Q.

A semicircular disc of radius R and mass M is pulled by a horizontal force F so that it moves with uniform velocity. The coefficient of friction between disc and ground is μ=43π . If the angle θ=π/n radians, then find the value of n .

Moderate

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By Expert Faculty of Sri Chaitanya

answer is 6.00.

(Detailed Solution Below)

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Detailed Solution

For uniform velocity, in horizontal direction, acceleration is zero. F=f ⇒F=μmgThe force F and friction force will form a couple.Balancing torque about point of contact with ground , we get μmg R(1−sinθ)=mg4R3πsinθ⇒sin θ=12 ⇒θ=30o=π6rad

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