Question

A horizontal 652 N merry-go-round of radius

1.85 m is started from rest by a constant

horizontal force of 46.6 N applied tangentially

to the merry-go-round.

Find the kinetic energy of the merry-goround after 2.07 s. The acceleration of gravity

is 9.8 m/s2. Assume the merry-go-round is a

solid cylinder.

Answer in units of J. Don't round answer.

Answer 1

$$I=0.5mr^2=0.5\frac{W}{g}r^2=0.5\frac{652}{9.8}(1.85)^2=113.85\ kg\cdot m^2$$

$$I\alpha=Fr\to 113.85\alpha=(46.6)(1.85)$$

$$\alpha=0.7572\frac{rad}{s^2}$$

Angular velocity:

$$\omega=\alpha t=0.7572(2.07)=1.567\frac{rad}{s}$$

$$K=0.5I\omega^2=0.5(113.85)(1.567)^2=140\ J$$