A ball with moment of inertia of 1.6, mass of 4 kg and radius of 1 m rolls without slipping down an incline which is 10 meters high. What is the speed of the ball when it reaches the bottom of the incline.
Conservation of energy is used to solve this problem of combined rotational and translational motion. Since the ball rolls without slipping, the kinetic energy can be expressed in terms of velocity v only.
Initially, the ball is at rest, so all energy is gravitational potential energy. When the ball reaches the bottom of the incline, all potential energy is converted into rotational and translational kinetic energy.
By conservation of energy:
Ei = Ef
⇒ (1/2) Mv² + (1/2) Iω² = Mgh
⇒ (1/2)(4)v² + (1/2)(1.6)v² = (4g)(10)
⇒ 2v² + 0.8v² = 40g
⇒ v = √(40g / 2.8) = 11.8 m/s
Angular velocity:
ω = 2π × (750 / 60) = 25π = 78.5 rad/s