Question

A uniform hollow cylinder has a density \( \rho \), a length \( L \), an inner radius \( a \), and an outer radius \( b \). Its moment of inertia about the axis of the cylinder is (Mass of the cylinder is \( M \)):

• A. \( M(b^2 + a^2) \)
• B. \( 2M(b^2 + a^2) \)
• C. \( \frac{M}{2} (b^2 + a^2) \)
• D. \( \frac{3}{4} M(b^2 + a^2) \)

Hint:

The moment of inertia of a hollow cylinder involves both the inner and outer radii and is influenced by the mass distribution.

Answer 1

The Correct Option is A

A uniform hollow cylinder is a cylindrical shell characterized by having both an inner and an outer radius. To find the moment of inertia \( I \) of the hollow cylinder about its axis, we integrate across the volume of the cylindrical shell. The moment of inertia for a differential mass element \( dm \) at a distance \( r \) from the axis is given by \( dI = r^2 \, dm \). 

The volume of the hollow cylinder is determined by subtracting the volume of the inner cylinder from the volume of the outer cylinder:

\[ V = \pi b^2 L - \pi a^2 L = \pi (b^2 - a^2) L \]

Given the density \( \rho \), the mass \( M \) of the hollow cylinder can be expressed as:

\[ M = \rho V = \rho \pi (b^2 - a^2) L \]

For a small ring element at radius \( r \), the differential area \( dA \) is the area of the ring: \( dA = 2\pi r \, dr \), and thus the differential volume of the ring \( dV = 2\pi r L \, dr \).

The differential mass is:

\[ dm = \rho \, dV = \rho (2\pi r L \, dr) \]

To find the moment of inertia of the entire cylinder, integrate over the thickness of the shell from \( a \) to \( b \):

\[ I = \int_{a}^{b} r^2 \, dm = \int_{a}^{b} r^2 \rho (2\pi r L \, dr) = 2\pi \rho L \int_{a}^{b} r^3 \, dr \]

The definite integral \( \int_{a}^{b} r^3 \, dr \) evaluates to:

\[ \int_{a}^{b} r^3 \, dr = \left[\frac{r^4}{4}\right]_{a}^{b} = \frac{b^4}{4} - \frac{a^4}{4} \]

Plug this result back into the expression for \( I \):

\[ I = 2\pi \rho L \left(\frac{b^4 - a^4}{4}\right) \]

Simplifying gives:

\[ I = \frac{\pi \rho L (b^4 - a^4)}{2} \]

Substituting \( M = \rho \pi (b^2 - a^2) L \), which implies \( \rho = \frac{M}{\pi (b^2 - a^2) L} \), we can write:

\[ I = \frac{M}{2} \frac{b^4 - a^4}{b^2 - a^2} \]

Notice that:

\[ \frac{b^4 - a^4}{b^2 - a^2} = (b^2 + a^2) \] (by polynomial identity)

So, finally:

\[ I = \frac{M}{2} (b^2 + a^2) \]

The correct answer is therefore \[ M(b^2 + a^2) \].