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

In summary, the radius of the merry-go-round is 2.2m and a force of 260N is applied for 12 seconds to make one complete rotation. The angular acceleration is 4π/144s^2, the torque exerted by the rope is 572Nm, and the moment of inertia is 6.6x103 m2kg. The unit for moment of inertia is m2kg and the answer should be rounded to 6.6x103 m2kg.
  • #1
cristina
53
0
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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  • #2
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).
 
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  • #3
472Nm*144s^2 = 4(pi)I
I=67968/4(pi) = 5408.72 (I feel there is something with the answer)
 
  • #4
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?
 
  • #5
I = 6554.64 ? I think this is the corect answer but I don't know the rounding if its correct.
 
  • #6
I also get 6,544. I'm not sure what you're asking, is this not the correct answer?
 
  • #7
Yes, is it the correct answer? Is the unite of I is the same as the unite of mass?
 
  • #8
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.
 
  • #9
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?
 
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  • #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
 

1. What is the concept of rotation in physics?

Rotation in physics refers to the circular movement of an object around an axis or a fixed point. It involves both the angular displacement and angular velocity of the object.

2. How is rotation different from revolution?

Rotation refers to the movement of an object around its own axis, while revolution refers to the movement of an object around a different object or point.

3. What are the units of measurement for angular displacement and angular velocity?

Angular displacement is measured in radians or degrees, while angular velocity is measured in radians per second or degrees per second.

4. How does the direction of rotation affect an object's angular displacement and angular velocity?

The direction of rotation determines whether the angular displacement and angular velocity are positive or negative. Clockwise rotation is considered negative, while counterclockwise rotation is considered positive.

5. Can an object rotate without changing its angular velocity?

No, an object cannot rotate without changing its angular velocity. Any circular movement involves both a change in angular displacement and angular velocity.

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