Question

A uniform solid sphere has a moment of inertia I about an axis tangent to its surface. What is the moment of inertia of this sphere about an axis through its center?a) 7/5 Ib) 3/5 Ic) 2/5 I
d) 1/7 I

Answer 1

Answer:

2/7 I

Explanation:

The theorem of parallel axis states that the moment of inertia of a body about a certain axis z' is equal to the moment of inertia of the body about the axis passing through the centre, z, plus the product between the mass of the body (M) and the square of the distance (r) between the two axis:

$I_z' = I_z + Mr^2$ (1)

For a solid sphere, the moment of inertia about the axis passing through the centre is

$I_z=\frac{2}{5}MR^2$ (2)

where R is the radius of the sphere.

The moment of inertia about an axis tangent to the surface then will be (applying (1) using r=R):

$I = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2$ (3)

The problem asks us to rewrite $I_z$, the moment of inertia about the centre, in terms of I, the moment of inertia about the axis tangent to the surface. We can do it by rewriting (2) as follows:

$MR^2 = \frac{5}{2}I_z$

And substituting this into (3):

$I=\frac{7}{5}(MR^2 )=\frac{7}{5}(\frac{5}{2} I_z) = \frac{7}{2}I_z\\I_z = \frac{2}{7}I$