Chapter 13, Problem 70PQ

A ball of mass M = 5.00 kg and radius r = 5.00 cm is attached to one end of a thin, cylindrical rod of length L = 15.0 cm and mass m = 0.600 kg. The ball and rod, initially at rest in a vertical position and free to rotate around the axis shown in Figure P13.70, are nudged into motion. a. What is the rotational kinetic energy of the system when the ball and rod reach a horizontal position? b. What is the angular speed of the ball and rod when they reach a horizontal position? c. What is the linear speed of the center of mass of the ball when the ball and rod reach a horizontal position? d. What is the ratio of the speed found in part (c) to the speed of a ball that falls freely through the same distance?

FIGURE P13.70

To determine

The rotational kinetic energy of the system when the ball and rod reach a horizontal position.

Answer to Problem 70PQ

The rotational kinetic energy of the system when the ball and rod reach a horizontal position is $\underset{\_}{10.3\text{J}}$.

Explanation of Solution

For the isolated rod-ball-Earth system with no friction, the mechanical energy is conserved.

  $K_i+U_i=K_f+U_i$                                                                                                      (I)

Here, $K_i$ is the initial kinetic energy, $U_i$ is the initial potential energy, $K_f$ is the final kinetic energy (rotational kinetic energy), and $U_f$ is the final potential energy of the system.

The initially the system has only potential energy and has no kinetic energy. At the final position, the entire potential energy is converted to kinetic energy. Thus, equation (I) can be modified as,

  $U_i=K_f$                                                                                                                     (II)

Write the expression for the initial potential energy of the given system.

  $U_i=m_{\text{rod}}g{y}_{\text{CM, rod}}+m_{\text{ball}}g{y}_{\text{CM, ball}}$                                                                                (III)

Here, $m_{\text{rod}}$ is the mass of rod, $m_{\text{ball}}$ is the mass of ball, $g$ is the acceleration due to gravity, $y_{\text{CM, rod}}$ is the height of at which the center of mass of the rod occupies, and $m_{\text{CM, ball}}$ is the height at which the center of mass of the ball occupies.

Use equation (III) in (II) and solve for $K_f$.

  $\begin{array}{c}{m}_{\text{rod}}g{y}_{\text{CM, rod}}+m_{\text{ball}}g{y}_{\text{CM, ball}}=K_f\\ K_f=\left(m_{\text{rod}}{y}_{\text{CM, rod}}+m_{\text{ball}}{y}_{\text{CM, ball}}\right)g\end{array}$                                         (IV)

For the uniform rod, the center of mass is at the mid-point. Since the length of the rod is $15.0\text{cm}$, initially, the center of mass of the rod is at $7.50\text{cm}$ above the ground. Since the ball has radius $5.00\text{cm}$, initially the center of mass of the ball is at $20.0\text{cm}$ above the ground since the ball is fixed on top of the rod.

Conclusion:

Substitute $0.600\text{kg}$ for $m_{\text{rod}}$, $7.50\text{cm}$ for $y_{\text{CM, rod}}$, $5.00\text{kg}$ for $m_{\text{ball}}$, $20.0\text{cm}$ for $y_{\text{CM, ball}}$, and $9.81{\text{m/s}}^2$ for $g$ in equation (IV) to find $K_f$.

  $\begin{array}{c}{K}_f=\left[\left(0.600\text{kg}\right)\left(7.50\text{cm}\right)+\left(5.00\text{kg}\right)\left(20.0\text{cm}\right)\right]\left(9.81{\text{m/s}}^2\right)\\ =\left[\left(0.600\text{kg}\right)\left(7.50\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)+\left(5.00\text{kg}\right)\left(20.0\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\right]\left(9.81{\text{m/s}}^2\right)\\ =10.3\text{J}\end{array}$

Therefore, the rotational kinetic energy of the system when the ball and rod reach a horizontal position is $\underset{\_}{10.3\text{J}}$.

To determine

The angular speed of the system when it reach the horizontal position.

Answer to Problem 70PQ

The angular speed of the system when it reach the horizontal position is $\underset{\_}{9.90\text{rad/s}}$.

Explanation of Solution

The system can be assumed as, the sphere as a point particle which occupies at a distance $h$ from the center of rotation.

Write the expression for the rotational inertia of the given rod-ball system.

  $I=\frac{1}{3}{m}_{\text{rod}}{L}^2+\left[\frac{2}{5}{m}_{\text{ball}}{R}^2+m_{\text{ball}}{h}^2\right]$                                                                            (V)

Here, $I$ is the rotational inertia, $L$ is the length of the rod, $R$ is the radius of the sphere, and $h$ is the distance to the center of mass of the sphere from the center of rotation.

Write the expression for the rotational kinetic energy of the system at the horizontal position..

  $K_f=\frac{1}{2}I{\omega }^2$                                                                                                             (VI)

Here, $\omega$ is the angular speed of the system.

Solve equation (VI) for $\omega$.

  $\omega =\sqrt{\frac{2K_f}{I}}$                                                                                                               (VII)

Conclusion:

Substitute $0.600\text{kg}$ for $m_{\text{rod}}$, $5.00\text{kg}$ for $m_{\text{ball}}$, $15.0\text{cm}$ for $L$, $5.00\text{cm}$ for $R$, and $20.0\text{cm}$ for $h$ in equation (VI) to find $I$.

  $\begin{array}{l}I=\frac{1}{3}\left(0.600\text{kg}\right){\left(15.0\text{cm}\right)}^2+\left[\frac{2}{5}\left(5.00\text{kg}\right){\left(5.00\text{cm}\right)}^2+\left(5.00\text{kg}\right){\left(20.0\text{cm}\right)}^2\right]\\ =\frac{1}{3}\left(0.600\text{kg}\right){\left(15.0\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2+\left[\frac{2}{5}\left(5.00\text{kg}\right){\left(5.00\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2+\left(5.00\text{kg}\right){\left(20.0\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2\right]\\ =0.210\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $0.210\text{kg}\cdot {\text{m}}^2$ for $I$, and $10.3\text{J}$ for $K_f$ in equation (VII) to find $\omega$.

  $\begin{array}{c}\omega =\sqrt{\frac{2\left(10.3\text{J}\right)}{0.210\text{kg}\cdot {\text{m}}^2}}\\ =9.90\text{rad/s}\end{array}$

Therefore, the angular speed of the system when it reach the horizontal position is $\underset{\_}{9.90\text{rad/s}}$.

To determine

The linear speed of the center of mass of the ball when the system reach the horizontal position.

Answer to Problem 70PQ

The linear speed of the center of mass of the ball when the system reach the horizontal position is $\underset{\_}{1.98\text{m/s}}$.

Explanation of Solution

It is obtained that the angular speed of the system when it reach the horizontal position is $9.90\text{rad/s}$. The distance to the center of the ball from the center of rotation is $20.0\text{cm}$.

Write the expression for the linear speed in terms of the angular speed.

  $v=r\omega$                                                                                                                     (VIII)

Here, $v$ is the linear speed, and $r$ is the perpendicular distance between the center of rotation and the rotating object.

Conclusion:

Substitute $20.0\text{cm}$ for $r$, and $9.90\text{rad/s}$ for $\omega$ in equation (VIII) to find $v$.

  $\begin{array}{c}v=\left(20.0\text{cm}\right)\left(9.90\text{rad/s}\right)\\ =\left(20.0\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\left(9.90\text{rad/s}\right)\\ =1.98\text{m/s}\end{array}$

Therefore, the linear speed of the center of mass of the ball when the system reach the horizontal position is $\underset{\_}{1.98\text{m/s}}$.

To determine

The ratio of the speed of the ball when the system is at the horizontal position and the speed of the ball that falls freely through the same distance.

Answer to Problem 70PQ

The ratio of the speed of the ball when the system is at the horizontal position and the speed of the ball that falls freely through the same distance is $\underset{\_}{1.00}$.

Explanation of Solution

It is obtained that the linear speed of the center of mass of the ball when the system reach the horizontal position is $1.98\text{m/s}$.

Write the expression for the speed of the freely balling ball.

  $v_f^2=v_i^2-2g\left(y_f-y_i\right)$                                                                                              (IX)

Here, $v_f$ is the final speed, $v_i$ is the initial speed, $y_f$ is the final height, and $y_i$ is the initial height.

Since the ball falls from rest, and reaches the ground, the initial speed is zero, and the final height is zero. Thus, equation (IX) can be modified and solved for $v_f$ as,

  $\begin{array}{c}{v}_f^2=2g{y}_i\\ v=\sqrt{2g{y}_i}\end{array}$                                                                                                                (X)

Conclusion:

Substitute $9.81{\text{m/s}}^2$ for $g$, and $20.0\text{cm}$ for $y_i$ in equation (X) to find $v_f$.

  $\begin{array}{c}{v}_f=\sqrt{2\left(9.81{\text{m/s}}^2\right)\left(20.0\text{cm}\right)}\\ =\sqrt{2\left(9.81{\text{m/s}}^2\right)\left(20.0\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}\\ =1.98\text{m/s}\end{array}$

This indicates that the speed of the ball when the system is at the horizontal position and the speed of the ball that falls freely through the same distance are the same so that their ratio is obtained as,

  $\frac{1.98\text{m/s}}{1.98\text{m/s}}=1.00$

Therefore, the ratio of the speed of the ball when the system is at the horizontal position and the speed of the ball that falls freely through the same distance is $\underset{\_}{1.00}$.