A small block is released from rest from the horizontal rim of a smooth fixed hemispherical bowl of radius R . The block slides down the inner surface of the bowl. At a certain position, the net acceleration vector of the block makes an angle of 45^ with its velocity vector. The angle made by the radius vector of the block with the horizontal at this position is :

Step-by-step solution

Let the radius vector of the block make an angle with the horizontal. The vertical distance descended by the block is h = R . By conservation of mechanical energy, the velocity v of the block is given by: 1 2 mv^2 = mgR v^2 = 2gR The centripetal acceleration of the block is: a_c = v^2 R = 2g The tangential acceleration of the block is the component of acceleration due to gravity along the tangent: a_t = g Since the velocity vector is directed along the tangent, the angle between the net acceleration vector and the