A body of mass 'm' attached at the end of a string is just completing the loop in a vertical circle. The apparent weight of the body at the lowest point in its path is ( g = gravitational acceleration)

Step-by-step solution

For a body to just complete a vertical circle, the tension in the string at the highest point is zero. T_H + mg = m v_H^2 r v_H = gr By conservation of mechanical energy between the highest and lowest points: 1 2 m v_L^2 = 1 2 m v_H^2 + mg(2r) v_L^2 = v_H^2 + 4gr = gr + 4gr = 5gr At the lowest point, the net force towards the center provides the centripetal acceleration: T_L - mg = m v_L^2 r T_L = mg + m(5gr) r = mg + 5mg = 6mg The apparent weight of the body is equal to the tension in the string at that point, whi