Question involving rotational motion

  • Thread starter gaobo9109
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Homework Statement


A ring of mass M is hung from the ceiling. At the top of the ring are two smaller rings of mass m. There is no friction between the big and small rings. If two smaller rings slip down the big ring, for what value of m will the big ring move upward. A picture is attached to make the question clearer.


Homework Equations


a = v(sqrt)/R


The Attempt at a Solution


I just don't understand how do all the forces react to give rise to the upward motion of the big ring.
 

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Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

Hi gaobo9109! Welcome to PF! :smile:

Hint: the only forces on the big ring are Mg downwards and the two normal forces, 2N, from the two little rings.

Assuming the big ring is fixed, what is N as a function of height (or of angle), and in which direction? :wink:
 
  • #3
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But there is still tension force from the rope. And from my understanding, i think that the tension force will vary accordingly when the small rings slip down the big ring. Downward force will always be balanced by upward tension force. How would the big ring actually move upward?
 
  • #4
tiny-tim
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No, you can ignore the tension, because …

when the big ring is in equilibrium, the tension automatically is equal and opposite to the other net downward forces,

but if the other net forces are upward, the tension is zero! :smile:

So all you have to find is whether the other net forces are upward, ie whether the resultant of the forces from the two little rings is greater than Mg. :wink:
 
  • #5
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Okay i know the general direction to sovlve this question. But half way through the solution, i was stuck.

Here is my working:

let the radius of the big ring be R.

mgR - mgRcosθ = 1/2mv2
mgcosθ - N = mv2/R
v2 = 2gR(1-cosθ)
N = mgcosθ - 2mgR(1-cosθ)/R
= mgcosθ - 2mg(1-cosθ)
= mg(3cosθ-2)

N(small ring on big ring) = mg(2-3cosθ)

I have no idea how to continue from here.
 
Last edited:
  • #6
tiny-tim
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Hi gaobo9109! :smile:
mgR - mgRcosθ = 1/2mv2
mgcosθ - N = mv2/R
v2 = 2gR(1-cosθ)
N = mgcosθ - 2mgR(1-cosθ)/R
= mgcosθ - 2mg(1-cosθ)
= mg(3cosθ-2)

N(small ring on big ring) = mg(2-3cosθ)

Fine :smile:, except that the PE is mgR - mgRsinθ.

ok, that was good ol' Newton's second law for one of the small rings.

Now you need Newton's second law for the big ring …

when will T be zero? :wink:
 
  • #7
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yeah I know that. But what I am not sure is how to express the y component of normal force in term of θ.

I don't know at what value of θ will the big ring move up. If θ is less than 90 degree, then below is the equation
2Ncosθ = mg
2mg(2-3cosθ)cosθ = Mg

But if θ is greater than 90 degree, the above equation would not apply.
 
  • #8
tiny-tim
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Yes, so θ can't be greater than 90º (for the big ring to start moving up). :smile:

(don't forget to change that cos θ to sin θ :wink:)
 

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