Question

Find the moment of inertia of a solid cylinder of mass M and radius R about a line parallel to the axis of the cylinder and on the surface of the cylinder.

• A. $\frac{3}{2}MR^2$
• B. $\frac{5}{2}MR^2$
• C. $\frac{1}{2}MR^2$
• D. $MR^2$

Answer 1

Answer: (A)

$\frac{3}{2}MR^2$

Answer 2

The moment of inertia of a solid cylinder of mass M and radius R about its own axis (passing through its center of mass) is given by:

$I_{CM} = \frac{1}{2}MR^2$

The problem asks for the moment of inertia about a line parallel to the axis of the cylinder and on the surface of the cylinder. This means the distance 'd' between the central axis and the new axis is equal to the radius R of the cylinder ($d=R$).

We can use the parallel axis theorem, which states that the moment of inertia $I$ about an axis parallel to an axis passing through the center of mass ($I_{CM}$) is given by:

$I = I_{CM} + Md^2$

Substitute the known values:

$I = \frac{1}{2}MR^2 + M(R)^2$ $I = \frac{1}{2}MR^2 + MR^2$ $I = \left(\frac{1}{2} + 1\right)MR^2$ $I = \frac{3}{2}MR^2$