Merry-go-round physics question

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In summary, we have a playground merry-go-round with a radius of 2.00m and a moment of inertia of 3000kgm² about a vertical axle through its center. A child applies an 18.0N force tangentially to the edge of the merry-go-round for 25.0s. The angular speed of the merry-go-round after this time interval is 0.3rad/s. The next question is to find the work done by the child on the merry-go-round. Using the torque equation and kinematic formulas, we can find the angular acceleration and then use it to find the angular velocity after a certain period of time. The correct answer for the work done is not 270J, as
  • #1
makeAwish
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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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  • #2


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


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


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]
 
  • #5


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


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.
 
  • #7


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


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.
 

1. How does a merry-go-round work?

The basic principle behind a merry-go-round is centripetal force. The force of the riders pushing against the outer edge of the spinning platform creates a centripetal force that keeps them moving in a circular path.

2. What is the relationship between speed and centripetal force on a merry-go-round?

The faster the merry-go-round spins, the greater the centripetal force needed to keep the riders moving in a circular path. This is because the centripetal force is directly proportional to the square of the speed.

3. What is the role of friction on a merry-go-round?

Friction plays a crucial role in keeping the riders on the merry-go-round. Without friction, the riders would slide off the platform due to the centrifugal force pulling them outwards.

4. How does the distance from the center of the merry-go-round affect the ride?

The farther away a rider is from the center of the merry-go-round, the faster they will travel in a circular path. This is because the distance affects the radius, which is directly proportional to the speed.

5. Can a merry-go-round ever reach a point of equilibrium?

No, a merry-go-round cannot reach a point of equilibrium because it is constantly moving and is subject to external forces such as friction and air resistance. These forces will always disrupt the balance and cause the merry-go-round to continue spinning.

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