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 .
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answer is 6.00.
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Detailed Solution
F=μMg The force F and friction force will form a couple.Balancing torque about centre of mass μmg R(1−sinθ)=mg4R3πsinθ⇒sin θ=12⇒θ=30o=π6rad
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