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A relationship between rotational and translational kinetic energy can be derived. That relationship is givenby the expression shown in Figure 2 .
Figure 2 . Relationship between rotational and translational kinetic energy. 

Krot = Q*Ktr where

A very interesting equation
The equation shown in Figure 2 is very interesting. It shows that the relationship that determines the distribution of kinetic energy betweentranslational kinetic energy and rotational kinetic depends solely on the constant Q, and is independent of the mass, the radius, etc.
The constant Q depends on how the mass is geometrically distributed in the rolling object. Generally speaking, the more the mass is distributed toward theouter edge of the object, the greater will be the value of Q.
The total kinetic energy is given by
K = Ktr + Krot, or
K = (1/2)*M*(Vcm)^2 + (1/2)*Icm*W^2
where
Through substitution we can write
K = Ktr + Krot, or
K = Ktr + Q*Ktr, or
K = (1 + Q)*Ktr, or
K = (1 + Q)*(1/2)*M*(Vcm)^2
Thus, an object with a given mass and a given translational velocity has a total kinetic energy that is proportional to (1+Q). The larger the value of Q,the greater will be the kinetic energy possessed by the object.
From Figure 1 , a rolling hollow cylinder with a Q value of 1 would have a greater total kinetic energy than a solid cylinder with the same mass and a Qvalue of 0.5 rolling with the same translational velocity. This is because the cylinder would have more rotational kinetic energy.
If there were no friction, an object would not roll down an incline. Instead it would simply slide down the incline. In the absence of torque, the rotationalinertia of the object would prevent it from experiencing angular acceleration.
The frictional force parallel to the incline and pointing up the incline produces a torque that causes the object to experience angular acceleration.
In this section, I will explain some example scenarios involving various aspects of rolling solid and hollow cylinders.
A cylindrical shell and a solid cylinder, each of unknown mass and unknown radii roll down an incline of unknown height and unknown angle.
Which object will have the greater translational velocity when they reach the bottom of the incline?
Solution:
From Figure 1 , the rotational inertia for a thin hollow cylinder is given by
I = M*R^2
The rotational inertia for a solid cylinder is given by
I = (1/2)*MR^2
Let the cylindrical shell be identified as M1 and the solid cylinder be identified as M2.
The total kinetic for each object at the bottom is equal to potential energy for that object at the top. Therefore,
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