# Kinetic Energy Ratio of an Eccentric Disk

1. Nov 5, 2012

### mc120

1. Let’s say we have a solid wheel. The wheel can be modeled as a disk. Imagine that instead, the wheel is rotated at a location location 0.47R from the center of the wheel, so that the wheel rolled around a kind of loop. Essentially, the CM goes around the dashed line in the drawing. R is the radius. What is the percentage of the total kinetic energy that must be rotational?

2. I am guessing at these being relevant:
Idisk=1/2mr2
Itotal=1/2mr2+md2
KErot=1/2Iω2
KEtrans=1/2mv2

3. Ratio = KErot/(KErot+KEtrans)
Substituting in: Itotal=1/2mr2+m(0.47r)2 into KErot=1/2Iω2 and KEtrans=1/2mω2 (since the CM will travel 2∏r in the same time period as rotation), I simplify to:
(1/2mr2+1/2m(0.47r)2)/(1/2mr2+1/2m(0.47r)2+m), which is the same as:
(1/2r2+(0.47r)2)/(1/2r2+(0.47r)2), which equals 1.

I think I have gone horribly awry in my assumptions. I'm not really sure I even understand why there is anything OTHER than rotational Kinetic energy in this equation. The parallel axis theory makes sense for calculating the Inertia of an eccentric disk, but isn't it all still rotational kinetic energy?

Thanks for any advice anyone can give. Cheers!

2. Nov 5, 2012

### SammyS

Staff Emeritus
Hello mc120. Welcome to PF !

Although, you could consider all the KE to be rotational, I think that the key here is the word must .

What is the percentage of the total kinetic energy that must be rotational?

3. Nov 5, 2012