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Sphere on a frictionless ramp

  1. May 2, 2013 #1
    1. The problem statement, all variables and given/known data
    As of the past few hours I've been trying to make sense one how to calculate the velocity of a sphere that moves down a frictionless ramp. My biggest problem with this seems to be that I'm confusing myself with the linear velocity and the angular velocity. Note that the sphere starts at rest.

    2. Relevant equations
    Conservation of energy:
    [tex]mgh=\frac{1}{2}mv^2 + \frac{1}{2}I \omega ^2 [/tex]

    Relation between linear velocity and angular velocity
    [tex] v= \omega r [/tex]

    3. The attempt at a solution
    I can get as far as equating the two and inserting the moment of inertia for a sphere.
    [tex] mgh=\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2 \omega ^2) [/tex]
    Unfortunately now I have no clue how to proceed. I'm in a sort of dilemma where I know if it starts sliding down the ramp then it doesn't seem like there should be any rotation. But if there is no rotation then is there any point in including the final part of the equation? Of course I could always refer this to the kinetic friction and come up with some sort of argument but I just don't seem to understand how it can make it make sense.
  2. jcsd
  3. May 2, 2013 #2

    Doc Al

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    Staff: Mentor

    Is the ramp frictionless or not? As you suspect, if there's no friction the sphere will not begin to rotate but will merely slide down.

    What's the exact statement of the problem?
  4. May 2, 2013 #3
    The ramp is frictionless.

    I'm trying to understand what is happening with the sphere. If it is not rolling is it necessary to include the angular velocity? I'm asking because it it's not rolling then wouldn't it be zero?

    I have to do this so I can isolate the velocity to see how it changes for the sphere when the height increases or decreases.
  5. May 2, 2013 #4

    Doc Al

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    Staff: Mentor

    Yes. Without friction ω = 0, so you can drop the rotational KE term from your energy equation.
  6. May 2, 2013 #5
    Thank you, that was just what I was looking for.
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