Chapter 8, Problem 47P

To determine

To find: Net work required to rotate marry-go-round.

Answer to Problem 47P

Net work required to rotate marry-go-round is $1.42\times {10}^4\text{J}$ .

Explanation of Solution

Given Information: Mass of marry-go-round $m=1640\text{kg}$

Radius of marry-go-round $r\text{= 7}\text{.50}\text{m}$

Rotation rate $\omega \text{=}\frac{1\text{rev}}{8\text{s}}=\frac{2\pi }{8}\text{rad/s}$

Concept used:

  $\begin{array}{l}w\text{=}\frac{1}{2}I{\omega }^2\\ \text{and}I=\frac{1}{2}m{r}^2\end{array}$

Thus, $w=\frac{1}{2}(\frac{1}{2}m{r}^2){\omega }^2$

Here,

  $w$ is the work,

  $I$ is the moment of inertia,

  $\omega$ is the angular velocity,

  $m$ is the mass,

  $r$ is the radius.

Calculation:

  $\begin{array}{l}w=\frac{1}{2}(\frac{1}{2}m{r}^2){\omega }^2\\ =\frac{1}{2}(\frac{1}{2}\times 1640\times {(7.5)}^2){(\frac{2\pi }{8})}^2\\ w=14226\text{J}\\ \text{or}\\ w=1.42\times {10}^4\text{J}\end{array}$

Conclusion: Net work required to rotate marry-go-round is $1.42\times {10}^4\text{J}$ .