Question

A horizontal 781 N merry-go-round of radius 1.75 m is started from rest by a constant

horizontal force of 75 N applied tangentially to the merry-go-round.

Find the kinetic energy of the merry-goround after 3.97 s. The acceleration of gravity

is 9.8 m/s2

Assume the merry-go-round is a solid cylinder.

Answer in units of J.

Answer 1

$$I=0.5mr^2=0.5\frac{W}{g}r^2=0.5\frac{781}{9.8}(1.75)^2=122\ kg\cdot m^2$$

$$I\alpha=Fr\to 122\alpha=(75)(1.75)$$

Angular acceleration:

$$\alpha=1.08\frac{rad}{s^2}$$

Angular velocity:

$$\omega=\alpha t=1.08(3.97)=4.29\frac{rad}{s}$$

$$K=0.5I\omega^2=0.5(122)(4.29)^2=1120\ J$$