Chapter 8, Problem 36P

To determine

To find: The torque that is required to produce acceleration.

Answer to Problem 36P

The force value is $169\text{N}$ .

Explanation of Solution

Given information:

A teenager pushes tangentially on a small hand-driven merry-go-round and is able to accelerate it from rest to a frequency of $15\text{rpm}$ in $10.0\text{s}$ . Assume the merry-go-round is a uniform disk of radius $2.5\text{m}$ and has a mass of $760\text{kg}$ , and two children (each with a mass of $25\text{kg}$ ) sit opposite each other on the edge.

Formula used:

  $I=\frac{1}{2}M{r}^2+m{r}^2$

Here,

  $I$ is the moment of inertia,

  $m,M$ is the mass,

  $r$ is the radius.

Calculation:

Frequency of rotation $f=15\text{rpm}$

Time taken $t=10.0\text{s}$

Radius of merry go round $r=2.5\text{m}$

Mass of merry go ground $M=760\text{kg}$

Mass of two children $m=\left(25\times 2\right)\text{kg}=\text{50}\text{kg}$

The moment of inertia for the system of merry go round and Children

  $\begin{array}{l}I=\frac{1}{2}M{r}^2+m{r}^2\\ =\left(\frac{1}{2}M+m\right)r^2\\ =\left(\frac{1}{2}\times 760+50\right){\left(2.5\right)}^2\text{kg}\cdot {\text{m}}^2\\ =\left(380+50\right){\left(2.5\right)}^2\text{kg}\cdot {\text{m}}^2\\ \text{or}=2.69\times {10}^3\text{kg}\cdot {\text{m}}^2\end{array}$

The torque required from $\tau =I\alpha$

  $\begin{array}{l}=\frac{I\left(\omega -{\omega }_0\right)}{t}\\ {\omega }_0=0\text{and}\\ \omega =\left(\frac{\text{15}\text{rpm}}{\text{60}\text{s/min}}\right)\left(\frac{\text{2π}\text{rad}}{\text{1}\text{rev}}\right)\\ \omega =1.57\text{rad/s}\end{array}$

Therefore, the Torque is

  $\begin{array}{l}\tau =\left(\frac{2.69\times {10}^3\times 1.57}{10}\right)\text{N}\cdot \text{m}\\ \tau =423\text{N}\cdot \text{m}\end{array}$

The force required at the edge $F=\tau /r$

  $\begin{array}{l}=\left(\frac{423}{2.5}\right)\text{N}\\ =169\text{N}\end{array}$

Force $F=169\text{N}$

Conclusion:

The force value is $169\text{N}$ .