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Rolling Motion Problem

  1. Jul 2, 2008 #1
    [/B]1. The problem statement, all variables and given/known data
    Hello:

    Can someone help with the following?

    A ball of mass M and radius R rolls smoothly from rest down a ramp and onto a circular loop of radius 0.48m. The initial height of the ball is h = 0.36m. At the loop bottom, the magnitude of the normal force on the ball is 2.00 Mg. The ball consists of an outer spherical shell (of certain uniform density) that is glued to a central sphere(of different uniform density). The rotational inertia of the ball can be expressed in terms of the general form bMR^2, but b is not 0.4 as it is for a ball of uniform density. Find the b for this ball.

    I thought to use conservation of energy? So this gives me:

    M*g*0.36 = 1/2*I*w^2 + 1/2*M*v^2, where I is the rotational inertia and w is the angular velocity and v is the linear velocity.

    But then I'm not sure what to do after like (for example, how do I get rid of the v's)? And is it true that the ball is not accelerating at the loop bottom with respect to the ground (since it is at the bottom) so that the normal force = force from gravity? But then what good is that since it only gives me M?

    Thanks!!
     
  2. jcsd
  3. Jul 2, 2008 #2

    Hootenanny

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    Looks good :approve:. Now if we assume that the ball rolls without slipping, can you relate the angular velocity to the linear velocity?
    In fact it doesn't even give M, the M's cancel. If the ball weren't accelerating, then the normal reaction force would simply be Mg would it not?

    HINT: The ball is undergoing circular motion so you should be able to relate the force acting on the ball to the linear velocity.
     
    Last edited: Jul 2, 2008
  4. Jul 2, 2008 #3
    Your equation,
    [tex]0.36Mg=\frac{1}{2}I\omega^2+\frac{1}{2}Mv^2[/tex]
    is correct.
    Just replace I with its general form bMv^2, eliminate M obviously and then recognize that the linear velocity is [itex]v=r\omega[/itex]. Imagine the angular velocity is 2pi/s (one rotation per second) then you get [itex]v=2\pi r[/itex] which is what you expect if it linearly translates the distance of the circumference in one second.
     
  5. Jul 2, 2008 #4
    Hello,

    Thanks. So I know that v = r*w (0.48*w) in this case. But I don't know how to find either w or v? If the ball is not slipping, then there is a frictional force that opposes the downward motion of the ball ... but I don't see where the normal force comes into use?
     
  6. Jul 2, 2008 #5

    Hootenanny

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    Correct. However, you are not looking to solve for w or v you are looking to eliminate them in order to find b.

    As I said previously, notice that the ball is moving in a circle. What is the requirement of the net force acting on any object undergoing circular motion?
     
  7. Jul 3, 2008 #6
    Um ... okay, there is a centripetal force, so then I know that the normal force, 2 = m*v^2/R (where R = 0.48). But then if I try to represent v using that equation I'm left with m's which I don't know the value of. Sorry.
     
  8. Jul 3, 2008 #7

    Hootenanny

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    Are you sure that that equation is correct?
    There's no need to apologise, not knowing how to answer a question straight away isn't anything to be ashamed of.
     
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