Question

A solid rectangular door of uniform thickness is $2m$ wide and has a mass of $60kg$. It is hinged about a vertical axis along one of its longer edges. Assume thickness of the door is small. If the moment of inertia about its hinges of the door is $10nkg{m}^2$, find $n$.

Answer 1

Moment of inertia about the Center of mass of door is $I_{cm}=\frac{M{b}^2}{12}=\frac{60\times 2^2}{12}=20kg{m}^2$

Here distance between cm and vertical axis, $d=1m$

Now using Parallel axis theorem $I=I_{cm}+M{d}^2=20+60\times 1^2=80kg{m}^2$

So, $n=8$