Question

1. Three children are riding on the edge of a merry‑go‑round that has a mass of 105 kg and a radius of 1.80 m. The merry‑go‑round is spinning at 22.0 rpm. The children have masses of 22.0, 28.0, and 33.0 kg. If the 28.0 kg child moves to the center of the merry‑go‑round, what is the new angular velocity in revolutions per minute? Ignore friction, and assume that the merry‑go‑round can be treated as a solid disk and the children as point masses.

2. On average, both arms and hands together account for 13%13% of a person's mass, while the head is 7.0%7.0% and the trunk and legs account for 80%.80%. We can model a spinning skater with her arms outstretched as a vertical cylinder (head, trunk, and legs) with two solid uniform rods (arms and hands) extended horizontally.

Suppose a 68.0 kg skater is 1.50 m tall, has arms that are each 64.0 cm long (including the hands), and a trunk that can be modeled as being 32.0 cm in diameter. If the skater is initially spinning at 72.0 rpm with her arms outstretched, what will her angular velocity ?2 be (in rpm) after she pulls in her arms and they are at her sides parallel to her trunk? Assume that friction between the skater and the ice is negligble.

w2=

3. A child's top is held in place upright on a frictionless surface. The axle has a radius of ?=2.96 mmTwo strings are wrapped around the axle, and the top is set spinning by applying ?=3.40 N of constant tension to each string. If it takes 0.440 s for the string to unwind, how much angular momentum does the top acquire? Assume that the strings do not slip as the tension is applied.

angular momentum:__________kg⋅m2s

Point P is located on the outer surface of the top, a distance ℎ=31.0 mm above the ground. The angle that the outer surface of the top makes with the rotation axis of the top is ?=27.0 . If the final tangential speed ?tv of point P is 1.85 m/s , what is the top's moment of inertia?

moment of inertia: ______________kg⋅m2

Answer 1

platform: I_p = ½mr² = ½ * 105kg * (1.8m)² = 170.1 kg·m²
children: initial I_c = (22 + 28 + 33)kg * (1.8m)² = 268.92 kg·m²
and final I_c = (22 + 33)kg * (1.8m)² = 178.2 kg·m²

Conserve momentum: initial L = final L
(170.1 + 268.92)kg·m² * 22rpm = (178.2 + 170.1)kg·m² * ω
ω = 27.73 rpm