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Chapter 20

Q. 20.9

Find the moment of inertia of a uniform solid cylinder of mass M, radius r and height h about an axis which is perpendicular to the axis of the cylinder and which passes through the centre of one end.

Step-by-Step

Verified Solution

Choose axes as shown in figure 20.24 so that the required moment of inertia is about the x axis.
Let ρ be the density of the cylinder, so that M = πr²hρ.
Subdivide the cylinder into elementary discs as shown.

A typical disc has thickness δy and centre at C(0, y).

Its mass δm is approximately ρπr²δy.

The moment of inertia of the element about an axis through C parallel to the x axis (i.e. a diameter) is \frac{1}{4}δmr².

C is the centre of mass of the disc, so by the parallel axes theorem, its moment of inertia about the x axis is

\frac{1}{4}δmr²  +  δmy² = \frac{1}{4}(ρπr²δy)r²  +  (ρπr²δy)y²

= \frac{1}{4}ρπr²(r²  +  4y²)δy.

For the whole cylinder, the moment of inertia about the x axis, I_{x}, is approximately

\sum{\frac{1}{4}ρπr²(r²  +  4y²)δy = \frac{1}{4}ρπr²} \sum{(r²  +  4y²)δy.}

In the limit as δy → 0, this gives

I_{x} = \frac{1}{4}ρπr² \int_{0}^{h}{(r²  +  4y²)dy}

= \frac{1}{4}ρπr²\left[r²y  +  4\frac{y^{3}}{3}\right]^{h}_{0}

= \frac{1}{4}ρπr^{4}h  +  \frac{1}{3}ρπr²h^{3}

= \frac{1}{4}Mr^{2}  +  \frac{1}{3}Mh^{2}         (M = ρπr²h).

fig 20.24