Chapter 10, Problem 55SP

A tiny solid ball $\left(I=2M{r}^2\text{/}5\right)$ rolls without slipping on the inside surface of a hemisphere as shown in Fig. 10-12. (The ball is much smaller than shown.) If the ball is released at A, how fast is it moving as it passes (a) point-B, and (b) point-C? Ignore friction losses. [Hint: Study the two previous questions. When it comes to the ball’s descent, its own radius is negligible.]

Fig. 10-12

To determine

The speed of the ball as it passes through the point B as shown in Fig. 10-12.

Answer to Problem 55SP

Solution:

$2.65\text{m}/\text{s}$

Explanation of Solution

Given data:

The moment of inertia of the solid ball is $\frac{2}{5}M{r}^2$.

The ball is released at point A. Therefore, the value of the initial velocity at that point is zero.

The movement of ball is as shown in Fig. 10-12;itfollows a circular path of radius $50.0\text{ cm}$.

Formula used:

Write the expression for calculating thepotential energy of an object:

$PE=mgh$

Here, $PE$ is the Potential Energy, $m$ is the mass of the object, $g$ is the acceleration due to gravity, and $h$ is the height of the object.

Write the expression for the total kinetic energy of a rolling object:

$K{E}_{\text{total}}=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$

Here, $K{E}_{\text{total}}$ is the total kinetic energy of the rolling object, $\frac{1}{2}m{v}^2$ is the translational kinetic energy, $m$ is the mass of the object, $v$ is the velocity of object, $\frac{1}{2}I{\omega }^2$ is the rotational kinetic energy of object, $I$ is the moment of inertia of object, and $\omega$ is the angular velocity of object.

Write the expression for total energy of system:

$TE=KE+PE$

Here, $TE$ is the total energy of system, $KE$ is the kinetic energy of system, and $PE$ is the potential energy of system.

Write the expression for conservation of energy:

$T{E}_i=T{E}_f$

Here, $T{E}_i$ is the initial total energy of the system and $T{E}_f$ is the final total energy of system.

Write the expression for linear velocity in terms of angular velocity:

$v=R\omega$

Here, $v$ is the linear velocity, $R$ is the radius, and $\omega$ is the angular velocity.

Explanation:

Draw the arrangement of the system shown in Fig 1:

Consider the point B be the reference point, that is, the height of the ball is measured with respect to this point.

The expression forpotential energy of the ball at point A is

$P{E}_A=mg{h}_A$

Here, $P{E}_A$ is the potential energy at point A and $h_A$ is the height of the ball at point A.

For point A, substitute $M$ for $m$, $9.8\text{m}/{\text{s}}^2$ for $g$, and $50.0\text{cm}$ for $h_A$

$\begin{array}{c}P{E}_A=M\left(9.8\text{m}/{\text{s}}^2\right)\left(50.0\text{ cm}\left(\frac{{10}^{-2}\text{ m}}{1\text{ cm}}\right)\right)\\ =4.9M\text{ J/kg}\end{array}$

The expression forpotential energy of ball at point B is

$P{E}_B=mg{h}_B$

Here, $P{E}_B$ is the potential energy at point B and $h_B$ is the height of the ball at point B.

For point B, substitute $M$ for $m$, $9.8\text{m}/{\text{s}}^2$ for $g$, and $0\text{cm}$ for $h_B$

$\begin{array}{c}P{E}_B=M\left(9.8\text{m}/{\text{s}}^2\right)\left(0\text{ cm}\left(\frac{{10}^{-2}\text{ m}}{1\text{ cm}}\right)\right)\\ =0\text{ J}\end{array}$

The expression for the total kinetic energy of theball at point A is

$K{E}_A=\frac{1}{2}m{v}_A{}^2+\frac{1}{2}I{\omega }_A{}^2$

Here, $K{E}_A$ is the total kinetic energy of the ball at point A, $v_A$ is the velocity of ball at point A, and ${\omega }_A$ is the angular velocity of ball at point A.

Since the ball is released from rest at point A, its initial velocity is zero.

$v_A=0\text{ m/s}$

For point A, substitute $M$ for $m$, $\frac{2}{5}M{r}^2$ for $I$, $0\text{ m/s}$ for $v_A$, and $0\text{ rad/s}$ for ${\omega }_A$

$\begin{array}{c}K{E}_A=\frac{1}{2}M{\left(0\text{ m/s}\right)}^2+\frac{1}{2}\left(\frac{2}{5}M{r}^2\right){\left(0\text{ rad/s}\right)}^2\\ =0\text{ J}\end{array}$

The expression for linear velocity in terms of angular velocity at point B is

$v_B=R{\omega }_B$

Here, $v_B$ is the linear velocity at point B and ${\omega }_B$ is the angular velocity at point B.

At point B, substitute $r$ for $R$

$v_B=r{\omega }_B$

Solve for ${\omega }_B$

${\omega }_B=\frac{v_B}{r}$

The expression for the total kinetic energy of ball at point B is

$K{E}_B=\frac{1}{2}m{v}_B{}^2+\frac{1}{2}I{\omega }_B{}^2$

Here, $K{E}_B$ is the total kinetic energy of the ball at point B, $v_B$ is the velocity of the ball at point B, and ${\omega }_B$ is the angular velocity of ball at point B.

For point B, substitute $M$ for $m$, $\frac{2}{5}M{r}^2$ for $I$, and $\frac{v_B}{r}\text{rad}/\text{s}$ for ${\omega }_B$

$\begin{array}{c}K{E}_B=\frac{1}{2}M{v}_B{}^2+\frac{1}{2}\left(\frac{2}{5}M{r}^2\right){\left(\frac{v_B}{r}\right)}^2\\ =\frac{1}{2}M{v}_B{}^2+\frac{1}{2}\left(\frac{2}{5}M{r}^2\right)\left(\frac{v_B{}^2}{r^2}\right)\\ =\frac{1}{2}M{v}_B{}^2+\frac{1}{5}M{v}_B{}^2\\ =\frac{7}{10}M{v}_B{}^2\end{array}$

The expression for total energy of the system at point A is

$T{E}_A=K{E}_A+P{E}_A$

Here, $T{E}_A$ is the total energy of the system at point A, $K{E}_A$ is the kinetic energy of the system at point A, and $P{E}_A$ is the potential energy of system at point A.

At point A, substitute $0\text{ J}$ for $K{E}_A$ and $4.9M\text{ J/kg}$ for $P{E}_A$

$\begin{array}{c}T{E}_A=0\text{ J}+4.9M\text{ J/kg}\\ =\text{4}\text{.9}M\text{ J/kg}\end{array}$

The expression for the total energy of the system at point B is

$T{E}_B=K{E}_B+P{E}_B$

Here, $T{E}_B$ is the total energy of system at point B, $K{E}_B$ is the kinetic energy of system at point B, and $P{E}_B$ is the potential energy of system at point B.

At point B, substitute $\frac{7}{10}M{v}_B{}^2$ for $K{E}_B$ and $\text{0 J}$ for $P{E}_B$

$\begin{array}{c}T{E}_B=\frac{7}{10}M{v}_B{}^2+0\text{ J}\\ \text{=}\frac{7}{10}M{v}_B{}^2\end{array}$

The expression for conservation of energy for the point of observation A and point B is

$T{E}_A=T{E}_B$

Substitute $4.9M\text{J/kg}$ for $T{E}_A$ and $\frac{7}{10}M{v}_B{}^2$ for $T{E}_B$

$4.9M\text{ J/kg}=\frac{7}{10}M{v}_B{}^2$

Solve for $v_B$

$\begin{array}{c}{v}_B{}^2=\frac{\left(4.9M\text{ J/kg}\right)}{\left(\frac{7}{10}M\right)}\\ =7{\text{ m}}^2/{\text{s}}^{\text{2}}\end{array}$

Take square root both sides:

$\begin{array}{c}{v}_B=\sqrt{7}\text{m/s}\\ =2.65\text{ m/s}\end{array}$

Conclusion:

The speed of the ball as it passes through the point B is $2.65\text{ m/s}$.

To determine

The speed of the ball as it passes through the point Cas shown in Fig. 10-12.

Answer to Problem 55SP

Solution:

$\text{2}\text{.32 }\text{m}/\text{s}$

Explanation of Solution

Given data:

The moment of inertia of the solid ball is $\frac{2}{5}M{r}^2$.

The ball is released at point A that is the velocity of the ball at this point is zero.

The movement of ball is shown in Fig. 10-12;it follows a circular path of radius $50.0\text{ cm}$.

Formula used:

Write the expression for calculating the potential energy of an object:

$PE=mgh$

Here, $m$ is the mass of the object, $g$ is the acceleration due to gravity, and $h$ is the height of the object.

Write the expression for total kinetic energy of a rolling object:

$K{E}_{\text{total}}=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$

Here, $K{E}_{\text{total}}$ is the total kinetic energy of the rolling object, $\frac{1}{2}m{v}^2$ is the translational kinetic energy, $m$ is the mass of the object, $v$ is the velocity of object, $\frac{1}{2}I{\omega }^2$ is the rotational kinetic energy of object, $I$ is the moment of inertia of object, and $\omega$ is the angular velocity of object.

Write the expression for total energy of a system:

$TE=KE+PE$

Here, $TE$ is the total energy of the system, $KE$ is the kinetic energy of the system, and $PE$ is the potential energy of the system.

Write the expression for conservation of energy:

$T{E}_i=T{E}_f$

Here, $T{E}_i$ is the initial total energy of the system and $T{E}_f$ is the final total energy of system.

Write the expression for linear velocity in terms of angular velocity:

$v=R\omega$

Here, $v$ is the linear velocity, $R$ is the radius, and $\omega$ is the angular velocity.

Explanation:

Let point B be the reference point, that is, the height of the ball is measured with respect to this point.

Recall the total energy at point A from subpart (a):

$T{E}_A=\text{4}\text{.9}M\text{ J/kg}$

The expression forpotential energy of ball at point C is

$P{E}_C=mg{h}_C$

Here, $P{E}_C$ is the potential energy at point C and $h_c$ is the height of point C.

The expression for height of point C is

$\begin{array}{c}{h}_C=r-r\cos \theta \\ =r\left(1-\cos \theta \right)\end{array}$

Here, $r$ is the radius of hemisphere and $\theta$ is the angle subtended by point C.

Substitute $50.0\text{ cm}$ for $r$ and $40°$ for $\theta$

$\begin{array}{c}{h}_C=\left(50.0\text{ cm}\left(\frac{{10}^{-2}\text{ m}}{1\text{ cm}}\right)\right)\left(1-\cos 40°\right)\\ =\left(50\times {10}^{-2}\text{ m}\right)\left(0.23395\right)\\ =0.11697\text{ m}\end{array}$

For point C, substitute $M$ for $m$, $9.8\text{m}/{\text{s}}^2$ for $g$, and $0.11697\text{ m}$ for $h$

$\begin{array}{c}P{E}_C=M\left(9.8\text{m}/{\text{s}}^2\right)\left(0.11697\text{ m}\right)\\ =1.1463M\text{ J/kg}\end{array}$

The expression for total kinetic energy of a rolling object at point C is

$K{E}_C=\frac{1}{2}m{v}_C{}^2+\frac{1}{2}I{\omega }_C{}^2$

Here, $K{E}_C$ is the total kinetic energy of rolling object at point C, $v_C$ is the velocity of the object at point C, and ${\omega }_C$ is the angular velocity of the object at point C.

The expression for linear velocity in terms of angular velocity is

$v_C=R{\omega }_c$

Here, $v_C$ is the linear velocity at point C and ${\omega }_C$ is the angular velocity at point C.

At point C, substitute $r$ for $R$

$v_C=r{\omega }_C$

Solve for ${\omega }_C$

${\omega }_C=\frac{v_C}{r}$

For point C, substitute $M$ for $m$, $\frac{2}{5}M{r}^2$ for $I$, and $\frac{v_C}{r}\text{rad}/\text{s}$ for ${\omega }_C$

$\begin{array}{c}K{E}_C=\frac{1}{2}M{v}_c{}^2+\frac{1}{2}\left(\frac{2}{5}M{r}^2\right){\left(\frac{v_C}{r}\right)}^2\\ =\frac{1}{2}M{v}_c{}^2+\frac{1}{2}\left(\frac{2}{5}M{r}^2\right)\left(\frac{v_C{}^2}{r^2}\right)\\ =\frac{1}{2}M{v}_C{}^2+\frac{1}{5}M{v}_C{}^2\\ =\frac{7}{10}M{v}_C{}^2\end{array}$

The expression for total energy of the system at point C is

$T{E}_C=K{E}_C+P{E}_C$

Here, $T{E}_C$ is the total energy of system at point C, $K{E}_C$ is the kinetic energy of the system at point C, and $P{E}_B$ is the potential energy of the system at point B.

At point C, substitute $\frac{7}{10}M{v}_c{}^2$ for $K{E}_C$ and $1.1463M\text{ J/kg}$ for $P{E}_C$

$T{E}_C=\frac{7}{10}M{v}_C{}^2+1.1463M\text{ J/kg}$

The expression for conservation of energy for the point of observation A and C is

$T{E}_A=T{E}_C$

Substitute $4.9M\text{J/kg}$ for $T{E}_A$ and $\left(\frac{7}{10}M{v}_C{}^2+1.1463M\text{ J/kg}\right)$ for $T{E}_C$

$4.9M\text{ J/kg}=\frac{7}{10}M{v}_C{}^2+1.1463M\text{ J/kg}$

Solve for $v_C$

$\begin{array}{c}4.9M\text{ J/kg}=M\left(\frac{7}{10}{v}_C{}^2+1.1463\text{ J/kg}\right)\\ \frac{7}{10}{v}_C{}^2=\left(4.9-1.1463\right){\text{ m}}^{\text{2}}{\text{/s}}^{\text{2}}\\ v_C{}^2=\frac{3.7537}{\left(\frac{7}{10}\right)}{\text{ m}}^{\text{2}}{\text{/s}}^{\text{2}}\\ v_C{}^2=\frac{37.537}{7}{\text{ m}}^{\text{2}}{\text{/s}}^{\text{2}}\end{array}$

Further, solve for $v_C$

$\begin{array}{c}{v}_C=\sqrt{5.3624}\text{m/s}\\ \text{=2}\text{.32 m/s}\end{array}$

Conclusion:

The speed of the ball as it passes through the point C is $2.32\text{ m/s}$.