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# Illustration 11.3: Translational and Rotational Kinetic Energy

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How do we describe rolling without slipping from an energy standpoint? We already know how to represent the kinetic energy of translation: (1/2) mv^{2}. We also know how to represent the kinetic energy of rotation: (1/2) Iω^{2}. But what if we have both? Restart.

As the ball rolls down the incline, the gravitational potential gets transformed into kinetic energy, but how much of each? With rolling without slipping, we found that there is a relationship between the linear velocity and the angular velocity: v = ωR. Given this relationship we know that KE_{trans} = (1/2) mv^{2}, while KE_{rot} = (1/2) I (v^{2}/R^{2}). But the moment of inertia always looks like *C*mR^{2}, so we find that KE_{rot} = (1/2) *C *mv^{2}. Therefore, we find that KE_{total} = (1+*C*) ( 1/2) mv^{2}. The gravitational potential energy gets transformed into the total kinetic energy, and what fraction goes into KE_{trans} or KE_{rot} is determined by the constant *C*. Specifically,

KE_{trans}/KE_{total} = 1/(1+*C*) and KE_{rot}/KE_{total} = *C*/(1+*C*).

A ball of radius 1.0 m and a mass of 0.25 kg rolls down an incline, as shown **(position is given in meters and time is given in seconds)**. The incline makes an angle θ = 20° with the horizontal. Watch the graph of gravitational potential energy and rotational and translational kinetic energy vs. time or position.

Why do you think that the energy vs. time graphs curve, while the energy vs. position graphs are straight lines?

Illustration authored by Steve Mellema, Chuck Niederriter, and Mario Belloni.

Script authored by Steve Mellema, Chuck Niederriter, and Mario Belloni.

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