- #1

vcsharp2003

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- Homework Statement
- The problem diagram is as given below.

A small disk with center ##O^{'}## is kept on a larger disk with center ##O## as shown in diagram.

The system is given an angular velocity ##\omega_ {0}## about a vertical axis passing through O'. The friction between disks is negligible and smaller disk is free to rotate about O'. Find the angular momentum of system at ##t = \frac {\pi } {\omega_0}## about a vertical axis passing through O.

- Relevant Equations
- ##I_{o'} = I_{o} + m r^2## (parallel axis theorem)

##\vec {L} = I \vec {\omega}## (angular momentum)

I think the the time given doesn't matter since no torque is acting on the system, but not sure. Therefore, all we need is to determine the angular momentum about the axis passing through O and perpendicular to the plane of disk. This will involve finding the moment of inertia of smaller disk about the axis through O by using the parallel axis theorem and also the angular velocity of the system about O. Does this reasoning sound correct or maybe I am missing something?

$$L_{o} \text { stands for the angular momentum of system} $$

$$L_{o} = L_{so} + L_{lo} \text { ,where s stands for smaller disk and l for larger disk} $$

$$L_{o} = I_{so} ~ \omega_{1} + I_{lo} ~ \omega_{1} \text { ,where } \omega_{1} \text { stands for angular velocity about O} $$

Also, the angular velocity about the axis through O should be different from angular velocity ##\omega_{0} ## about the axis through O'. I get ##\omega_{1}## in terms of ##\omega_{0}## as below, after equating the linear velocity of a point along OO' and on the edge of larger disk.

$$\omega_{0} ~ R = \omega_{1} ~ 2R$$

$$\therefore \omega_{1} = \frac {\omega_{0}} {2}$$

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