# Homework Help: Moment of Inertia of a Sphere

1. Dec 10, 2011

### cc2hende

1. The problem statement, all variables and given/known data
A small sphere of mass 2.0kg revolves at the end of a 1.2 m long string in a horizontal plane around a vertical axis. Determine its moment of inertia with respect to that axis.

2. Relevant equations

Moment of inertia for sphere= (2/5)mr^2

3. The attempt at a solution
I=(2/5)(2.0kg)(1.2m)^2
I=1.2 kgm^2

However, my book says to use the equation: I=mr^2. As a result they got I=2.9kgm^2. Why do they do this?? Isn't that the equation for Moment of inertia of a hollow cylinder?

2. Dec 10, 2011

### ehild

The sphere is small, 1.2 m is the radius of its orbit around an axis. You calculated the moment of inertia of a big sphere of radius 1.2 m.

Think of the definition of moment of inertia with respect an axis: for a very small body, a point mass, it is I=mr^2 where r is the distance of the body from the axis.

The moment of inertia of an extended body is the sum (integral) of the moment of inertia of its parts. Imagine you have a very small sphere with radius r=1 cm, for example, and it revolves around the axis of a circle of radius R=1.2 m. How far are the parts of the sphere from the axis? The closest point is 1.19 m, the farthest is at 1.21 m. So the moment of inertia must be close to mR2.
The moment of inertia of a sphere with respect to an axis going through its own centre is I(CM)=2/5 mr2. The Parallel Axis Theorem establishes the relation between the moment of inertia about an arbitrary axis and that about a parallel axis at distance R, going through the centre of mass of an object: I=I(CM)+mR2.

ehild