Question

A hollow ball and a solid sphere are released from rest and roll down an inclined plane. Both have a mass of 1 kg, a radius of 5 cm, and experience a torque of 0.01 N m. Assume both balls roll without slipping. What is the difference in their linear velocities after the balls have both lost 4 m of elevation? I’d like you to use energy to solve this problem.

Answer 1

by conservation of energy

initial energy = final energy

initial energy = potential energy

initial energy of both the balls will be equal since mass and height are same

initial energy = mgh

initial energy = 1 * 9.8 * 4

final energy = linear kinetic energy + rotational kinetic energy

rotational kinetic energy = 0.5 * moment of inertia * angular velocity

angular velocity = linear velocity / radius

moment of inertia of solid sphere = 2 * mass * radius^2 / 5

moment of inertia of hollow sphere = 2 * mass * radius^2 / 3

so,

rotational kinetic energy of solid sphere = 0.5 * 2 * mass * radius^2 * (linear velocity / radius)^2 / 5

rotational kinetic energy of solid sphere = mass * linear velocity^2 / 5

rotational kinetic energy of hollow sphere = 0.5 * 2 * mass * radius^2 * (linear velocity / radius)^2 / 3

rotational kinetic energy of hollow sphere = mass * linear velocity^2 / 3

so,

for solid sphere

1 * 9.8 * 4 = 0.5 * 1 * v^2 + 1 * v^2 / 5

v = 7.483 m/s

for hollow sphere

1 * 9.8 * 4 = 0.5 * 1 * v^2 + 1 * v^2 / 3

v = 6.858 m/s

difference in their linear velocity = 7.483 - 6.858

difference in their linear velocity = 0.625 m/s