Rolling motion and rotational energy

Motion and energy of a rolling ballWhen a ball rolls across the ground without slipping, the faster it moves across the ground, the faster it rotates about its center. The linear velocity v and angular velocity ω of a rolling ball are directly related to each other by the equation v = ωr, as we learned on page 414. Every time the ball rotates one complete turn (or 2π radians), it has moved forward by a distance of 2πr (the ball’s circumference). Read the text aloud
When the ball is rolling, it has both kinetic energy and rotational energy, which is sometimes called the kinetic energy of rotation. Linear kinetic energy is given by ½mv2, while rotational (kinetic) energy is given by ½2. The equations look similar: You just substitute I for m and ω for v! Read the text aloud
(13.4) E r = 1 2 I ω 2
Er  = rotational energy (J)
I  = moment of inertia (kg m2)
ω  = angular velocity (rad/s)
Rotational energy
When a ball is rolling across the ground, it has both kinetic energy and rotational energy. Its total energy is the sum of the two:
E tot = E k + E r = 1 2 m v 2 + 1 2 I ω 2
How does the energy of a rolling ball compare with the energy of a skidding ball—i.e., a ball that is sliding across a frozen lake—moving at the same velocity? When the ball is sliding, it has no rotational energy, so its total energy is less than the rolling ball that moves at the same velocity. Read the text aloud
Linear motion and rotational motion may appear different, but the mathematical representations for both are very similar. Whereas linear motion uses distances and velocities, rotational motion uses angles and angular velocities. Mass is used in linear motion but moment of inertia is used in rotational motion.

Comparison between linear and rotational motion
Linear motionRotational motion
QuantityVariable or
UnitsQuantityVariable or
PositionxmPosition angleθrad
Velocityv = Δxtm/sAngular velocityω = Δθtrad/s
MassmkgMoment of inertiaIkg m2
Momentump = mvkg m/sAngular momentumL = Iωkg m2/s
Kinetic energyEk = ½mv2JRotational energyEr = ½Iω2J
Read the text aloud Show Advanced rotational dynamics
A diver jumps off a diving board into the water without spinning, entering the water feet first. He then repeats the dive, but this time he does several somersaults before entering the water. If he followed the exact same trajectory in both dives, did his energy differ between the two dives? Explain. Show

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