Question

How much Kinetic energy would a 10.0 kg thin-walled hollow sphere with a diameter of 1.60 m have if it was rotated at a speed of 45.0 rpm about an axis passing through its diameter?

Answer 1

1 rpm = 2π/60 rad/s = π/30 rad/s

M = 10 kg, D = 2R = 1.6 m => R = 0.8 m, ω = 45 rpm = 45 × (π/30) rad/s = 4.71 rad/s

Where, M is mass, R is radius, D is diameter and ω is Angular velocity

Moment of Inertia of Hollow Sphere (I) = MR^{​​​​​2}​ = (10 kg)*(0.8 m)² = 6.4 kg-m²

Kinetic Energy = (1/2)Iω² = (1/2)*(6.4 kg-m²)*(4.71 rad/s)² = 71.0 J

Hence, Kinetic Energy of Hollow Sphere is 71.0 J

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