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

  • Thread starter Thread starter cristina
  • Start date Start date
  • Tags Tags
    Homework Rotation
AI Thread Summary
To calculate the angular acceleration of a merry-go-round with a radius of 2.2m and a force of 260N applied for 12 seconds, the torque exerted is 572 Nm. The angular acceleration can be derived from the relationship between torque and moment of inertia, which is assumed to be that of a solid cylinder. The moment of inertia is calculated using the formula I = 1/2(MR^2), but the mass M needs to be determined from the angular acceleration. The final moment of inertia is debated, with calculations suggesting values around 6554.64 m²kg, and rounding rules indicate it should be expressed as 6.6 x 10^3 m²kg. The discussion emphasizes the importance of significant figures in reporting the final answer.
cristina
Messages
53
Reaction score
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
 
Physics news on Phys.org
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:
d\alpha = \alpha _0 + \omega _0t + \frac{1}{2}a_rt^2
2\pi = \frac{1}{2}a_r(12s)^2
a_r = \frac{4\pi }{144s^2}

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
 
Back
Top