Question

A child pushes her friend (m = 25 kg) located at a radius r = 1.5 m on a merry-go-round (r_mgr = 2.0 m, I_mgr = 1000 kg*m²) with a constant force F = 90 N applied tangentially to the edge of the merry-go-round (i.e., the force is perpendicular to the radius). The merry-go-round resists spinning with a frictional force of f = 10 N acting at a radius of 1 m and a frictional torque τ = 15 N*m acting at the axle of the merry-go-round, and the merry-go-round is initially at rest.

1. What is the magnitude of the torque due to the 90 N force at 2 m?
2. In what direction is the torque due to the 90 N force at 2 m?
3. What is the magnitude of the torque due to the 10 N force at 1 m?
4. In what direction is the torque due to the 10 N force at 1 m?
5. What is the magnitude of the torque due to the friction at the merry-go-round axle?
6. In what direction is the torque due to the friction at the merry-go-round axle?
7. (1A) What is the magnitude of the net torque acting on the merry-go-round about its axle?
8. What is the moment of inertia of merry-go-round apparatus itself?
9. What is the moment of inertia of the child on the merry-go-round?
10. (1B) What is moment of inertia of the merry-go-round with the child on it?
11. Is the merry-go-round speeding up or slowing down?
12. (1C) What is the magnitude of the rotational acceleration of the merry-go-round?
13. In what direction is the merry-go-round accelerating?
14. (1D) If the child pushes for 15 s, what is the magnitude of rotational velocity of the merry-go-round at 15 s?
15. If the child stops pushing her friend on the merry-go-round with the 90 N force, will the merry-go-round speed up or slow down?
16. If the child stops pushing her friend on the merry-go-round with the 90 N force, what is the new net torque acting on the merry-go-round?
17. If the child stops pushing her friend on the merry-go-round with the 90 N force, in what direction is the new net torque acting on the merry-go-round?
18. If the child stops pushing her friend on the merry-go-round with the 90 N force, what is the magnitude of the deceleration of the merry-go-round?
19. (1E) Once the child stops pushing at 15 s, how long does it take for the merry-go-round to come to a stop?

Answer 1

Ans

i) The magnitude of the torque due to the 90 N force at 2 m = 90 N x 2m = 180 N.m.

ii) The direction of the torque due to 90N force is counterclockwise. (Assuming the force is acting to the left ).

iii) The magnitude of the torque due to the 10 N force at 1 m = 10 N x 1m = 10 N.m

iv) The direction of the torque due to 10N force is clockwise. (opposing the motion of the merry go round)

v) The magnitude of the torque due to the friction at the merry-go-round axle is 15 N.m

vi) The direction of the torque due to the friction at the merry-go-round axle is clockwise.

1A) The magnitude of the net torque acting on the merry-go-round about its axle = 180 - 10 - 15 = 155 N.m

The moment of inertia of merry-go-round apparatus itself = 1000 kg.m²

The moment of inertia of the child on the merry-go-round = mr² = 25kg x (1.5m) ² = 56.25 kg.m²

1B) The moment of inertia of the merry-go-round with the child on it = 1000 + 56.25 = 1056.25 kg.m²

The merry-go-round speeding up in counterclockwise direction.