This set of questions uses an interesting formula. A torque applied for a certain time causes a change in angular momentum. Δt=ΔL. It is just like the impulse momentum theorem which says that a force applied for a certain time causes a change in momentum, FΔt=Δp. Lindsey is on the merry-go-round again. Her mass is 33.0kg . The merry-go-round has a mass of 78.0kg and a radius of 2.20m . Lindsey is standing 0.150m from the center and has an initial angular velocity of 3.45radsec . Her older brother Mike applies a force of 200N tangent to the outer edge for 0.800s causing the merry-go-round to spin faster. What was the initial angular momentum before Mike pushed, in kgm2s ? 2nd Question: How much torque did Mike apply?

a.  653.79 kgm²/s b. 440 Nm

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

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speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.