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.
Derivation Of Moment Of Inertia Of Common Shapes:
- Uniform Rigid Rod (“Beginners’ Lesson”)
- Uniform Solid Sphere
- Thin Spherical Shell
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.
dude ur the plug thanks soo much
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
Thanx its help me easily
Thanks it helped me alot
its help me very easily. Thanks
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)
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?
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.
Shouldn’t it be R2+R1 in the end?
Yes, you’re right. My apologies.
So p (rho) = M/(pi*(R2^2+R1^2)*L)? Or what?
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.