=}\mathbf{45}\mathbf{.}\mathbf{0}\mathbf{cm}$. As the ball nears the bottom of the loop, the shape of the track deviates from a perfect circle so that the ball leaves the track at a point $\mathbf{h}\mathbf{=}\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{cm}$ below the horizontal section. (a) Find the ball’s speed at the top of the loop. (b) Demonstrate that the ball will not fall from the track at the top of the loop. (c) Find the ball’s speed as it leaves the track at the bottom. (d) What If? Suppose that static friction between ball and track were negligible so that the ball slid instead of rolling. Would its speed then be higher, lower, or the same at the top of the loop? (e) Explain your answer to part (d)

Short answer

(a) The speed of the ball at the top of the loop is $2.37\text{\hspace{0.17em}m}/\text{s}$.

(b) It is showed that the ball will not fall from the track at the top of the loop.

(c) The speed of the ball when it leaves the track is $4.31\text{\hspace{0.17em}m}/\text{s}$.

(d) The speed of the ball when it slides on the track is$\sqrt{-1.399}\text{\hspace{0.17em}m}/\text{s}$ .

(e) The ball will never go top of the loop.

Step-by-step solution

Write the given data from the question:

The speed of the ball,${\mathrm{v}}_1=4.03\text{\hspace{0.17em}m}/\text{s}$

The radius of the circular loop,$\mathrm{r}=45\text{\hspace{0.17em}cm}$

The distance between the point from where ball leave the track and horizontal section,$\mathrm{h}=20\text{\hspace{0.17em}cm}$.

Determine the formulas to calculate the speed of the ball, and acceleration of the ball.

The expression to calculate the angular velocity is given as follows.

$\mathbf{\omega }\mathbf{=}\frac{\mathbf{v}}{\mathbf{R}}$

The expression to calculate the inertia about the centre of mass of hollow cylinder is given as follows.

${\mathbf{I}}_{\mathbf{c}}\mathbf{=}{\mathbf{mR}}^{\mathbf{2}}$

Here, $\mathbf{m}$is the mass of the hollow cylinder and$\mathbf{R}$ is the radius of the hollow cylinder.

The expression to calculate the inertia of the hollow sphere is given as follows.

${\mathbf{I}}_{\mathbf{s}}\mathbf{=}\frac{\mathbf{2}}{\mathbf{3}}{\mathbf{mR}}^{\mathbf{2}}$

The expression to calculate the rotational kinetic energy is given as follows.

$\mathbf{KE}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{I\omega }}^{\mathbf{2}}$

Calculate the speed of the ball at the top of the loop.

(a)

According to the conservation of the law the initial energy of the system remains the same until unless any external force disturbs the system.

$\frac{1}{2}{\mathrm{mv}}_2^2+\frac{1}{2}{\mathrm{I\omega }}_2^2+{\mathrm{mgy}}_2=\frac{1}{2}{\mathrm{mv}}_1^2+\frac{1}{2}{\mathrm{I\omega }}_1^2+{\mathrm{mgy}}_1$

Substitute $\frac{2}{3}{\mathrm{mR}}^2$for $I$, $\frac{{\mathrm{v}}_1}{\mathrm{R}}$for ${\mathrm{\omega }}_1$, $\frac{{\mathrm{v}}_2}{\mathrm{R}}$for ${\mathrm{\omega }}_2$,$2r$for $y_2$and $0$for${\mathrm{y}}_1$into above equation.

$\begin{array}{c}\frac{1}{2}{\mathrm{mv}}_2^2+\frac{1}{2}\left(\frac{2}{3}{\mathrm{mR}}^2\right){\left(\frac{{\mathrm{v}}_2}{\mathrm{R}}\right)}^2+\mathrm{mg}\left(2\mathrm{r}\right)=\frac{1}{2}{\mathrm{mv}}_1^2+\frac{1}{2}\left(\frac{2}{3}{\mathrm{mR}}^2\right){\left(\frac{{\mathrm{v}}_2}{\mathrm{R}}\right)}^2+\mathrm{mg}\left(0\right)\\ \frac{1}{2}{\mathrm{mv}}_2^2+\frac{1}{3}{\mathrm{mv}}_2^2+2\mathrm{rmg}=\frac{1}{2}{\mathrm{mv}}_1^2+\frac{1}{3}{\mathrm{mv}}_1^2\\ \frac{5}{6}{\mathrm{mv}}_2^2+2\mathrm{rmg}=\frac{5}{6}{\mathrm{mv}}_1^2\\ \frac{5}{6}{\mathrm{mv}}_2^2=\frac{5}{6}{\mathrm{mv}}_1^2-2\mathrm{rmg}\end{array}$

Solve further as,

$\begin{array}{c}\frac{5}{6}{\mathrm{v}}_2^2=\frac{5}{6}{\mathrm{v}}_1^2-2\mathrm{rg}\\ {\mathrm{v}}_2^2={\mathrm{v}}_1^2-\frac{12}{5}\mathrm{rg}\end{array}$

Substitute$4.03\text{\hspace{0.17em}m}/\text{s}$ for ${\mathrm{v}}_1$,$45\text{\hspace{0.17em}cm}$ for $\mathrm{r}$,and$9.8\text{\hspace{0.17em}m}/{\text{s}}^2$ for$g$ into above equation.

$\begin{array}{l}{\mathrm{v}}_2^2={4.03}^2-\frac{12}{5}\times 9.9\times 45\times {10}^{-2}\\ {\mathrm{v}}_2^2=16.2409-10.584\\ {\mathrm{v}}_2^2=5.6569\\ {\mathrm{v}}_2=2.37\text{\hspace{0.17em}m}/\text{s}\end{array}$

Hence the speed of the ball at the top of the loop is $2.37\text{\hspace{0.17em}m}/\text{s}$.

Demonstrate that the ball will not fall from the track at the top of the loop.

(b)

The ball will not fall from the tract at the top of the loop when the acceleration of the ball along the vertical is greater than the acceleration of the gravity.

Calculate the acceleration of the ball,

${\mathrm{a}}_{\mathrm{c}}=\frac{{\mathrm{v}}_2^2}{\mathrm{r}}$

Substitute $2.37\text{\hspace{0.17em}m}/\text{s}$for $v_2$and $45\text{\hspace{0.17em}cm}$for $r$into above equation.

$\begin{array}{c}{\mathrm{a}}_{\mathrm{c}}=\frac{{2.37}^2}{45\times {10}^{-2}}\\ {\mathrm{a}}_{\mathrm{c}}=\frac{5.6169}{0.45}\\ {\mathrm{a}}_{\mathrm{c}}=12.482\text{\hspace{0.17em}m}/{\text{s}}^2\end{array}$

Since the acceleration of the ball is $12.482\text{\hspace{0.17em}m}/{\text{s}}^2$along the vertical of the loop which is greater than the acceleration due to gravity therefore the ball will not fall from the track at the top of the loop.

Hence the ball will not fall from the track at the top of the loop .

Calculate the ball’s speed as it leaves the track at the bottom.

(c)

To calculate the ball’s speed when it leaves the track use the conservation of energy between the starting and end of the track.

$\frac{1}{2}{\mathrm{mv}}_3^2+\frac{1}{2}{\mathrm{I\omega }}_3^2+{\mathrm{mgy}}_3=\frac{1}{2}{\mathrm{mv}}_1^2+\frac{1}{2}{\mathrm{I\omega }}_1^2+{\mathrm{mgy}}_1$

Substitute $\frac{2}{3}m{R}^2$for $I$, $\frac{v_3}{R}$for ${\omega }_3$,$\frac{{\mathrm{v}}_1}{\mathrm{R}}$for ${\mathrm{\omega }}_1$, $-h$for $y_3$and $0$for ${\mathrm{y}}_1$into above equation.

$\begin{array}{c}\frac{1}{2}{\mathrm{mv}}_3^2+\frac{1}{2}\left(\frac{2}{3}{\mathrm{mR}}^2\right){\left(\frac{{\mathrm{v}}_3}{\mathrm{R}}\right)}^2+\mathrm{mg}(-\mathrm{h})=\frac{1}{2}{\mathrm{mv}}_1^2+\frac{1}{2}\left(\frac{2}{3}{\mathrm{mR}}^2\right){\left(\frac{{\mathrm{v}}_2}{\mathrm{R}}\right)}^2+\mathrm{mg}\left(0\right)\\ \frac{1}{2}{\mathrm{mv}}_3^2+\frac{1}{3}{\mathrm{mv}}_3^2-\mathrm{mgh}=\frac{1}{2}{\mathrm{mv}}_1^2+\frac{1}{3}{\mathrm{mv}}_1^2\\ \frac{5}{6}{\mathrm{mv}}_3^2-\mathrm{mgh}=\frac{5}{6}{\mathrm{mv}}_1^2\\ \frac{5}{6}{\mathrm{mv}}_3^2=\frac{5}{6}{\mathrm{mv}}_1^2+\mathrm{mgh}\end{array}$

Solve further as,

$\begin{array}{c}\frac{5}{6}{\mathrm{v}}_3^2=\frac{5}{6}{\mathrm{v}}_1^2+\mathrm{gh}\\ {\mathrm{v}}_3^2={\mathrm{v}}_1^2+\frac{6}{5}\mathrm{gh}\end{array}$

Substitute$4.03\text{\hspace{0.17em}m}/\text{s}$ for $v_1$,$20\text{\hspace{0.17em}cm}$ for$h$ ,and $9.8\text{\hspace{0.17em}m}/{\text{s}}^2$for$\mathrm{g}$ into above equation.

$\begin{array}{c}{\mathrm{v}}_3^2={4.03}^2+\frac{6}{5}\times 9.8\times 20\times {10}^{-2}\\ {\mathrm{v}}_3^2=16.2409+2.352\\ {\mathrm{v}}_3^2=18.5929\\ {\mathrm{v}}_3=4.31\text{\hspace{0.17em}m}/\text{s}\end{array}$

Hence the speed of the ball when it leaves the track is $4.31\text{\hspace{0.17em}m}/\text{s}$.

Calculate the speed of the ball when it slides on track.

d.

Since the friction between the track and ball is zero and it slide instead of rolling, therefore the rotational kinetic energy would be zero.

$\frac{1}{2}{\mathrm{mv}}_2^2+{\mathrm{mgy}}_2=\frac{1}{2}{\mathrm{mv}}_1^2+{\mathrm{mgy}}_1$

Substitute$2\mathrm{r}$ for $y_2$and $0$for${\mathrm{y}}_1$ into above equation.

$\begin{array}{c}\frac{1}{2}{\mathrm{v}}_2^2+\mathrm{g}\left(2\mathrm{r}\right)=\frac{1}{2}{\mathrm{v}}_1^2+\mathrm{g}\left(0\right)\\ \frac{1}{2}{\mathrm{v}}_2^2+2\mathrm{gr}=\frac{1}{2}{\mathrm{v}}_1^2\\ \frac{1}{2}{\mathrm{v}}_2^2=\frac{1}{2}{\mathrm{v}}_1^2-2\mathrm{gr}\\ {\mathrm{v}}_2^2={\mathrm{v}}_1^2-4\mathrm{g}\end{array}$

Substitute $4.03\text{\hspace{0.17em}m}/\text{s}$for ${\mathrm{v}}_1$,$9.8\text{\hspace{0.17em}m}/{\text{s}}^2$for $g$and $45\text{\hspace{0.17em}cm}$for role="math" localid="1663752612086" $\mathrm{r}$into above equation.

$\begin{array}{c}{\mathrm{v}}_2^2={4.03}^2-4\times 9.8\times 45\times {10}^{-2}\\ {\mathrm{v}}_2^2=16.2409-17.64\\ {\mathrm{v}}_2^2=-1.399\\ {\mathrm{v}}_2=\sqrt{-1.399}\text{\hspace{0.17em}m}/\text{s}\end{array}$

Hence the speed of the ball when it slides on the track is $\sqrt{-1.399}\text{\hspace{0.17em}m}/\text{s}$.

Explanation of the part (d).

(e)

Since the speed of the ball when it slied on the tract is$\sqrt{-1.399}\text{\hspace{0.17em}m}/\text{s}$ which is an imaginary speed therefore the ball will never go top of the loop.

Hence the ball will never go top of the loop.