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

 

 

hollow-solid cylinder

A hollow cylinder has an inner radius R1, mass M, outer radius R2 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: (R1=R2=R)

hoop

$$I = M R^{2}$$

Disk or solid cylinder: (R1=0)

Disk

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

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: