Rene is playing with a large spherical rock at the edge of the cliff.

Question

Rene is playing with a large spherical rock at the edge of the cliff.

The spherical rock, which has a mass of 215 kilograms and a radius of 0.55 meters, is falling off a cliff that is 110 meters tall. It is rotating at 3.2 radians/second.

What is the total energy of the rock when it reaches the bottom of the cliff?

Answer 1

M*g*h = (1/2) * I * w^2 + (1/2)*M*V^2

M*g*h = 215*9.81*110 = 232006.5 J

Mass Moment of Inertia of Sphere: 2/5 * M * R^2 = 2/5 * 215 * 0.55^2 = 26.015 Kg*m^2

M*g*h = (1/2) * I * w^2 + (1/2)*M*V^2

232006.5 = (1/2) * 26.015*3.2^2 + (1/2)*215*V^2

V = 46.443 m/s

Total KE = (1/2) * I * w^2 + (1/2)*M*V^2 = (1/2) * 26.015*3.2^2 + (1/2)*215*46.44^2 = 232006 J

Please note logically PE = KE, therefore for this problem the KE is equal to the M*g*h

Major assumptions

Spin of the rock remains constant at all points

No air resistance

Answer 2