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Find the speed of the descending block

  1. Jun 9, 2017 #1
    1. The problem statement, all variables and given/known data
    19059384_748454021995385_4693005147032055119_n.jpg
    So, we have a ring 'A' having mass 'm' that can slide on the horizontal rod. It is attached to a massless string, whose other end is attached to a block of mass 'M = 2m'. 'B' is a massless and frictionless pulley.
    The problem states that at an instant, the string between ring and the pulley makes an angle 'θ' with the horizontal rod. If the speed of the ring at that instant is 'v', what will be the speed with which the block C descends.

    2. Relevant equations
    Σ Fx = max
    ∑ Fy = may
    ∑ Fz = maz

    3. The attempt at a solution
    I have drawn the free body diagrams listing all the forces.
    The forces acting on the ring 'C' are:
    Σ Fx = Tcosθ = ma
    ∑ Fy = N - mg - Tsinθ = 0

    Forces acting on the block are:
    ∑ Fy = T - Mg = Ma' .

    That's all I did.

    What I don't understand is, will the acceleration due to the X component of tension force, Tcosθ be the same in magnitude as the acceleration a' of the descending block?
     
  2. jcsd
  3. Jun 9, 2017 #2

    berkeman

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    I don't think you need to use forces for this problem (but I could be wrong). They give you the speed of the ring horizontally, and as long as the string stays taut, that will just translate into a change in the length of the string between the pulley and M. Can you try approaching the problem that way?
     
    Last edited: Jun 9, 2017
  4. Jun 9, 2017 #3

    haruspex

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    Quite so.
    @NoahCygnus , these kinematics problems can be confusing. The best way is to think in terms of the component of relative velocity in the direction of the string (or bar, in some questions). How fast is the ring moving towards the pulley?
     
  5. Jun 9, 2017 #4
    I'll try to approach it the way you mentioned. Though I face only one problem , does the acceleration of the ring in x direction have the same magnitude as the acceleration of the block ? Or is it more?
     
  6. Jun 9, 2017 #5

    berkeman

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    That's for you to figure out, but I checked my answer by thinking about it intuitively. Say θ is small, so the top part of the string is almost parallel with the bar. What is the speed of the string in relation to the ring at that small angle? And what happens as the ring gets close to directly over M, so the string is vertical and θ = 90 degrees? What is the vertical speed of the string at that point? Does this function remind you of any trig functions....? :smile:
     
  7. Jun 9, 2017 #6

    haruspex

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    The question as posted involves speeds, not accelerations. But maybe there are more parts to the question?
     
  8. Jun 10, 2017 #7
    Yes, I also has to find the acceleration , which is the easier part. But I can't figure out if the acceleration on the ring in x direction should be the same as that of the block in y direction .
     
  9. Jun 10, 2017 #8

    haruspex

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    Did you try the approach I recommended in post #3? Did you understand it?
     
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