- #26

- 75

- 3

woops you're right, hit square root instead of cube root I think. As I said, I'm no expert. When I did these kinds of problems in my mechanics class we never took rotation into account. If we took rotation into account, wouldn't the energy be the exact same anyways?I get 2.08 m/s. The ball is 2.2 cm in radius, yes?

But I agree, the book answer is way off. I can't see what simple error leads to it.

I don't understand your reasoning. Even at the lowest point, the ball is rotating as well as moving horizontally. However, it turns out that the energy taken up by that is only 0.01% of the total, so it can be safely ignored. It would be a very different story if the radius were similar in magnitude to the length of the string.

## \frac {1}{2} Iω^2 = \frac {1}{2}(mr^2) \frac {v^2}{r^2} = \frac {1}{2} mv^2## no?