A hollow cylinder has an inner radius R_{1}, mass M, outer radius R_{2} and length L. Calculate/derive its moment of inertia about its central axis.

**Guide:**

– The cylinder is cut into infinitesimally thin rings centered at the middle. The thickness of each ring is dr, with length L.

We write our moment of inertia equation:

$dI = r^{2} \: dm$

Now, we have to find dm, (which is just density multiplied by the volume occupied by one ring)

$dm = \rho \: dV$

We’ve introduced dV in the above equation, so, we have to find out what dV is:

$dV = dA \: L$

The dA is just the area of the top of the ring, which is the area of the big (radius: r + dr) ring minus that of the smaller (radius: r) ring. We have:

$dA = \pi (r + dr)^{2} – \pi r^{2}$

$dA = \pi (r^{2} + 2rdr + (dr)^{2}) – \pi r^{2}$

Note: (dr)^{2} is equal to 0. An infinitesimally small number multiplied by another infinitesimally small number = 0.

$dA = 2 \pi r \: dr$

Note: Another way of obtaining dA is by differentiating.

$$A = \pi r^{2}$$

Differentiating wrt r, $$dA = 2 \pi r dr$$

Substituting dA into dV,

$dV = 2 \pi r L \: dr$

Using the above equation, substitute into dm,

$dm = 2 \rho \pi r L \: dr$

Finally, we have an expression for dm. We substitute that into the dI equation,

$dI = 2 \rho \pi r^{3} L \: dr$

Now, we can integrate to find the moment of inertia, (Note: I did not substitute in the expression for density because it is quite messy and it is not needed in the integration process)

$I = 2 \rho \pi L \: \int_{R_{1}}^{R_{2}} r^{3} \: dr$

I’m sure you are able to do this integration by yourself. Now, we can find the expression for density.

Recall:

$\rho = \frac{M}{V}$

Hence,

$\rho = \frac{M}{\pi (R_{2}^{2} – R_{1}^{2}) L}$

Substituting this back into the integrated solution, we have:

$$I = \frac{1}{2}M(R_{2}^{2}+R_{1}^{2})$$

**Special Cases:**

**Hoop or thin cylindrical shell: (R _{1}=R_{2}=R)**

$I = M R^{2}$

**Disk or solid cylinder: (R _{1}=0)**

$I = \frac{1}{2} M R^{2}$

**Sanity check:** I is expected to be highest for hoop or cylindrical shell since all the mass are furthest away from the axis of rotation.

Back To Rotation And Moment Of Inertia

**Derivation Of Moment Of Inertia Of Common Shapes:**

- Uniform Rigid Rod (“Beginners’ Lesson”)
- Uniform Solid Sphere
- Thin Spherical Shell