Question

A playground merry-go-round has a mass of 120 kg and a radius of 1.80 m and it is rotating with an angular velocity of 0.600 rev/s. What is its angular velocity (in rev/s) after a 25.0 kg child gets onto it by grabbing its outer edge

Answer 1

Answer:

The final velocity $\omega_f = 0.4235 \frac{rev}{s}$

Explanation:

Given data

Mass of merry go round $M_m$ = 120 kg

Radius = 1.8 m

Initial angular velocity $\omega_i$ = 0.6 $\frac{rev}{sec}$

Mass of boy $M_{boy}$ = 25 kg

We know that the final velocity is given by

$\omega_f = \frac{\frac{1}{2}M_m \omega_i }{M_{boy} + \frac{1}{2} M_m }$

Put all the values in above formula we get

$\omega_f = \frac{\frac{1}{2}(120) 0.6}{25 + \frac{1}{2} (120) }$

$\omega_f = 0.4235 \frac{rev}{s}$

This is the final velocity.