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Question

# A sphere of mass M and radius R is released from the top of an inclined plane of inclination θ. The minimum coefficient of friction between the plane and the sphere so that it rolls down the plane without sliding is given by

A
μ=tan θ
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B
μ=23tan θ
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C
μ=25tan θ
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D
μ=27tan θ
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Solution

## The correct option is D μ=27tan θWhen the sphere rolls down the plane, its acceleration is given by a=g sin θ1+K2R2 Where K is the radius of gyration of the sphere about its diameter. Now, the moment of inertia of the sphere about its diameter is I=25MR2 ∴K2=IM=25R2 Therefore, a=g sin θ1+25R2R2=57 g sin θ For rolling without sliding, the frictional force f provides the necessary torque τ which is given by τ=force×moment arm=fR But τ=Iα, where α is the angular acceleration of the sphere. Thus, Iα=fR. Also, linear acceleration a=α R. Therefore, f=IαR=IαR2=25 Ma (∵I=25MR2) Now Force of friction =μ×normal reaction=μ Mg cos θ Or a=52 μg cos θ ⋯(ii) Equation (i) and (ii) we have 75g sin θ=52μ g cos θ or μ=27 tan θ

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