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Please help me with this qn on SHM

  1. Sep 4, 2005 #1


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    A small thin disk of radius r and mass m is attached rigidly to the face of a second thin disk of radius R and mass M. The centre of the small disk is located at the edge of the large disk. The large disk is mounted on its centre of a frictionless axle. the assembly is rotated through a small angle theetre from its equilibrium position and then released. Show that the speed of the small disk as it passes through the equilibrium position is

    v=2{Rg(1-cos theetre)/[ M/m+ (r/R)^2+2]}^0.5

    and the period is

    T=2pi {[(M+2m)R^2+mr^2]/[2mgR]}^0.5

    I am stuck in the first part already.
    My method was: Since the axis of rotation is through the COM of the larger disk, hence the nett torque is mgRsin theetre.
    This torque is equal to I alpha which is equal to (1/2MR^2+mR^2) alpha.
    angular aceleration is not constant so i must carry out an integration when using the eqn V^2=U^2+2aS

    i have been struggling with this prob. for a long time . pl help me.

    Attached Files:

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  3. Sep 4, 2005 #2


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    Gold Member

    You don't have to do an integration. You can regognize the differential equation [itex]\theta = k\ddot{\theta}[/itex] as the harmonic equation. It has a general solution of [itex]\theta (t)= A\cos{\sqrt{k}t} + B\sin{\sqrt{k}t}[/itex] for constants A and B to be determined by initial conditions. Remember that it was specified that the angle was small, so the approximation [itex]\sin{\theta} \approx \theta[/itex] is valid. Also, you don't have the moment of inertia correct. I is equal to the moment of inertia of the big disk plus that of the small disk. The moment of inertia of the large one is simply [itex]\frac{MR^2}{2}[/itex], while the second can be computed with the use of the parallel axis theorem. I am getting a slightly different answer than the one you gave. I get the 2 in v=2{...}^.5 to be inside the square root. One of us made a mistake, so just check and make sure it isn't you.
  4. Sep 4, 2005 #3


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    no it was correct.2 is outside of the sq. rt.
    But the angular a is not constant, so i thought a integration must be carried out?

    I was using the parallel axis theorem: 1/2MR^2 + mR^2
  5. Sep 4, 2005 #4

    Doc Al

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    Staff: Mentor

    Where did you use the parallel axis theorem? Here you are treating the small disk as a point mass, which it is not. Use the parallel axis theorem to find the rotational inertia of the small disk about the axis of the large disk. (You neglected the rotational inertia of the small disk about its own center.)
  6. Sep 4, 2005 #5


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    No, there is no need for integration. Once you use the approximation [itex]\sin{\theta} \approx \theta[/itex], then it simply becomes the harmonic equation, for which the solution is known. The parallel axis theorem says that the moment of inertia of the small disk is [itex]\frac{mr^2}{2} + mR^2[/itex]. Combine that with the moment of inertial of the larger disk to get [itex]\frac{mr^2}{2} + mR^2 +\frac{MR^2}{2} = (\frac{M}{2} + m)R^2 + \frac{mr^2}{2}[/itex].
  7. Sep 4, 2005 #6


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    Homework Helper

    The Restoring torque is given by

    tau = - mgR@

    hence torque constant K is mgR
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