A resourceful physics student decides to build a device to lower him from his 3 story window safely. He begins with a heavy flywheel mounted on a frictionless axle which is attached to the roof. The flywheel has a diameter 82.0 cm He wraps a long rope around the outside of the flywheel and attaches a handle to the free end. Grabbing the free end he descends from rest a distance 7.00 m in a time of 6.10 s. If the mass of the physics student is 72.0 kg, what is the moment of inertia of the flywheel

I ≈ 316 kg•m²

Explanation:

I = τ/α

τ = FR = mgR

α = a/R = (2h/t²)/R =  2h/Rt²

I = mgR / 2h/Rt² = mgR²t²/2h

I = 72.0(9.81)(0.82/2)²(6.10²) / (2(7.00))

I = 315.5737...≈ 316 kg•m²

t = 1/400 h   ⇔ 0.0025 h ⇔ 9s ⇔ 0.15 m

any of the answers are correct.

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