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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) mv2. We also know how to represent the kinetic energy of rotation: (1/2) Iω2. 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 KEtrans = (1/2) mv2, while KErot = (1/2) I (v2/R2). But the moment of inertia always looks like CmR2, so we find that KErot = (1/2) C mv2. Therefore, we find that KEtotal = (1+C) ( 1/2) mv2. The gravitational potential energy gets transformed into the total kinetic energy, and what fraction goes into KEtrans or KErot is determined by the constant C. Specifically,
KEtrans/KEtotal = 1/(1+C) and KErot/KEtotal = 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.