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Sphere slipping against block

  1. Jan 7, 2014 #1
    Sphere slipping against blocks

    1. The problem statement, all variables and given/known data

    In the given arrangement ,a sphere of radius R is placed on the two blocks A and B where A is fixed and B is moving at a constant speed ‘v’ towards left .Find the speed of the sphere’s center when the distance between the blocks is √R and the sphere’s center is at a vertical distance √R from the top of the blocks .All surfaces are frictionless .

    Ans : v/2

    2. Relevant equations



    3. The attempt at a solution

    Let the distance between the blocks be 'y' at any instant.
    Let the distance between the sphere’s center be at a vertical distance 'x' from the top of the blocks .
    Let the horizontal distance between B and sphere's center O be 'l' .

    Then l=y/2 .

    Differentiating we get,dl/dt = (1/2)dy/dt = v/2

    From the geometry of the setup , x2 + l2 = R2

    Differentiating , we get dx/dt = -(l/x)dl/dt .

    Now when x=√R and y=√R (i.e l=√R/2) ,we have dx/dt = -v/4 i.e the horizontal speed of O is v/4 downwards .

    The center also has a horizontal speed v/2 towards left.

    The net speed of the center is (√5/4)v which is incorrect .

    I would be grateful if someone could help me with the problem .
     

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    Last edited: Jan 7, 2014
  2. jcsd
  3. Jan 7, 2014 #2
    Hmmm, maybe I'm not understanding the problem correctly, but the geometry for this problem seems to be over-constrained.

    From the problem statement:

    • The distance from the center of the sphere to the contact point on block B (or A for that matter) is (√R)/2. We'll call this B.
    • The distance from the center of the sphere to the top of the blocks is √R. We'll call this C.
    • This distance from the contact point on block B to the center of the sphere is the radius R. We'll call this A.

    These three constraints form a right triangle, and the Pythagorean theorem tells us that A = √(B^2+C^2). Plugging the above in gives us:

    R = √(((√R)/2)^2+(√R)^2) = √((R/4)+√(R)) = (√5/4)R

    This shows that the given dimensions cannot result in a right triangle, suggesting an ill-defined problem.
     
  4. Jan 7, 2014 #3

    TSny

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    Hello, Tanya.

    I'm wondering if there's a mistake in the wording of the problem. It doesn't seem dimensionally correct to take the distance between the blocks to be √R. Anyway, if you let x = √R and l = (√R)/2, then you certainly wouldn't satisfy x2 + l2 = R2.

    [EDIT: Incorrect statement deleted.]
     
    Last edited: Jan 7, 2014
  5. Jan 7, 2014 #4

    haruspex

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    I tried working backwards from the v/2 answer to figure out the misprint. I deduced that the blocks are touching! And if you think about that arrangement you can see it does indeed give v/2. The centre of the sphere will be moving left at half the speed of the block.
    Before I did that, I would have guessed it was supposed to be a separation of R√2, with the sphere's centre initially R/√2 above the blocks. That gives a speed of v/√2. Could this be a record for misprints in one question?

    Tanya, to get some benefit from the question, try taking the separation to be some unknown x and post what you get.
     
  6. Jan 7, 2014 #5
    Hi TSny…

    It seems the new year hasn’t started well as far as putting up a question on PF is concerned.I regret putting up an ill defined problem :redface:.

    Anyways I feel I have figured out the problem with the problem .Interestingly,all the radical signs have gone topsy turvy .

    Here is what I believe the question should have been :

    "In the given arrangement ,a sphere of radius R is placed on the two blocks A and B where A is fixed and B is moving at a constant speed ‘v’ towards left .Find the speed of the sphere’s center when the distance between the blocks is √2R and the sphere’s center is at a vertical distance R/√2 from the top of the blocks .All surfaces are frictionless .

    Ans:v/√2 "


    What is your opinion ?
     
  7. Jan 7, 2014 #6
    I feel the same .

    Yes... But this record has left me :redface:.
     
    Last edited: Jan 8, 2014
  8. Jan 7, 2014 #7

    TSny

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    I like it. It's the same as haruspex's prognosis!
     
  9. Jan 7, 2014 #8
    Thanks TSny...

    Actually I figured out the problem with the problem the moment I saw the initial response .But found it quite hard to believe that the question had so many mistakes .I even responded just after you have replied , only to delete the post .
     
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