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A uniform disc of mas M and radius R is ...

A uniform disc of mas `M` and radius `R` is smoothly pivoted at `O`. A light inextensible string wrapped over the disc hangs a particle of mass `m`. If the system is released from rest, assuming that the string does not slide on the disc, find the angular speed of the disc as the function of time using impulse momentum equation.

Text Solution

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The net torque about `O` is
`vectau=mgRhatk` ………i
The angular momentum of the disc particle system about `O` is `L=I_(0)omega+mvR`, where `v=Romega`
`L=(MR^(2))/2omega+mR^(2)omega=((MR^(2))/2+mR^(2))omega`

`vecL=((M+2m)/2)R^(2)omegahatk`.........ii
Impulse momentum equation is
`/_\vecOL=intvectau dt`
Using eqn i , ii and ii we have
`((M+2m)/2)R^(2)omega=mgRint_(0)^(1)dt`
`omega=(2m"gt")/((M+2m)R)`
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