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A uniform ring of mass m is placed on a ...

A uniform ring of mass `m` is placed on a rough horizontal fixed surface as shown in the figure. The coefficient of friction between the left part of the ring and left part of the horizontal surface is `mu_(1)=0.6pi` and between right half and the surface is `mu_(2)=0.2pi`. At the instant shown, now the ring has been imparted an angular velocity in clockwise sense in the figure shown. At this moment magnitude of acceleration of centre `O` of the ring (in `m//s^(2)`) is (take `g=10m//s^(2)`)

Text Solution

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The correct Answer is:
4
`2[int_(0)^(pi//2)(mu_(1)-mu_(2))Rglamdacosthetad theta]=2(mu_(1)-mu_(2))Rlamdag`
`:.a=(2(mu_(1)-mu_(2))Rlamdag)/(2piRlamda)=4`
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Knowledge Check

  • A uniform ring placed on a rough horizontal surface is given a sharp impulse as shown in the figure-5.130. As a consequence, it acquires a linear velocity of 2m/s. If coefficient of friction between the ring and the horizontal surface is 0.4 :

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    Ring wil lstart pure rolling afler O.25s
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  • A particle of mass m is pulled along an arbitrary curved surface without acceleration as shown in figure. Coefficient of friction between surface and particle is mu . Then codes:

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