Two beads of mass m each are positioned at the top of a frictionless hoop of mass M and radius R, which stands vertically on the ground. The beads are given a small push, and they slide down the hoop,

Question

Two beads of mass m each are positioned at the top of a frictionless hoop of mass M and radius R, which stands vertically on the ground. The beads are given a small push, and they slide down the hoop, one to the right and other-one to the left, as shown. The smallest value of m/M for which the hoop will rise up off the ground at some time during the motion, is :

TestHub · Accepted Answer

Newton's second law for a bead: N + m g cos θ = R m v 2 ​ ; energy: v 2 = 2 g R ( 1 − cos θ ) , so N = m g ( 2 − 3 cos θ ) . The hoop rises when the beads pull it up against its weight: 2 N cos θ = M g ⇒ 2 m g ( 2 − 3 cos θ ) cos θ = M g . Maximum upward force: d θ d ​ ( 2 cos θ − 3 cos 2 θ ) = 0 ⇒ cos θ = 3 1 ​ . 2 m g ( 2 − 1 ) × 3 1 ​ = M g ⇒ M m ​ = 2 3 ​ minimum - (C)

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