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Homework Help: Rotational to linear kinetic energy

  1. Apr 7, 2010 #1
    1. The problem statement, all variables and given/known data
    A 5kg hollow metal cylinder with a radius 0.1m rolls off the edge of a roof with an incline of 30degrees and height of 1metre.

    If the cylinder was initially at rest, what is the ratio of the rotational to linear kinetic energy when it reaches the end of the roof?

    and the ration when it hits the ground?


    I don't understand what "The ratio of rotational to linear kinetic energy" means
    does it mean the angular velocity? or the velocity?
     
  2. jcsd
  3. Apr 7, 2010 #2

    Doc Al

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    Staff: Mentor

    I'm sure you know to find linear KE. It's just 1/2mv^2. How is rotational KE defined? How are they related? Hint: Assume it rolls without slipping.
     
  4. Apr 7, 2010 #3
    so
    KEr= 1/2 iw^2 (w being omega) and i(cylinder) = MR^2

    do i plug in i = mr and then w = v/r? and make it = KE

    i wrote it all out but it looked silly so i deleted it
    is MR^2 the centre of mass of the cylinder? i read I = Icom + Mh^2
     
  5. Apr 7, 2010 #4

    Doc Al

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    Good.

    I don't understand what you're saying here. But w = v/r is the condition for rolling without slipping.


    That's the rotational inertia about an axis through the center of mass, which is what you need.
    That's the parallel axis theorem. Not needed here.
     
  6. Apr 7, 2010 #5
    I think i understand now
    I have to find the KE and KEr at the end of the slope, and show the difference as a ratio?

    In air when it leaves the edge, would its angular velocity decrease? So i'd treat it as a projectile to find the time it takes to hit the ground and somhow find the angular velocity?
     
  7. Apr 8, 2010 #6

    Doc Al

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    Right.

    What do you think? What's required for the angular velocity to change?
    Yes. As soon as it leaves the roof, it's a projectile.
     
  8. Apr 8, 2010 #7
    sorry for being a pest
    i think, then over think, then make afew solutions, then have 0 confidence
    at first i thought that in the air the angular velocity would decrease because theres no surface to roll on, but it'd be the angular acceleration decreasnig wouldnt it?
    so the angular velocity doesnt change, so KEr is the same and and KE is different?
    thanks for your patients ;)
     
  9. Apr 8, 2010 #8

    Doc Al

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    Not quite sure what you're saying here. Try and rephrase it.
    That's true. Once it leaves the surface, there's no more friction to provide any torque, so the angular velocity--and thus rotational KE--doesn't change. Translational KE certainly does.
     
  10. Apr 8, 2010 #9
    Thanks heaps for your help Doc Al.
    I ended up getting for the edge of the roof the ratio 1:1 and then 7:1 7 being KE (linear)
    Could i just double check my working in these two questions?
    I prnt screened and wrote under the questions.
    When i did the ski one i found the range of jump then found the t for that range, then halved t to find the highest point in the jump, and with that t found h

    The other one seems to make sense, then smaller angle moves faster than the neutron at 60 degrees

    Do you think i did it right?

    [PLAIN]http://img199.imageshack.us/img199/8657/eeebb.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
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