# Merry-Go-Round physics homework

## Homework Statement

A merry-go-round in a playground has a radius of about 2 m and a mass of about 400 kg. Starting from rest, a parent begins to rotate it with constant angular acceleration. After 5 s, it reaches its final angular velocity of about pi/2 rad/s, and continues to rotate at that angular velocity

a.
Treat the merry-go-round as a uniform disk (the moment of inertia of a uniform disk is mr2/2); the final rotational kinetic energy of the merry-go-round is about ___ J.

b.
5 children, each with a mass of about 20 kg, step on the merry-go-round and stay near its edge. As they step on it, its angular velocity drops to about ____ rad/s. Ignore friction.

## Homework Equations

I_disk=0.5mr^2
KE_rotational=0.5*I*w^2

## The Attempt at a Solution

For part a I did...

KE=(0.5)(0.5)(400)(2^2)(pi/2)^2=987 J

For part b I did...

KE_i=KE_f

987 = 0.5(100)(2^2)(w^2) + 0.5(0.5)(400)(2^2)(w^2)

and get w=1.3 but the answer is supposed to be 1.0. Is mr^2 the correct way to do the kid's intertia?

Last edited:

## Answers and Replies

Gyroscope
Why do you think that kinetic energy will remain the same?

What physical quantity you are sure that will remain constant? Maybe angular momentum, who knows!?

Yes, think about what gyro is stating.
Also, your KE eqn is incorrect, there should be an omega squared. You seemed to take this into account in your soln. though.*edit* it is fixed now

Last edited:
Used conservation of angular momemtum and got the correct answer. Though, why is KE not conserved?

Thanks,
Brandon

Gyroscope
When the kids jump to the merry-go-round they exert forces on it. Forces cause change in kinetic energy, do not they?