# Angular Velocity of a swinging Sphere

## Homework Statement

A uniform solid sphere of mass M and radius R hangs from a string of length R/2. Suppose that the sphere is released from an initial position making an angle of 45 degrees with the vertical. Calculate the angular velocity of the sphere when it swings through the vertical position.

## Homework Equations

Conservation of Energy. U= Mgh = I ($$\omega$$^2)
Parallel Axis Theorem I= I$$_{cm}$$+Md$$^{2}$$

## The Attempt at a Solution

I'm in an odd situation in that I have the solution, but am trying to wrap my head around what exactly it means. I first rearranged mgh=(I)Omega^2 to get Omega=$$\sqrt{mgh/I}$$. I figured that I would then have my answer as soon as I solved for I. Using the Parallel Axis Theorem, and my book's given moment of inertia for a sphere being rotated about its diameter (2/5)MR^2, gave me I= (2/5)MR^2 + Md^2. However, my professor's solutions give a d (distance from axis of rotation to axis through center of mass) as 3R/2. I am having trouble figuring out where this number comes from. As far as I can tell, the only distance which is 3R/2 is the distance from the pivot point (where the string is mounted) to the bottom of the sphere. Wouldn't using this distance as "d" imply that the sphere is rotating about an axis through the end of the ball? Shouldn't it be rotating through its center of mass?

Any help would be appreciated, as I know I'm somehow looking at this motion wrong. And I apologize for the messy lack of formatting, I'm new to this system of symbol representation and was having trouble with it.

## Answers and Replies

Related Introductory Physics Homework Help News on Phys.org
alphysicist
Homework Helper
Hi zylaxice,

The distance (3/2)R is the distance from the pivot point to the center of the sphere. This is the distance from the center of mass of the sphere to the axis of rotation which you need in the parallel axis theorem.

The distance from the pivot point to the bottom of the ball would be the length of the string plus the diameter of the ball, which would be (R/2) + 2R = (5/2)R

tiny-tim
Science Advisor
Homework Helper
Welcome to PF!

Hi zylaxice! Welcome to PF!

You're confusing radius with diameter.

(Did you draw a diagram? Always draw a diagram!)

Thank you both, that's exactly what I did. No matter how many times I rewrote the problem, I guess I always made the same mistake. I think because my original diagram was so horrendously out of scale I somehow associated R with being the diameter of the sphere instead of the radius, and it took someone else pointing it out to make me realize how blind I was.