Moment of inertia and torque of a yoyo

  • Thread starter JulienB
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Homework Statement



Hi everybody! I know that is a classical problem, but I haven't been able to find an answer to my questions in the other threads so here we go:

We have a yoyo made of 3 aluminium disks (density ρ). The two side disks have a radius R and thickness D, and the middle disk has a radius R0 and a thickness D0. The string has length l (see picture).

a) Calculate the moment of inertia of the yoyo. Derive then the moment of inertia of a disk of mass M and radius r about its axis of symmetry.
b) Calculate the torque when the yoyo is "falling". In which direction does it point?
c) What is the angular frequency ω when the string is half unrolled?

Homework Equations



Moment of inertia, torque, tangential acceleration and so on.

The Attempt at a Solution



a) My first issue attempting to solve this problem is that I am unsure whether the radius R0 should be considered when calculating the moment of inertia of the yoyo. I am confused, because the string is responsible for the rotation but yet the yoyo is rotating about its axis of symmetry. Therefore I would assume the moment of inertia of the yoyo is:

Iyoyo = 2⋅Ibig disk + Ismall disk
= 2⋅(ρπDR4) + ½ρπD0R04
= ρπ⋅(DR4 + ½⋅D0R04)

I quickly add the calculation for the moment of inertia of a disk:
Idisk = ρ∫ r2dV = ρDπr4/2 = ½ ⋅Mr2

Is that correct, or is there any need to use the parallel axis theorem?

b) Here it becomes complicated. I think the only force playing a role on the rotation is the tension force of the string, therefore:

Στ = I⋅α = F⋅R0 where α is of course the angular acceleration ⇒ α = aT/R0.

I also think aT = ay and Myoyo⋅g - I⋅α/R0 = Myoyo⋅g - I⋅ay/R02 = Myoyo⋅ay

ay = Myoyo⋅g / (Myoyo + I/R0)

But when I insert that in my torque equation, it gets crazy:

Στ = I⋅α = (I/r0)⋅(Myoyo⋅g / M + I/R0)

When I plug my moment of inertia in, nothing good comes out and it gets very messy. Have I made a mistake or should I also put up an equation of energy?


Thank you very much in advance.


EDIT: I just saw a similar problem on internet where the guy takes the point of contact of the string with the disk as the torque axis. Is that really allowed?? If so, would I have to use the parallel axis theorem because we don't take the center of the yoyo as axis of rotation anymore?


Julien.
 

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Answers and Replies

  • #2
haruspex
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You seem to have forgotten the factor 1/2 in Ibig disk.
You can approach it in either of two ways. Use the MoI about mass centre and torque about mass centre (radius x tension); or use the parallel axis theorem to find the MoI about point of contact of the string and torque about that point (radius x mg). They should lead to the same answer. You just have to pick your axis and be consistent.
 
  • #3
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@haruspex I don't think I forgot the 1/2, because there are two big disks. No? I'll give the problem another try tomorrow and post again :)
 
  • #4
haruspex
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@haruspex I don't think I forgot the 1/2, because there are two big disks. No? I'll give the problem another try tomorrow and post again :)
Well, you forgot it here:
= 2⋅(ρπDR4) + ½ρπD0R04
But I see you found it again here:
= ρπ⋅(DR4 + ½⋅D0R04)
 
  • #5
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@haruspex My mistake I forgot it by copying my notes. Thank you for your attention :)
 
  • #6
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I gave it another go but still get a crazy result for the torque:
 

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  • #7
haruspex
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I gave it another go but still get a crazy result for the torque:
Why do you think it's crazy. I would not bother multiplying out the factors in the numerator. You have an error in the denominator going from the second line to the third line... a factor of 2.
 
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@haruspex Thank you for your answer. I'm doing the problem again, and I'm wondering: shouldn't I have taken aT = g (instead of the whole acceleration (m⋅g - T)/m) in my torque equation?

Julien.
 
  • #9
haruspex
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@haruspex Thank you for your answer. I'm doing the problem again, and I'm wondering: shouldn't I have taken aT = g (instead of the whole acceleration (m⋅g - T)/m) in my torque equation?

Julien.
How are you defining aT? Certainly the sum of vertical forces is mg-T downwards.
 
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  • #10
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@haruspex Nevermind I got it wrong. Sorry for that :)
 

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