Chapter 8, Problem 1SP

(a)

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 about its axle is $132\text{ N}\cdot \text{m}$.

Explanation of Solution

Given info: The radius of the merry-go-round is $1.8\text{ m}$. The rotational inertia is $900\text{ kg}\cdot {\text{m}}^2$. The force applied is $80\text{ N}$ and the frictional torque of the axle of the merry-go-round is $12\text{ N}\cdot \text{m}$.

Write the expression to find the torque acting due to the force.

$\tau =Fl$

Here,

$F$ is the applied force

$l$ is the radius of the merry-go-round

Substitute $80\text{ N}$ for $F$ and $1.8\text{ m}$ for $l$ in the above equation.

$\begin{array}{c}\tau =\left(80\text{ N}\right)\left(1.8\text{ m}\right)\\ =144\text{ N}\cdot \text{m}\end{array}$

Write the expression to find the net torque acting on the axle of the merry-go-round.

${\tau }_{\text{net}}=\tau -{\tau }_f$

Here,

$\tau$ is the torque due to the force

${\tau }_f$ is the frictional torque at the axle

Substitute $144\text{ N}\cdot \text{m}$ for $\tau$ and $12\text{ N}\cdot \text{m}$ for ${\tau }_f$ in the above equation.

$\begin{array}{c}{\tau }_{\text{net}}=144\text{ N}\cdot \text{m}-12\text{ N}\cdot \text{m}\\ =132\text{ N}\cdot \text{m}\end{array}$

Conclusion:

Therefore, the net torque acting on the merry-go-round about its axle is $132\text{ N}\cdot \text{m}$.

(b)

To determine

The rotational acceleration of the merry-go-round.

Answer to Problem 1SP

The rotational acceleration of the merry-go-round is $0.147\text{rad}/{\text{s}}^2$.

Explanation of Solution

Write the expression to find the rotational acceleration of the merry-go-round.

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

Here,

$\alpha$ is the rotational acceleration

$\tau$ is the torque

$I$ is the rotational inertia

Substitute $132\text{ N}\cdot \text{m}$ for $\tau$ and $900\text{ kg}\cdot {\text{m}}^2$ for $I$ in the above equation.

$\begin{array}{c}\alpha =\frac{132\text{ N}\cdot \text{m}}{900\text{ kg}\cdot {\text{m}}^2}\\ =0.147\text{rad}/{\text{s}}^2\end{array}$

Conclusion:

Therefore, the rotational acceleration of the merry-go-round is $0.147\text{rad}/{\text{s}}^2$.

(c)

To determine

The rotational velocity of the merry-go-round after $15\text{ s}$ if it starts from rest.

Answer to Problem 1SP

The rotational velocity of the merry-go-round after $15\text{ s}$ if it starts from rest is $2.2\text{rad}/\text{s}$.

Explanation of Solution

Write the expression to find the rotational velocity of the merry-go-round.

$\omega =\alpha t$

Here,

$\alpha$ is the rotational acceleration

$\omega$ is the rotational velocity

$t$ is the time taken

Substitute $0.147\text{rad}/{\text{s}}^2$ for $\alpha$ and $15\text{ s}$ for $t$ in the above equation.

$\begin{array}{c}\omega =\left(0.147\text{rad}/{\text{s}}^2\right)\left(15\text{ s}\right)\\ =2.2\text{rad}/\text{s}\end{array}$

Conclusion:

Therefore, the rotational velocity of the merry-go-round after $15\text{ s}$ if it starts from rest is $2.2\text{rad}/\text{s}$.

(d)

To determine

The rotational acceleration of the merry-go-round after the child stops pushing after $15\text{ s}$ and the time taken for the merry-go-round to stop turning.

Answer to Problem 1SP

The rotational acceleration of the merry-go-round after the child stops pushing after $15\text{ s}$ is $-1.3\times {10}^{-2}\text{rad}/{\text{s}}^2$ and the time taken for the merry-go-round to stop turning is $165\text{ s}$.

Explanation of Solution

Write the expression for rotational acceleration of the merry-go-round after the child stops pushing.

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

Substitute $12\text{ N}\cdot \text{m}$ for ${\tau }_f$ and $900\text{ kg}\cdot {\text{m}}^2$ for $I$ in the above equation.

$\begin{array}{c}\alpha =\frac{-12\text{ N}\cdot \text{m}}{900\text{ kg}\cdot {\text{m}}^2}\\ =-1.3\times {10}^{-2}\text{rad}/{\text{s}}^2\end{array}$

Write the expression to find the time taken for the merry-go-round to stop turning.

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

Substitute $-1.3\times {10}^{-2}\text{rad}/{\text{s}}^2$ for $\alpha$ and $2.2\text{rad}/\text{s}$ for $\omega$ in the above equation.

$\begin{array}{c}t=\frac{2.2\text{rad}/\text{s}}{1.3\times {10}^{-2}\text{rad}/{\text{s}}^2}\\ =165\text{ s}\end{array}$

Conclusion:

Therefore, the rotational acceleration of the merry-go-round after the child stops pushing after $15\text{ s}$ is $-1.3\times {10}^{-2}\text{rad}/{\text{s}}^2$ and the time taken for the merry-go-round to stop turning is $165\text{ s}$.