Chapter 8, Problem 11Q

The moment of inertia of a rotating solid disk about an axis through its cm is $\frac{1}{2}$MR²(Fig. 8-20c). Suppose instead that a parallel axis of rotation passes through a point on the edge of the disk. Will the moment of inertia be the same, larger, or smaller? Explain why.

To determine

The moment of inertia for the disk about the axis parallel to the central axis

Answer to Problem 11Q

Solution:

The required moment of inertia is $\frac{3}{4}M{R}^2$ , which is greater than the moment of inertia about the axis passing through the center of mass.

Explanation of Solution

Given:

Moment of inertia of the disk about the axis through its center of mass is $\frac{1}{2}M{R}^2$

Formula used:

The parallel axes theorem tells us that the moment of inertia about an axis parallel to an axis passing through an object’s center of mass is equal to the sum of the moment of inertia about the axis passing through the center of mass and the product of mass and square of the distance between the two parallel axes.

If I’ is the moment of inertia about the parallel axis passing through a point on the edge of the disk and I is the moment of inertia about the axis passing through the center of mass:

  $I’=I+M{R}^2$

Calculation:

  $I’=I+M{R}^2$

Since the radius is the distance between the two axes

Therefore,

  $\begin{array}{c}I’=\frac{1}{2}M{R}^2+M{R}^2\\ =\frac{3}{2}M{R}^2\end{array}$

Conclusion

Therefore, the moment of inertia about the parallel axis passing through the edge is greater than the moment of inertia about the axis through the center of mass.