Suppose a child gets off a rotating merry-go-round. Does the angular velocity of the merry-go-round increase, decrease, or remain the same if: (a) He jumps off radially? (b) He jumps backward to land motionless? (c) He jumps straight up and hangs onto an overhead tree branch? (d) He jumps off forward, tangential to the edge? Explain your answers.

Question

Physics

Zackery

9 July, 03:12

Suppose a child gets off a rotating merry-go-round. Does the angular velocity of the merry-go-round increase, decrease, or remain the same if: (a) He jumps off radially? (b) He jumps backward to land motionless? (c) He jumps straight up and hangs onto an overhead tree branch? (d) He jumps off forward, tangential to the edge? Explain your answers.

+3

Answers (2)

Know the Answer?

Juliana Rowland · Answer 1

a. Same : If he jumps off radially, no torque is applied, this is because the angle between the force and radius is Zero sin0=0.

b. Increase : For him to jump backward, force is needed to act on him backwards. From Newton's 3rd law of motion, a forward force is would have acted on the merry ground, therefore the torque would be in the direction of its angular speed.

c. Same : The Axile would not permit its rotation in a different direction.

d. Increase : Jumping forward requires a force which will act on him forward. Again from Newton's 3rd law of motion, a backward force will act of the merry ground which will make torque to be at a direction opposite to its angular speed.

Tess Valdez · Answer 2

See explanation

Explanation:

a. same. Zero applied torque.

- When he jumps radially the moment of inertia of the system (child + merry-go-round) remains the same. Since, Torque T = I*α, then the angular acceleration is zero. Hence, angular speed remains same.

b. increases. Since he applies torque along the direction of motion.

- A torque is applied by the feet of the child on to the merry-go-round in the direction of motion. The moment of inertia of the system (child + merry-go-round) remains the same. Since, Torque T = I*α, then the angular acceleration increases as Torque applied is in the same direction. Hence, angular speed also increases.

c. increases. Since the moment of inertia decreases.

- The moment of inertia of the system (child + merry-go-round) as child is no longer part of the system. The angular speed (squared) is inversely proportional to moment of inertia. Hence, as inertia decreases the angular speed increases.

d. decreases. Since he applies torque opposite to direction of motion.

- A torque is applied by the feet of the child on to the merry-go-round against the direction of motion. The moment of inertia of the system (child + merry-go-round) remains the same. Since, Torque T = I*α, then the angular acceleration decreases as Torque applied is in the opposite direction. Hence, angular speed also decreases. (Braking Torque)