### Derivation of the moment of inertia of a hollow/solid cylinder

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 – since the density is not dependent on r)

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

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.

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

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

Thank you very much sir. Now everythink is clear im my mind. 🙂

Thanks.I love your simplified explanation.you are among the best but you did not derived that of a rectangle

inertia for both differ….

Heck of a simplification at the end! After plugging in rho, after the 1/2 M I’m left with (R1^4 – R2^4) / (R1^2 – R2^2), I’ll assume that equals R1+R2 because that’s way too much algebra!

(R1^4-R2^4) = (R1^2 – R2^2)(R1^2 + R2^2) and therefore cancels with denominator.

From what you presented, you just proved that your assumption, R1+R2, was incorrect.