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Homework Help: Find speed of cylinder

  1. Apr 25, 2014 #1


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    1. The problem statement, all variables and given/known data
    Two identical uniform cylinders of radius R each are placed on top of each other next to a wall as shown. After a disturbance, the bottom cylinder slightly moves to the right and the system comes into motion. Find the maximum subsequent speed of the bottom cylinder. Neglect friction between all surfaces.

    See attachment for diagram

    3. The attempt at a solution

    Let the final speed of the lower and upper cylinders be vb and vu respectively and their angular velocity be ωb and ωu.
    Conserving energy

    [itex]3mgR + mgR=\dfrac{mv_b ^2}{2} + \dfrac{mv_u ^2}{2}+\dfrac{mv_b ^2}{4}+\dfrac{mv_u ^2}{4} + 2mgR \\
    v_b ^2 + v_u ^2=\dfrac{8gr}{3} [/itex]

    I have assumed that the cylinders do not slip on the surface and perform pure rolling. I need another equation as there are two variables.

    Attached Files:

  2. jcsd
  3. Apr 25, 2014 #2
    I don't see why you are worried about the pure rolling motion when the problem clearly states that the surfaces are frictionless. This should be a simple constraint problem.

    Assume that lower cylinder shifts by a distance ##x## towards right and the top cylinder descends by ##y##. Can you relate the two?
  4. Apr 25, 2014 #3


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    Thanks a lot! :smile:
  5. Apr 26, 2014 #4
    If at any time , the center of the upper cylinder is at a distance 'y' from the floor and the center of the lower cylinder is at a distance 'x' from the wall ,then

    (x-R)2+(y-R)2 = (2R)2

    dx/dt = -[(y-R)/(x-R)]dy/dt

    Is this the correct relation ?If yes,what should be the next step ?
    Last edited: Apr 26, 2014
  6. Apr 26, 2014 #5


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    Applying conservation of energy :smile: and then eliminate all y and find the x coordinate where dx/dt is maximum.

  7. Apr 26, 2014 #6


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    Denote the initial position of COM of upper cylinder with A and that of lower cylinder with B. Assume that the bottom cylinder shifts a distance x from its initial position and consequently, the upper cylinder descends a distance y from its initial position. Let their new position of COM be C and D respectively.
    BC=x and AD=y. Let ∠DCB = θ.

    Now try to relate y and x using suitable equations. (Hint: Drop a perpendicular from A to ground. You already know its length).
  8. Apr 26, 2014 #7

    x = (2R-y)/tanθ . Here all x,y and θ vary . So what next ?

    By the way what is the answer ?
  9. Apr 26, 2014 #8
    Your previous relation is OK. (I think you got a sign error there)

    As suggested by ehild, you can then use energy conservation.
  10. Apr 26, 2014 #9


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    The correct answer is [itex]\sqrt{\dfrac{16gR}{27}} [/itex]
  11. Apr 26, 2014 #10
    Do you mean the relation in post#4 ? What is the error ?

    Did I get the relation right in post#7 ? Do I need to conserve energy along with this relation ?
  12. Apr 26, 2014 #11
    Yes, it should be -dy/dt instead of dy/dt.

    But I would suggest a different approach. Let the CM of lower cylinder shift by ##x## and the CM of upper cylinder descends by ##y##. Let ##O## denote the initial position of CM of lower cylinder and ##A## and ##B## denote the position of lower cylinder and top cylinder at a later time respectively. Denote ##\angle BAO=\theta##. You get the following relations:
    $$x=2R\cos\theta \Rightarrow \frac{dx}{dt}=-2R\sin\theta \frac{d\theta}{dt}$$
    $$y=2R(1-\sin\theta) \Rightarrow \frac{dy}{dt}=-2R\cos\theta \frac{d\theta}{dt}$$
    From energy conservation:
    $$\Rightarrow 4gR(1-\sin\theta)=4R^2\sin^2\theta \left(\frac{d\theta}{dt}\right)^2+4R^2\cos^2\theta \left(\frac{d\theta}{dt}\right)^2$$
    $$\Rightarrow \frac{d\theta}{dt}=\sqrt{\frac{g}{R}}\sqrt{1-\sin\theta}$$
    $$\Rightarrow \frac{d^2\theta}{dt^2}=\sqrt{\frac{g}{R}}\frac{-\cos\theta}{2\sqrt{1-\sin\theta}}$$
    For maximum speed, you have to equate ##d^2x/dt^2=0##. Can you proceed now?
  13. Apr 26, 2014 #12
    Hi Pranav...

    Thank you very much for the nice explaination.Please forgive me if I am wrong with the reasoning .

    dy/dt is negative and there is already a negative sign in the RHS . Are you sure there should be an additional negative sign ?

    I think this should be R(1-2sinθ) in the LHS .

    I think this should be mg(R+y)/2 in the LHS . Shouldn't that be loss of PE of the CM of the system of the two spheres ?
  14. Apr 26, 2014 #13
    I have attached a sketch. Let me know if that clears up the above issues.

    Sorry, I was being hasty, you are correct in your analysis.

    But I am little unsure about the sign issue. The problem with the signs and the messy algebra was the reason I switched to expressing everything with ##\theta##. I hope somebody else can clear it up. :P

    Attached Files:

    Last edited: Apr 26, 2014
  15. Apr 26, 2014 #14
    Don't you think 'y' is the distance by which upper cylinder descends whereas we need distance by which COM of the system of two spheres descends ?
  16. Apr 26, 2014 #15
    Why do you need the distance by which COM descends? :confused:

    You can calculate the change in gravitational potential energy of individual spheres, no?
  17. Apr 26, 2014 #16

    You are right . What I am saying is also correct .Just that I wrongly calculated location of COM .Either way it comes out to be mgy .
    Last edited: Apr 26, 2014
  18. Apr 26, 2014 #17


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    I chose to work with "x" instead of θ and here's what I got.

    [itex]y=2R-\sqrt{4R^2-x^2} [/itex]

    If you differentiate both sides you will get a relation between vx and vy. Next step is to use energy conservation.

    [itex]3mgR = \frac{m}{2} (v_x^2 + v_y^2) + mg(R+\sqrt{4R^2-x^2}) [/itex]

    Note that I've not taken into account the GPE of the bottom cylinder since it does not change. Replace vy in terms of vx(using the relation derived earlier) and you will get an expression for vx in terms of x. For maximum velocity d(vx)/dx = 0.

    I hope this helps. :smile:
  19. Apr 26, 2014 #18
    Thanks a lot :)
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