# Merry-go-round physics question

1. Mar 20, 2009

### makeAwish

The problem statement, all variables and given/known data

A playground merry-go-round has radius 2.00m and moment of inertia 3000kgm² about a vertical axle through its center, and it turns with negligible friction.
A child applies an 18.0N force tangentially to the edge of the merry-go-round for 25.0s . If the merry-go-round is initially at rest, what is its angular speed after this 25.0s interval?

The attempt at a solution

I have no idea what to apply actually. I only thinking of treating the child as a particle.
Moment of inertia of child = mr²

Can apply consv of angular momentum? But is the child considered to be acting external force?

Can someone help me pls? Thanks!!

2. Mar 20, 2009

### JaWiB

Re: Merry-go-round

Based on what you wrote, it sounds like the child isn't actually on the merry-go-round (especially since the child's weight is not given) so I would assume the child's push is simply an external force.

I'd think you could use torque/angular acceleration or angular impulse/momentum to solve

3. Mar 20, 2009

### Staff: Mentor

Re: Merry-go-round

The child isn't riding the merry-go-round, she's just exerting an external force on it. Angular momentum is not conserved. What torque does she exert? Apply Newton's 2nd law for rotation.

4. Mar 20, 2009

### Delphi51

Re: Merry-go-round

The child is not on the merry-go-round so it does not make sense to calcuate his/her moment of inertia. This question is analogous to applying a force to a mass and asking how fast it would go after a given time. You would use F = ma to find the acceleration and then use v = at to find the velocity. In the merry-go-round problem, use the analogous circular motion formulas.
F = ma ---> $$Torque = I\omega$$
v = vi + at ----> $$\omega = \omega i + \alpha t$$

5. Mar 20, 2009

### makeAwish

Re: Merry-go-round

Okay, i got it. thanks! angular speed = 0.3rad/s

The next part is actually:

How much work did the child do on the merry-go-round?

I tried working out.
W = Fs = Frθ = Fr(integration of ω) = 18*2*0.3*25 = 270J

But its wrong..

6. Mar 21, 2009

### Staff: Mentor

Re: Merry-go-round

It's wrong because the angular speed is not constant, so θ ≠ ωt. Use your kinematic formulas to find the angle.

7. Mar 21, 2009

### sArGe99

Re: Merry-go-round

Write the torque equation and find out angular acceleration.
Use it to find the angular velocity after a certain period of time. (Use kinematic relationship)

8. Mar 21, 2009

### sArGe99

Re: Merry-go-round

The angle through which it has rotated can be found out by kinematics again.
The angle won't be of the form $$\theta = \omega t$$ cause there is acceleration.