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.

hollow cylinder

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.

Back To Rotation (UY1)

Derivation Of Moment Of Inertia Of Common Shapes:

What Others Are Saying:

  1. Luiz scribbled

    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)

  2. John Adex scribbled

    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.

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