Chapter 8, Problem 1SP

A merry-go-round in the park has a radius of 1.5 m and a rotational inertia of 800 kg·m². A child pushes the merry-go-round with a constant force of 92 N applied at the edge and parallel to the edge. A frictional torque of 14 N·m acts at the axle of the merry-go-round.

1. a. What is the net torque acting on the merry-go-round about its axle?
2. b. What is the rotational acceleration of the merry-go-round?
3. c. At this rate, what will the rotational velocity of the merry-go-round be after 16 s if it starts from rest?
4. d. If the child stops pushing after 16 s, the net torque is now due solely to the friction. What then is the rotational acceleration of the merry-go-round? How long will it take for the merry-go-round to stop turning?

To determine

The net torque acting on the merry –go-round about its axle.

Answer to Problem 1SP

The net torque acting on the merry-go-round is $124\text{ N}\cdot \text{m}$ directed along the merry-go-round.

Explanation of Solution

Given info: Radius is $1.5\text{ m}$, rotational inertia is $800\text{ kg}\cdot {\text{m}}^2$, constant force is $92\text{ N}$ and the frictional torque is $14\text{ N}\cdot \text{m}$.

Write the expression for the net torque.

$\tau =-Fl$

Here,

$\tau$ is the torque

$F$ is the force

$l$ is the length

Substitute $92\text{ N}$ for $F$ and $1.5\text{ m}$ for $l$ to find $\tau$.

$\begin{array}{c}\tau =-92\text{ N}\times \text{1}\text{.5 m}\\ \text{=138 N}\cdot \text{m}\end{array}$

The net torque acting on the merry-go round is given by subtracting from the frictional torque.

$\begin{array}{c}\tau =138\text{ N}-14\text{ N}\cdot \text{m}\\ \text{=124 N}\cdot \text{m}\end{array}$

Conclusion:

Therefore, the net torque acting on the merry-go-round is $124\text{ N}\cdot \text{m}$ directed along the merry-go-round.

To determine

The rotational acceleration of the merry-go-round.

Answer to Problem 1SP

The rotational acceleration of the merry-go-round is $0.155\text{rad}/{\text{s}}^2$.

Explanation of Solution

Write the expression for the rotational acceleration.

$\alpha =\frac{\tau }{I}$

Here,

$I$ is the rotational inertia

$\alpha$ is the rotational acceleration

Substitute $124\text{ N}\cdot \text{m}$ for $\tau$ and $800\text{ kg}\cdot {\text{m}}^2$ for $I$ to find $\alpha$.

$\begin{array}{c}\alpha =\frac{124\text{ N}\cdot \text{m}}{800\text{ kg}\cdot {\text{m}}^2}\\ =0.155\text{rad}/{\text{s}}^2\end{array}$

Conclusion:

Therefore, the rotational acceleration of the merry-go-round is $0.155\text{rad}/{\text{s}}^2$.

To determine

The rotational velocity of the merry-go-round after $16\text{ s}$.

Answer to Problem 1SP

The rotational velocity of the merry-go-round will be $2.325\text{rad}/\text{s}$.

Explanation of Solution

Write the expression for the rotational velocity.

$\omega =\alpha t$

Here,

$\omega$ is the rotational velocity

$\alpha$ is the rotational acceleration

$t$ is the time

Substitute $0.155\text{rad}/{\text{s}}^2$ for $\alpha$ and $16\text{ s}$ for $t$ to find $\omega$.

$\begin{array}{c}\omega =0.155\text{rad}/{\text{s}}^2\times 16\text{ s}\\ =2.325\text{rad}/\text{s}\end{array}$

Conclusion:

Therefore, the rotational velocity of the merry-go-round will be $2.325\text{rad}/\text{s}$.

To determine

Rotational acceleration of the merry-go-round and the time taken for the merry-go-round to stop running.

Answer to Problem 1SP

The rotational acceleration will be $0.0175\text{rad}/{\text{s}}^2$ and the time taken will be $132\text{ s}$.

Explanation of Solution

The net torque is only due to the friction.

Write the expression for the rotational acceleration.

$\alpha =\frac{\tau }{I}$

Substitute $14\text{ N}\cdot \text{m}$ for $\tau$ and $800\text{ kg}\cdot {\text{m}}^2$ for $I$ to find $\alpha$.

$\begin{array}{c}\alpha =\frac{14\text{ N}\cdot \text{m}}{800\text{ Kg}\cdot {\text{m}}^2}\\ =0.0175\text{rad}/{\text{s}}^2\end{array}$

Write the expression for the rotational velocity.

$\omega =\alpha t$

Rearrange the above equation to find $t$.

$t=\frac{\omega }{\alpha }$

Substitute $2.325\text{rad}/\text{s}$ for $\omega$ and $0.0175\text{rad}/{\text{s}}^2$ for $\alpha$ to find $t$.

$\begin{array}{c}t=\frac{2.325\text{rad}/\text{s}}{0.0175\text{rad}/{\text{s}}^2}\\ =132\text{ s}\end{array}$

Conclusion:

Therefore, rotational acceleration will be $0.0175\text{rad}/{\text{s}}^2$ and the time taken will be $132\text{ s}$.