Chapter 13, Problem 73PQ

A uniform disk of mass m = 10.0 kg and radius r = 34.0 cm mounted on a frictionless axle through its center, and initially at rest, is acted upon by two tangential forces of equal magnitude F, acting on opposite sides of its rim until a point on the rim experiences a centripetal acceleration of 4.00 m/s² (Fig. P13.73). a. What is the angular momentum of the disk at this time? b. If F = 2.00 N, how long do the forces have to be applied to the disk to achieve this centripetal acceleration?

FIGURE P13.73

To determine

Angular momentum of the disk.

Answer to Problem 73PQ

Angular momentum of the disk is $\underset{\_}{1.98\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}}$.

Explanation of Solution

The disk in the question rotates about an axis passing through the center of the disk. Also the axis is perpendicular to the surface of the disk.

Write the equation to find the moment of inertia of the disk about an axis passing through its center.

  $I=\frac{1}{2}M{r}^2$                                                                                                          (I)

Here, $I$ is the moment of inertia of the disk about the axis passing through the center, $M$ is the mass of the disk , and $r$ is the radius of the disk.

Write the equation to find the centripetal acceleration felt by the disk.

  $a=\frac{v^2}{r}$                                                                                                                (II)

Here, $a$ is the centripetal acceleration, $v$ is the linear velocity, and $r$ is the radius of the disk.

Write the equation to find the linear speed.

  $v=r\omega$                                                                                                                 (III)

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

Substitute equation (III) in (II).

  $\begin{array}{c}a=\frac{{\left(r\omega \right)}^2}{r}\\ ={\omega }^{2}r\end{array}$                                                                                                        (IV)

Rewrite equation (IV) to get $\omega$.

  $\omega =\sqrt{\frac{a}{r}}$                                                                                                                (V)

Write the equation to find the angular momentum of the disk.

  $L=I\omega$                                                                                                               (VI)

Here, $L$ is the angular momentum.

Substitute (V) to (VI) to get $L$.

  $L=I\sqrt{\frac{a}{r}}$                                                                                                           (VII)

Conclusion:

Substitute $10.0\text{kg}$ for $M$ and $34\text{cm}$ for $r$ in equation (I) to get $I$.

  $\begin{array}{c}I=\frac{1}{2}\left(10.0\text{kg}\right){\left(34\text{cm}\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right)}^2\\ =0.578\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $4.00\text{m}/{\text{s}}^{\text{2}}$ for $a$ , $0.578\text{kg}\cdot {\text{m}}^2$ for $I$ and $34\text{cm}$ for $r$ in equation (VII) to get $L$.

  $\begin{array}{c}L=0.578\text{kg}\cdot {\text{m}}^2\sqrt{\frac{\left(4.00\text{m}/{\text{s}}^{\text{2}}\right)}{34\text{cm}\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)}}\\ =1.98\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}\end{array}$

Therefore, angular momentum of the disk is $\underset{\_}{1.98\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}}$.

To determine

The time duration for which the force have to be applied so that the disk achieves the centripetal acceleration.

Answer to Problem 73PQ

The forces have to act for a time duration of $\underset{\_}{1.46\text{s}}$.

Explanation of Solution

Torque is the rate of change of angular momentum.

Write the equation to find the torque acting on the disk.

  $\tau =\frac{\Delta L}{\Delta t}$                                                                                                               (VIII)

Here, $\tau$ is the torque acting on the disk, $\Delta L$ is the change in angular momentum and $\Delta t$ is the time duration.

Rewrite equation (VIII) to get $\Delta t$.

  $\Delta t=\frac{\Delta L}{\tau }$                                                                                                                (IX)

Write the equation to find $\tau$ in both the sides of the disk.

  $\tau =2Fr$

Here, $F$ is the force acting on the disk.

Substitute above equation in (IX).

  $\Delta t=\frac{\Delta L}{2Fr}$                                                                                                             (X)

Change in angular momentum is the difference between the final angular momentum and initial angular momentum of the disk.

Rewrite equation (IX).

  $\Delta t=\frac{L_f-L_i}{2Fr}$                                                                                                       (XI)

Here, $L_f$ is the final angular momentum and $L_i$ is the initial angular momentum.

Conclusion:

The disk was initially at rest. Therefore $L_i$ is zero.

Substitute $1.98\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}$ for $L_f$, $0\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}$ for $L_i$, $2.00\text{N}$ for $F$ , and $34\text{cm}$ for $r$ in equation (XI) to get $\Delta t$.

  $\begin{array}{c}\Delta t=\frac{1.98\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}}{2\left(2.00\text{N}\right)\left(34\text{cm}\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right)}\\ =1.46\text{s}\end{array}$

Therefore, the forces have to act for a time duration of $\underset{\_}{1.46\text{s}}$.