Question

A small bead can slide without friction on a circular hoop that is in a vertical plane and has a radius of 0.100 m. The hoop rotates at a constant rate of 3.90 rev>s about a vertical diameter. (a) Find the angle b at which the bead is in vertical equilibrium. (It has a radial acceleration toward the axis.) (b) Is it possible for the bead to ""ride"" at the same elevation as the center of the hoop? (c) What will happen if the hoop rotates at 1.00 rev>s?

Answer 1

The balance of centripetal and gravitational forces determines the bead's equilibrium on a rotating hoop. The capability for a bead to remain at the same elevation as the hoop's center depends on the hoop's rotational speed.

Step-by-step explanation:

Vertically Equilibrium and Centripetal Acceleration are crucial concepts when analyzing the motion of an object in a rotating system. The scenario depicts a bead on a rotating hoop requiring the equilibrium of centripetal and gravitational forces to maintain a position. As the hoop's rotation changes, the dynamics of the bead's motion will also change accordingly. In particular, for a bead to be in vertical equilibrium on a rotating hoop, the centripetal force provided by the hoop's rotation must balance out the gravitational force pulling the bead downwards. However, whether a bead can 'ride' at the same elevation as the center is dependent on the precise balance of forces, which is determined by the hoop's rotational speed. If the hoop's rotational speed decreases, the balancing point shifts and the bead may not remain at the same elevation.

Answer 2

The bead on a rapidly rotating hoop will be in vertical equilibrium at a specific angle where gravitational and centripetal forces balance. If the rotational speed changes, the centripetal force changes, altering the bead's equilibrium position. The equilibrium height can theoretically match the hoop center at precise conditions.

Step-by-step explanation:

Addressing the given student's question about the circular motion of a bead on a rotating hoop, we need to consider forces acting on the bead. For part (a), if the hoop is rotating at a constant rate of 3.90 revolutions per second, the bead will experience a centripetal acceleration due to rotational motion. The angle at which the bead will be in vertical equilibrium (β) can be found using Newton's second law for circular motion and solving for the angle that satisfies the radial force balance, which includes gravitational and centripetal components.

For part (b), in theory, the bead could "ride" at the same elevation as the center of the hoop if the centripetal force due to rotation exactly balances the gravitational force. However, this would require precise conditions dependent on the rotational speed of the hoop.

For part (c), if the hoop rotates at 1.00 revolution per second, the centripetal acceleration will reduce. This means the bead will likely settle at a position closer to the bottom of the hoop, where the required centripetal force is less than what is provided by the gravitational component of the force acting on the bead. We can conclude what might happen based on the balance of forces and Newton's laws of motion.