How Do You Calculate the Angular Acceleration and Torque of a Merry-Go-Round?

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Homework Help Overview

The discussion revolves around calculating the angular acceleration and torque of a merry-go-round, with a specific focus on the relationships between force, torque, and moment of inertia. The problem involves applying principles of rotational motion and dynamics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between torque, angular acceleration, and moment of inertia, questioning how to isolate variables such as mass. Some suggest using motion equations to derive angular acceleration from rotational displacement.

Discussion Status

There is ongoing exploration of calculations related to moment of inertia, with participants sharing differing results and questioning the correctness of their answers. Some guidance on significant figures and units is provided, but no consensus on the final answer has been reached.

Contextual Notes

Participants are working under the constraints of the problem's parameters, including the assumption of the merry-go-round's shape and the implications of using specific equations of motion. There is uncertainty regarding the rounding of final answers and the appropriate units for moment of inertia.

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The radius of a park merry-go-round is 2.2m. To start it rotating, you wrap a rope around it and pull with a force of 260N for 12s. During this time, the merry-go-round makes one complete rotation. a) Find the angular acceleration of the merry-go-round. b) What torque is exerted by the rope on the merry-go-round? c) What is the moment of inertia of the merry-go-round?

A)
The angular acceleration is alpha = Torque/I
I did conceder the merry-go-round as a solid cylinder so I =1/2(MR^2)
Torque = Fr
Acceleration = (260N * 2.2m)/(1/2(MR^2))
I don’t know how to get rid of the M?

B)
Torque = Fr = (260N * 2.2m) = 572 Nm

C) The moment of inertia of the merry-go-round is I =1/2(MR^2)
Can I get the M in the torque from F = ma (after I know the acceleration from a) )?

Thank you
 
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The merry-go-round completes one rotation, i.e 2π radians, in 12 seconds. I think it's valid to apply the normal motion equations here, so:
[tex]d\alpha = \alpha _0 + \omega _0t + \frac{1}{2}a_rt^2[/tex]
[tex]2\pi = \frac{1}{2}a_r(12s)^2[/tex]
[tex]a_r = \frac{4\pi }{144s^2}[/tex]

Now that you know the angular acceleration you can find the merry-go-round's moment of inertia since a = Torque/I as you said. With this method you don't even need to know the shape of the device (and I don't think you are allowed to assume it is a cylinder).
 
Last edited:
472Nm*144s^2 = 4(pi)I
I=67968/4(pi) = 5408.72 (I feel there is something with the answer)
 
It's 572N, not 472N. According to this the mass of the merry-go-round would be 2.7 tons... do you know what the answer should be?
 
I = 6554.64 ? I think this is the corect answer but I don't know the rounding if its correct.
 
I also get 6,544. I'm not sure what you're asking, is this not the correct answer?
 
Yes, is it the correct answer? Is the unite of I is the same as the unite of mass?
 
No, the unit of I, the moment of inertia, is m^2kg. I don't know if it's the right answer or not, but as far as I'm concerned my calculations are correct.
 
I really appreciate your help. I think the answer is correct But I am not sure if I should round it to 6555 m^2.kg?
 
Last edited:
  • #10
like in this case what would be the significant figures rule?
 
  • #11
The general rule would be that you cannot have more significant digits then the numbers used in the calcultion. I would say the answer should be

6.6x 103 m2kg
 

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