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A smooth ring of mass m and radius R = 1...

A smooth ring of mass `m` and radius `R = 1 m` is pulled at `P` with a constant acceleration `a= 4 ms^(-2)` on a horizontal surface such that the plane of the ring lies on the surface. Find the angular acceleration of the ring at the given position. (in `rad//s^(2)`)

Text Solution

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The correct Answer is:
2
The pseudo force acting at the `CM` of the ring `vecF_(ps)=-mahati` which produces a torque about `P` given as
`tau_(ps)=-(ma)(R)hatk`

This produce an angular acceleration
`vecalpha=vectau_(ps)/(I_(P))=(-maRhatk)/(2mR^(2))` (`:' I_(P)=mR^(2)+mR^(2)=2mR^(2))`
` =-a/(2R)hatk=4/(2xx1)=2rad//s^(2)`
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