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

  • Thread starter Thread starter makeAwish
  • Start date Start date
  • Tags Tags
    Physics
makeAwish
Messages
128
Reaction score
0
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!
 
on Phys.org


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 ---> [tex]Torque = I\omega[/tex]
v = vi + at ----> [tex]\omega = \omega i + \alpha t[/tex]
 


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 [tex]\theta = \omega t[/tex] cause there is acceleration.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
Replies
8
Views
3K
Replies
18
Views
8K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
7K
  • · Replies 9 ·
Replies
9
Views
7K