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March 31, 2015

March 31, 2015

Posted by **Bernadette** on Friday, February 8, 2008 at 6:41pm.

- AP Physics -
**bobpursley**, Friday, February 8, 2008 at 6:53pmI will be happy to critique your thinking on this.

You do not need the radius, because you are dealing with fractions of energy. When you work out the KE, radius will be there, and the same radius term will be in the rotational energy part, and the radius terms divide out.

- AP Physics -
**Bernadette**, Friday, February 8, 2008 at 7:06pmThe equation I came up with to use was KE(rotational)/KE(total)=4(.5mr^2w^2)/(4(.5mr^2w^2)+(.5mr^2w^2)) and got an answer of 5.4%. My physics teacher gives us answers to problems in our book that are almost the same as our homework problems, but when I used this equation with the corresponding question, I did not get the right answer. Where am I going wrong?

- AP Physics -
- AP Physics -
**bobpursley**, Friday, February 8, 2008 at 7:24pmI agree with your formula, except how you used it. The masses of the wheels are not the same as the mass of the car, and the portion of KE translational has to use the mass of the cars.

- AP Physics -
**Buddy**, Tuesday, November 3, 2009 at 11:33pmThe formula for the moment of inertia is actually incorrect for you.

I = (1/2)mr^2

so

K = (1/2)((1/2)mr^2)(w^2)

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