Chapter 8, Problem 4SP

(a)

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 $12\text{kg}\cdot {\text{m}}^2/\text{s}$, upward from the plane of the wheel.

Explanation of Solution

Given info: The rotational velocity of the wheel is $6\text{rev}/\text{s}$ and its rotational inertia is $2\text{ kg}\cdot {\text{m}}^2$, the rotational inertia of the student and the wheel is $5\text{ kg}\cdot {\text{m}}^2$.

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 $2\text{ kg}\cdot {\text{m}}^2$ for $I$ and $6\text{rev}/\text{s}$ for $\omega$ to find $L$.

$\begin{array}{c}L=\left(2\text{ kg}\cdot {\text{m}}^2\right)\left(6\text{rev}/\text{s}\right)\\ =12\text{kg}\cdot {\text{m}}^2/\text{s}\end{array}$

Conclusion:

Therefore, the angular momentum of the system is $12\text{kg}\cdot {\text{m}}^2/\text{s}$, upward from the plane of the wheel.

(b)

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 $4.8\text{rev}/\text{s}$.

Explanation of Solution

The angular momentum gained by the student, stool, and wheel is exactly twice the original angular momentum of the wheel. The student can stop the rotation of the stool by flipping the wheel axis back to its original direction.

Write the expression for conservation of angular momentum.

$L_s-L_w=L_w$

Here,

$L_s$ is the angular momentum of the student and the stool

$L_w$ is the final angular momentum of the wheel

Write the expression for the angular momentum of the wheel.

$L{w}_{}=I\omega$

Here,

$I$ is the rotational inertia of the wheel

$\omega$ is the angular velocity of the wheel

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

$\begin{array}{c}{L}_w=\left(2\text{ kg}\cdot {\text{m}}^2\right)\left(6\text{rev}/\text{s}\right)\\ =12\text{kg}\cdot {\text{m}}^2/\text{s}\end{array}$

Equating with the angular momentum of the wheel and the stool,

$\begin{array}{c}{L}_s-12\text{kg}\cdot {\text{m}}^2/\text{s}=12\text{kg}\cdot {\text{m}}^2/\text{s}\\ L_s=24\text{kg}\cdot {\text{m}}^2/\text{s}\end{array}$

Write the expression for the angular momentum of the stool.

$L{s}_{}=I_s{\omega }_s$

Here,

$I_s$ is the rotational inertia of the stool

${\omega }_s$ is the angular velocity of the stool

Substitute $24\text{kg}\cdot {\text{m}}^2/\text{s}$ for $L_s$ and $5\text{ kg}\cdot {\text{m}}^2$ for $I_s$ in the above expression to find ${\omega }_s$.

$\begin{array}{c}{\omega }_s=\frac{24\text{kg}\cdot {\text{m}}^2/\text{s}}{5\text{ kg}\cdot {\text{m}}^2}\\ =\text{4}\text{.8 }\text{rev}/\text{s}\end{array}$

Conclusion:

Therefore, the rotational velocity of the student and the stool about their axis is $\text{4}\text{.8 }\text{rev}/\text{s}$.

(c)

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

The sum of the angular-momentum vector of the wheel about its axis of rotation and the angular-momentum vector of the student, stool, and wheel about the axis of rotation of the stool add to yield the original angular momentum. This will be true if the angular momentum gained by the student, stool, and wheel is exactly twice the original angular momentum of system.

The student will exert some forces on the handles to produce a torque on the wheel which then produces an equal and opposite torque on the student.

Conclusion:

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