How Does Angular Momentum Affect Merry-Go-Round Speed?

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The discussion focuses on calculating the angular speed of a merry-go-round after a child applies a tangential force. The child exerts an 18.0N force for 25 seconds, leading to a final angular speed of 0.3 rad/s. Participants clarify that the child is considered an external force, and angular momentum is not conserved in this scenario. To determine the work done by the child, it is noted that the angular speed is not constant, requiring kinematic equations to find the angle of rotation. The correct approach involves calculating torque and angular acceleration to accurately assess the work done.
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Homework Statement

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!
 
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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
 


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.
 


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
 


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..
 


makeAwish said:
I tried working out.
W = Fs = Frθ = Fr(integration of ω) = 18*2*0.3*25 = 270J
It's wrong because the angular speed is not constant, so θ ≠ ωt. Use your kinematic formulas to find the angle.
 


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)
 


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