The solid semi-cylinder of mass m and radius R is rolled slowly through an angle θ by the horizontal force P. If the coefficient of static friction is μₛ, the angle θ at which the cylinder begins to slip on the horizontal surface as P gradually increased is θ = sin⁻¹[3πμₛ/(n−3πμₛ)]. Find n.
Physics · Mixed Chapter
JEE Mains

Question

The solid semi-cylinder of mass m and radius R is rolled slowly through an angle θ by the horizontal force P. If the coefficient of static friction is μₛ, the angle θ at which the cylinder begins to slip on the horizontal surface as P gradually increased is θ = sin⁻¹[3πμₛ/(n−3πμₛ)]. Find n.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: For a uniform semicylinder, its centre of mass is displaced by 4R/(3π) from the circle centre towards the curved part. Step 2: As the body is moved quasistatically, impose net force and net torque equal to zero; friction reaches its limiting value f=μₛN at the onset of slip. Step 3: Taking moments about the current contact point and resolving forces along horizontal and vertical axes gives sinθ=3πμₛ/(4−3πμₛ). Step 4: Comparing with the form in the question, n=4.