Question

# A uniform solid cylinder of mass 'm' can rotate freely about its axis which is kept horizontal. A particle of mass m0 hangs from the end of a light string wound round the cylinder. When the system is allowed to move, the acceleration with which the particle descends is

A
2m0gm0+2m
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B
2m0gm+2m0
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C
m0gm+m0
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D
2m0gm+2m0
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Solution

## The correct option is B 2m0gm+2m0Suppose that the tension in the string is T. Then, =m0g−T=m0a ∴T=m0(g−a) where a = acceleration Further, T.r = I α α = angular acceleration and T.r = moment of force acting on cylinder or T=Iαr =(mr22)(αr)=mrα2=ma2 ∴a=2Tm=2m0(g−a)m Solving for a, we get a=(2m0gm+2m0).

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