a 1m radius cylinder with a mass of 852.9 kg rolls without slipping down a hill...

a 1m radius cylinder with a mass of 852.9 kg rolls without slipping down a hill which is 82.1 meters high. at the bottom of the hill what friction of its total kinetic energy is invested in rotational kinetic energy

Solutions

Expert Solution

Gravitational acceleration = g = 9.81 m/s²

Mass of the cylinder = M = 852.9 kg

Radius of the cylinder = R = 1 m

Moment of inertia of the cylinder = I

Height of the hill = H = 82.1 m

Velocity of the cylinder at the bottom of the hill = V

Angular speed of the cylinder at the bottom of the hill = = V/R

Rotational kinetic energy of the cylinder at the bottom of the hill = KE_R

Translational kinetic energy of the cylinder at the bottom of the hill = KE_T

Total kinetic energy of the cylinder at the bottom of the hill = KE

Ratio of the rotational kinetic energy and the total kinetic energy = f

f = 1/3

Fraction of the total kinetic energy of the cylinder invested in rotational kinetic energy at the bottom of the hill = 1/3

genius_generous answered 1 year ago

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