Question

In: Physics

A solid disk with a mass of 0.85 kg starts at the top of a 0.79...

A solid disk with a mass of 0.85 kg starts at the top of a 0.79 m tall hill and rolls down without slipping. What is the translational speed of the disk at the bottom of the hill? What fraction of the disk's kinetic energy is rotational at the bottom of the hill?

Expert Solution

Gravitational acceleration = g = 9.81 m/s²

Height of the hill = H = 0.79 m

Mass of the solid disk = M = 0.85 kg

Radius of the solid disk = R

Moment of inertia of the solid disk = I = MR²/2

Translational speed of the disk at the bottom of the hill = V

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

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

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

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

KE = KE_T + KE_R

The initial potential energy of the disk at the top of the hill is converted into the total kinetic energy of the disk at the bottom of the hill.

MgH = KE

MgH = 3MV²/4

gH = 3V²/4

(9.81)(0.79) = 3V²/4

V = 3.21 m/s

Ratio of the rotational kinetic energy and the total kinetic energy is given by,

A) Translational speed of the disk at the bottom of the hill = 3.21 m/s

B) Fraction of the disk's kinetic energy is rotational at the bottom of the hill = 1/3

genius_generous answered 8 months ago

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