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 \[ M(b^2 + a^2) \].

Answer 2

Instead of integrating over thin rings directly, we can find the moment of inertia of the hollow cylinder by treating it as a solid cylinder of radius \( b \) with a solid cylinder of radius \( a \) removed from its center — both cut from material of the same density \( \rho \) and length \( L \).

The mass of a solid cylinder of radius \( b \) (same density and length) would be \( M_b = \rho \pi b^2 L \), and its moment of inertia about the central axis would be \( I_b = \frac{1}{2}M_b b^2 = \frac{1}{2}\rho \pi b^4 L \). Similarly, the solid cylinder of radius \( a \) that is removed has mass \( M_a = \rho \pi a^2 L \) and moment of inertia \( I_a = \frac{1}{2}M_a a^2 = \frac{1}{2}\rho \pi a^4 L \).

The hollow cylinder's moment of inertia is the difference:

\[ I = I_b - I_a = \frac{1}{2}\rho \pi L (b^4 - a^4) \]

The actual mass of the hollow cylinder is \( M = M_b - M_a = \rho \pi L (b^2 - a^2) \), so \( \rho \pi L = \frac{M}{b^2-a^2} \). Substituting:

\[ I = \frac{1}{2}\frac{M}{b^2-a^2}(b^4-a^4) = \frac{M}{2}\cdot\frac{b^4-a^4}{b^2-a^2} = \frac{M}{2}(b^2+a^2) \]

since \( b^4 - a^4 = (b^2-a^2)(b^2+a^2) \).

Checking each option against this superposition result:

1. Option A, \( M(b^2+a^2) \): this is twice as large as what the subtraction of the two solid cylinders gives.
2. Option B, \( 2M(b^2+a^2) \): four times too large.
3. Option C, \( \frac{M}{2}(b^2+a^2) \): matches exactly the result from subtracting the inner solid cylinder's moment of inertia from the outer one.
4. Option D, \( \frac{3}{4}M(b^2+a^2) \): does not arise from this superposition method under any combination of \( a \) and \( b \).

The superposition method confirms option C.

Therefore, the correct answer is \( \frac{M}{2}(b^2+a^2) \).