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Rotational Motion of a marble

  1. Nov 4, 2011 #1
    1. The problem statement, all variables and given/known data

    A solid uniform marble and a block of ice, each with the same mass, start from rest at the same height H above the bottom of a hill and move down it. The marble rolls without slipping, but the ice slides without friction.

    Find the speed of each of these objects when it reaches the bottom of the hill.
    Vice:?
    Vmarble:?

    2. Relevant equations

    1/m v^2 = mgh

    3. The attempt at a solution

    Ice:
    1/2 mv^2=mgh

    v= squareroot 2gh

    Marble:
    mgh=1/2 mv^2 + 1/2 I ω^2

    =1/2 mv^2 +1/2(2/5 mr^2) (v/r)^2

    The answers i entered in were marked incorrect by masteringphysics.

    Please advise!!!!!!!! : (
     
  2. jcsd
  3. Nov 4, 2011 #2

    I like Serena

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    Hi valerie! x0x :smile:

    Your v for the block of ice is correct.
    And your equation for the marble is correct too.

    The only thing missing is that you didn't give v of the marble.
    Did you solve your equation for v?
     
  4. Nov 4, 2011 #3
    well i got stuck.... and i dont know why its marking me wrong. for the marble i got v: squareroot 2/4 mgh


    anythoughts?
     
  5. Nov 4, 2011 #4

    I like Serena

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    I'm afraid that in that case you did not solve your own equation correctly.

    For starters, there should not be an "m" in there.
    There isn't an "m" for the block of ice either!

    But your fraction of 2/4 is wrong too.

    How did you arrive at your solution for v?
    Can you show some (major) steps?
     
  6. Nov 4, 2011 #5
    oh never mind .......... this stupid mastering physics wanted a capital H instead of h.

    aughhhhhh!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! the marble was squareroot 10 gh/7
     
  7. Nov 4, 2011 #6
    thanks for your help!!!!!! i caught that at the last min!
     
  8. Nov 4, 2011 #7

    I like Serena

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    Congrats with your right answer! :smile:
     
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