# UY1: Calculation of moment of inertia of a hollow/solid cylinder

### Derivation of the moment of inertia of a 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)

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

Disk or solid cylinder: (R1=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.

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Derivation Of Moment Of Inertia Of Common Shapes: ### 25 thoughts on “UY1: Calculation of moment of inertia of a hollow/solid cylinder”

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

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

3. inertia for both differ….

• they are meant to, no?

4. 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.

5. Why not go from dV=dA* L directly to dV = 2*Pi*r*dr*L by considering a loop of thickness dr?

Overall the moment of inertia derivations are very well done.

6. In the beginning you have already supposed that for an infinitesimally thin cylindrical shell is I=mr^2 in the first equation I=r^2 *dm. No need to mention it again at the end.

7. dude ur the plug thanks soo much

8. Thanks a lot, it helped me substantially

Just one question: In the picture on beginning, the variables should not be L, R1 and R2 instead of h, r1 and r2?

thanks again

• it doesn’t matter at all

9. Thanx its help me easily

10. Thanks it helped me alot

• its help me very easily. Thanks

11. Observe that in my comment below, something like a^2, means the variable ‘a’ to the exponent ‘2’ and R1, R2, the variables R index 2 and 1 respectively.

I=(1/2)M(R2^2+R1^2) is still incorrect! Because the integration works out to be

I = (1/2)M(R2^4 – R1^4),

you could rewrite it as

I = (1/2)M[R2^2 – R1^2 – 2*(R2^2)(R1^2)/(R2^2 – R1^2)]

In the case R1 = 0, you’ll get

I = (1/2)MR2^2. However, you CAN NOT use R1 = R2, for it would mean infinite material density. For that case, I is always ZERO, for only real objects (those with mass) can have I > 0.

Please, let me know if you have trouble with the algebraic development. :0)

• But refer to wikipedia article “List of moments of inertia” which says :-
“This expression assumes that the shell thickness is negligible. It is a special case of the thick-walled cylindrical tube for r1 = r2.”

12. Please can you simplify a little because it’s kinda confusing. If not for you I wouldn’t have been able to do my assignment. Thanks.

• Could you point out the parts to be simplified?

13. While cosidering small element ring , thikness of ring is dr but what does it mean that length is L?

• The rings are actually very thin hollow cylinders, with height L.

14. Shouldn’t it be R2+R1 in the end?

• Yes, you’re right. My apologies.

• • The density equation is correct in my post, which is $R_{2}^{2} – R_{1}^{2}$. The previous moment of inertia equation is wrong. I’ve corrected it.