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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.

HTML updated for JavaScript by Ricky Davidson.

Physlets were developed at Davidson College and converted from Java to JavaScript
using the
SwingJS
system developed at St. Olaf College.