Chapter 8, Problem 4SP

A student sitting on a stool that is free to rotate, but is initially at rest, holds a bicycle wheel. The wheel has a rotational velocity of 8 rev/s about a vertical axis, as shown in the SP4 diagram. The rotational inertia of the wheel is 2.5 kg·m² about its center, and the rotational inertia of the student and wheel and stool about the rotational axis of the stool is 6 kg·m².

1. a. What is the rotational velocity of the wheel in rad/s?
2. b. What are the magnitude and direction of the initial angular momentum of the system?
3. c. If the student flips the axis of the wheel, reversing the direction of its angular-momentum vector, what is the rotational velocity (magnitude and direction) of the student and the stool about their axis after the wheel is flipped? (Hint: See fig. 8.24.)
4. d. Where does the torque come from that accelerates the student and the stool? Explain.

To determine

The rotational velocity of the wheel in $\text{ rad}/\text{s}$.

Answer to Problem 4SP

The rotational velocity of the wheel is $50.24\text{rad}/\text{s}$.

Explanation of Solution

Given info: The rotational velocity is $8\text{rev}/\text{s}$.

Write the expression for conversion relation connecting $\text{rev}/\text{s}$ and $\text{rad}/\text{s}$.

$1\frac{\text{rev}}{\text{s}}=2\pi \frac{\text{rad}}{\text{s}}$

Convert $8\text{rev}/\text{s}$ into $\text{rad}/\text{s}$.

$\begin{array}{c}8\frac{\text{rev}}{\text{s}}=8\left(2\pi \frac{\text{rad}}{\text{s}}\right)\\ =50.27\text{rad}/\text{s}\\ \approx 50.3\text{rad/s}\end{array}$

Conclusion:

Therefore, the rotational velocity of the wheel is $50.3\text{rad}/\text{s}$.

To determine

The magnitude and the direction of the initial angular momentum of the system.

Answer to Problem 4SP

The angular momentum of the system is $125.75\text{Kg}\cdot {\text{m}}^2/\text{s}$.

Explanation of Solution

Write the expression for the angular momentum.

$L=I\omega$

Here,

$L$ is the angular momentum

$I$ is the rotational inertia

$\omega$ is the angular velocity

Substitute $2.5\text{ Kg}\cdot {\text{m}}^2$ for $I$ and $50.3\text{Kg}\cdot {\text{m}}^2/\text{s}$ for $\omega$ to find $L$.

$\begin{array}{c}L=2.5\text{ Kg}\cdot {\text{m}}^2\times 50.3\text{rad}/\text{s}\\ =125.75\text{Kg}\cdot {\text{m}}^2/\text{s}\end{array}$

Conclusion:

Therefore, the angular momentum of the system is $125.75\text{Kg}\cdot {\text{m}}^2/\text{s}$.

To determine

The rotational velocity of the student and the stool about their axis after the wheel is flipped.

Answer to Problem 4SP

The rotational velocity of the student and the stool about their axis is $6.67\text{rev}/\text{s}$.

Explanation of Solution

From the conservation of angular momentum, the angular velocity of the student and the stool is,

$L_w=-L_w+L_s$

Here, $L_w$ is angular momentum of the wheel, $-L_w$ is angular momentum of the wheel after its flip, and $L_s$ is angular momentum of the student and the stool.

Rewrite the relation of the angular momentum then rearrange it for the rotational velocity of the student and the stool.

$\begin{array}{l}{L}_w=-L_w+L_s\\ L_w+L_w=I_s{\omega }_s\\ {\omega }_s=\frac{2L_w}{I_s}\end{array}$ (1)

Rewrite the relation for the rotational velocity of the student and the stool.

${\omega }_s=\frac{2L_w}{I_s}$

Substitute $125.75\text{kg}\cdot {\text{m}}^2\text{/s}$ for $L_w$ and $6\text{kg}\cdot {\text{m}}^{\text{2}}$ for $I_s$ to find ${\omega }_s$.

$\begin{array}{c}{\omega }_s=\frac{2\left(125.75\text{kg}\cdot {\text{m}}^2\text{/s}\right)}{6\text{kg}\cdot {\text{m}}^{\text{2}}}\\ =41.9\text{rad/s}\\ =6.67\text{rev/s}\end{array}$

The direction of the rotational velocity of the student and the stool would be the direction of initial rotational velocity direction of the wheel.

Conclusion:

Therefore, the rotational velocity of the student and the stool about their axis is $41.9\text{rad/s}$ or $6.67\text{rev}/\text{s}$ and the direction of the rotational velocity of the student and the stool would be the direction of initial rotational velocity direction of the wheel.

To determine

Where will be the torque come from that accelerates the student and the stool.

Answer to Problem 4SP

The student exerts forces on the handles when he flips the wheel.

Explanation of Solution

For the flip of the wheel, the student exerts a certain amount force which creates the torque on the wheel then this torque produce the equal amount of opposite torque on the student and the stool. This happens for the system to be conserved.

Conclusion:

Therefore, the student exerts forces on the handles when he flips the wheel.