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Show that the ball undergoes simple harmonic motion

  1. Nov 1, 2005 #1
    The problem goes as following:

    Solid ball (Solid sphere) of mass M and radius r is kept at the equilibrium position on a concave surface of radius R , for small displacement up the concave surface , show that the ball shows SHM motion assuming the ball rolls without slipping.

    I have been trying this problem by using the energy method , since force method would be too tedious I suppose.

    The nete nergy which I think should be , assuming that the ball is displaced up a distance x:

    U= [tex]\frac{1}{2} mv^2[/tex] + [tex]\frac{1}{2} I \omega ^2[/tex] + mgx

    Then I differentiate it wrt T and I get an expression whihc nowhere helps me.

    Please someone help me do this thing through proper energy method.
     
    Last edited by a moderator: Nov 1, 2005
  2. jcsd
  3. Nov 1, 2005 #2

    siddharth

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    Doing this by the force method is not that tedious.
    Consider the two angular velocities (w1 and w2) as due to the motion of
    (i) The sphere about it's center
    (ii) The center of the sphere about the center of the concave surface.
    Then you will get
    v=rw1 (This is due to condition of pure rolling)
    and
    w2=v/(R-r).

    From this you have the relation betweem the two angular velocities and hence between the angular accelerations.
    Now, if you draw the FBD diagram and find the restoring Torque for a small angular displacement (about the center of the concave surface), you will be able to find the time period for small oscillations.
     
  4. Nov 1, 2005 #3
    Please tell me the energy method .

    BJ
     
  5. Nov 1, 2005 #4

    siddharth

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    It's going to be very similar. Write down the total energy and diff wrt to time. I can't see your Latex post for some reason so I can't see what total energy you have written. It should involve the KE, Rotational KE about the center of the sphere and Potential Energy. (Try writing the velocity of the center of the sphere in terms of the angular velocity about the center of the concave surface)
    Then, if you use the relation between the two angular velocities and angular accelerations and eliminate the common term, you should get the time period from the resulting equation.
     
  6. Nov 1, 2005 #5

    Astronuc

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    Staff: Mentor

    Some background on SHM -

    http://www.plus2physics.com/vibrations/study_material.asp?chapter=1&page=2

    Think of the form of the differential equation for SHM

    [itex]\ddot{x} + \lambda^2 x = 0[/itex]

    and for a rotating object (ball), how [itex]\omega[/itex]t relates to x, y where x is the lateral displacement from equilibrium and y is the vertical displacment.

    The circular bowl provides a constraint on x and y.
     
  7. Nov 3, 2005 #6
    I think the force method would be easiest. Just show that for small oscillations the restoring force is of the form F = -k * x, where k is some constant. To do this, note that the restoring force is mg*sin theta, then taylor expand sin and drop higher order terms.

    Then by definition you have SHM, since F = ma = - k * x so a = -k/m * x
     
  8. Nov 5, 2005 #7

    Astronuc

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    Draw a diagram of a small circle (spherical ball of mass m), radius r, and a large circular curve of radius, R, such that the ball has minimum potential energy at the lowest point on the large circle. The center of mass of the ball then is at distance R'=(R-r) from the axis of the curve or radius R.

    Take the angle [itex]\theta[/itex] to be the rotational angle for the ball from point of minimum potential energy and [itex]\phi[/itex] to be the angle from vertical to a line or radius R which passes from the axis of concave curve R through the ball's center of mass to the point of contact of the ball and concave surface; [itex]\theta[/itex] = 0 corresponds to [itex]\phi[/itex] = 0.

    We now observe that when the ball rotates by angle [itex]\theta[/itex] it travels a distance [itex]r\,\theta[/itex] and this distance equals [itex]R\,\phi[/itex], which assumes no slipping. Therefore [itex]r\,\theta[/itex] = [itex]R\,\phi[/itex], and because r and R are constant,

    [itex]r\,\dot{\theta}[/itex] = [itex]R\,\dot{\phi}[/itex], and

    [itex]r\,\ddot{\theta}[/itex] = [itex]R\,\ddot{\phi}[/itex]

    Also recognize that when the ball traverses a distance [itex]R\,\phi[/itex] on the concave surface, the center (of mass) of the ball traverses a distance [itex](R-r)\,\phi[/itex] = [itex]R'\,\phi[/itex].

    Now, let [itex]\omega[/itex]=[itex]\dot{\theta}[/itex], which is the conventional definition of angular velocity, and recognize that

    [itex]\omega\,r[/itex] = vt = v, which is the tangential velocity of the spherical ball on the concave surface of radius R, so v can be written in terms of [itex]\dot{\theta}[/itex] or [itex]\dot{\phi}[/itex].

    Also, as the ball moves through some distance [itex]R\,\phi[/itex], the center of mass rises some distance h, which is given by h = [itex]R'\,-\,R'\,cos{\phi}[/itex] = [itex]R'\,(1\,-\,cos{\phi})[/itex] and the gravitational potential energy increases by mgh.

    So one then needs to write an equation for this system, which for SHM would be of the form,

    [itex]\ddot{\theta}[/itex] + [itex]{\lambda}^2\theta[/itex] = 0

    one can also find x([itex]\theta[/itex]) and write a similar equation in terms of x.
     
  9. Nov 8, 2005 #8
    Thanks Astronuc I have got the right answer.

    BJ
     
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