Answer:
a. 653.79 kgm²/s b. 440 Nm
Step-by-step explanation:
a. The initial angular momentum
The initial angular momentum L of the boy and merry-go-round is
L = Iω where I = moment of inertia of merry-go-round + moment of inertia of Lindsey about merry-go-round and ω = initial angular velocity = 3.45 rad/s
moment of inertia of merry-go-round I₁ = 1/2MR² where M = mass of merry-go-round = 78 kg and R = radius of merry-go-round = 2.20 m
I₁ = 1/2MR² = 1/2 × 78 kg ×(2.20 m)² = 188.76 kgm²
moment of inertia of Lindsey about merry-go-round I₂ = mh² where m = mass of boy = 33.0 kg and h = distance of Lindsey from center of merry-go-round = 0.150 m
I₂ = mh² = 33.0 kg × (0.150 m)² = 0.743 kgm²
I = I₁ + I₂ = 188.76 kgm² + 0.743 kgm² = 189.503 kgm²
So, L = Iω = 189.503 kgm² × 3.45 rad/s = 653.79 kgm²/s
b. How much torque does Mike apply?
Since torque τ = FRsinθ where F = force = 200 N, R = radius of merry-go-round = 2.20 m and θ = 90° (since the force is applied tangentially to the merry-go-round)
τ = FRcosθ
= 200 N × 2.20 m × sin90°
= 440 Nm