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Situation

If i want to conserve angular momentum from the point of contact of the solid sphere and the ground how should I do it?

I know we can use $mvh = I\omega + MvR$ ,if the collision is elastic. But in this case the particle sticks to the sphere after hitting at a height h above the ground. I am not getting how should i be able to conserve the momentum of this system about the point of contact of sphere and ground. To be precise the inelastic collision part of the particle is creating a problem.

$m$ is mass of particle, $M$ is mass of sphere, $R$ is the radius. All surfaces are frictionless.

Thanks in advance.

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  • $\begingroup$ Please note that the Homework-and-Exercises tag indicates not that the question is a homework problem, rather that it needs problem-solving and mathematics more than theoretical physics. $\endgroup$ Oct 28, 2015 at 14:15
  • $\begingroup$ oh no, i didnt purposely reject it, i edited the question thats why it got rejected, forgot to add homewok tag @TamoghnaChowdhury $\endgroup$
    – user94848
    Oct 28, 2015 at 14:19
  • $\begingroup$ its all right, part of the learning curve. $\endgroup$ Oct 28, 2015 at 14:30

1 Answer 1

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If the collision is inelastic, try this:

$$mvh = I_m\omega_m + I_M\omega_M$$

$I_M$, the moment of inertia of the solid sphere, is taken as that about its tangent at the point of contact with the ground.

$$I_M = \frac {7MR^2}{5}$$

and, since the particle of mass $m$ revolves about the centre of the sphere,

$$I_m = \frac {mR^2} {2}$$

Here, the angular momentum has been conserved about the Instantaneous Axis of Rotation, which in this case is the point where the sphere touches the ground.

IAoR I hope the notation is self-explanatory (I used what you provided).

Here, treat the particle and the sphere separately, even if they are attached together.

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  • $\begingroup$ about which point have you conserved angular momentum ? What is IM = ? $\endgroup$
    – user94848
    Oct 28, 2015 at 14:36
  • $\begingroup$ Sorry.edited the answer. $\endgroup$ Oct 28, 2015 at 14:38
  • $\begingroup$ Where is the Instantaneous Axis of Rotation ? $\endgroup$
    – user94848
    Oct 28, 2015 at 14:40
  • $\begingroup$ I mentioned that, wait a minute, I'll add a picture. $\endgroup$ Oct 28, 2015 at 14:40
  • $\begingroup$ oh yeah, didnt see it, right $\endgroup$
    – user94848
    Oct 28, 2015 at 14:41

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