A smooth inclined track ends in a vertical circular loop of radius R, as shown in the figure. A small body is released from rest from a height h above the lowest point of the loop. At the highest point of the loop, the body exerts a normal force on the track equal to three times its weight. If h = αR, what is the value of α?

Question

G1 Physics · Accepted Answer

At the highest point of the loop, both the weight mg and the normal reaction N act toward the centre. Given, N = 3mg. Therefore the centripetal-force equation is mg + N = mv² ÷ R. Hence, mg + 3mg = mv² ÷ R, so v² = 4gR. Now apply conservation of mechanical energy between the release point and the highest point of the loop. The top of the loop is at height 2R above the lowest point: mgh = mg(2R) + (1 ÷ 2)mv². Substitute v² = 4gR: mgh = 2mgR + (1 ÷ 2)m(4gR) = 4mgR. Therefore, h = 4R. Thus α = 4.…