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## Q. 20.8

Use the parallel axes theorem to find the moments of inertia of
i) a thin uniform rod of mass M and length h about a perpendicular axis through its end
ii) a thin uniform solid disc of mass M and radius r about
a) an axis perpendicular to its plane through a point on its circumference
b) a tangent.
iii) Which of the above is equally applicable to a solid cylinder?

## Verified Solution

i) For the rod $I_{G} = \frac{1}{3}M\left(\frac{h}{2}\right)^{2}$ and the axes are a distance $\frac{h}{2}$ apart. Hence

$I_{A} = I_{G} + M\left(\frac{h}{2}\right)^{2}$

$= \frac{1}{3}M\left(\frac{h}{2}\right)^{2} + M\left(\frac{h}{2}\right)^{2}$

$= \frac{1}{3}Mh^{2}.$

ii)   a)   In this case $I_{G} = \frac{1}{2}Mr^{2}$ and the axes

are a distance r  apart.

Hence          $I_{A} = I_{G} + Mr^{2}$

$= \frac{1}{2}Mr^{2} + Mr$

$= \frac{3}{2}Mr^{2}.$

b)   Now  $I_{G} = \frac{1}{4}Mr^{2}$

⇒       $I_{A} = \frac{1}{2}Mr^{2} + Mr$

$= \frac{5}{4}Mr^{2}.$

iii) The result is applicable to a cylinder so long as this is formed by elongating the body in a direction parallel to the axis of rotation. It therefore applies in part ii) a) but not in the other cases.