A block of mass m travels in a horizontal circle against the wall of a cylindrical room of radius R . The floor of the room is smooth, while the friction coefficient between the wall and the block is . The block is imparted an initial speed v₀ . By integrating the tangential acceleration ( dv dt = v dv ds ) , obtain the speed of the block after exactly one revolution.

Step-by-step solution

The normal force exerted by the wall on the block provides the necessary centripetal force for circular motion: N = mv^2 R The frictional force acting tangentially opposite to the direction of motion is: f = N = mv^2 R The tangential acceleration of the block is given by Newton's second law: m a_t = -f m (v dv ds ) = - mv^2 R v dv ds = - v^2 R Separating the variables to integrate: dv v = - R ds Integrating both sides for one complete revolution, where the distance covered is s = 2 R , and the velocity changes from