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I'm completly stuck here: my system consist of a sphere rotating in a ring.

How do I calculate the moment of inertia of this system? I know the MOI of a sphere rotating around it's cog, but here, I can't even define the rotational axis. I imagine the MOI is different if the sphere rolls or glides.

Moving the MOI (Steiner) to the contact point appears to be false, that way I completly ignore the ring size.

The ring is stationary!

schematics

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  • $\begingroup$ Is the ring stationary? The sphere rolls around inside it? Then the axis of rotation is a line perpendicular to the page through the center of the ring. $\endgroup$
    – mmesser314
    Feb 19, 2016 at 15:23
  • $\begingroup$ @mmesser314: The ring is stationary. $\endgroup$
    – rst
    Feb 19, 2016 at 15:49

2 Answers 2

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The way I would approach this as follows:

  1. Establish the kinematics given the orbit rotational speed $\Omega$ find the spin $\omega$ subject to the no-slip constraint $\Omega (R-r) = \omega r $

    $$\omega = \left( \frac{R-r}{r} \right) \Omega $$

    pic

  2. Calculate the total kinetic energy. Add the linear and angular energy components of the ball, given the MMOI of $I_{\rm ball}$ and mass $m$. Note that linear velocity is $ v = \omega r$

    $$ KE = \tfrac{1}{2} I_{\rm ball} \omega^2 + \tfrac{1}{2} m v^2 = \tfrac{1}{2} \left( I_{\rm ball} + m r^2\right) \omega^2 $$

    $$ KE = \tfrac{1}{2} \left( I_{\rm ball} + m r^2\right) \left( \frac{R-r}{r} \right)^2 \Omega^2 $$

  3. Extract the effective MMOI. Equate the kinetic energy to the one calculated by spinning the effective mass moment of inertia only

    $$ KE = \tfrac{1}{2} I_{\rm eff} \Omega^2 $$

    $$ \boxed{ I_{\rm eff} = \left( \frac{R-r}{r} \right)^2 \left(I_{\rm ball} + m r^2\right) }$$

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Since you does not give any information regarding the motion of ring, I will consider the ring to be stationary. Then, obviously the axis of motion is the axis passing through the centre of ring and perpendicular to the plane of paper. For finding this moment of inertia, if the M.I of sphere throug its diameter is $I_0$ , we can use parellell axis theorem: $$I=I_0+Ma^2$$ where $M$ is the mass of sphere, and $a$ is the distance b/w the two axes. If $R$ is the radius of ring and $r$ , that of sphere, then $$a=R-r$$

But from your question, you say that "double rotating" sphere, and so I think you have a confusion to find its M.I because the sphere has rotational motion about its own axis, and revolution along the ring. But, it does not affect moment of inertia of body about a fixed axis, because it is just analogous to mass in translational motion.

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  • $\begingroup$ I can't imagine that this is correct. It completly ignores the "own" rotation of the sphere. $\endgroup$
    – rst
    Feb 20, 2016 at 19:33

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