Question

A rigid body of mass $2$ kg can be hinged about any point on the x-axis and the axis of rotation is always along y-axis. When it is hinged such that the hinge is at x, the moment of inertia is given by $I=(2x^2-12x+27)kg{m}^2$. What is the x-coordinate (in m) of centre of mass?

Answer 1

Uing paralell axis theorm, $I_y=I_{CM}+M{x}_1^2$ where $x_1$ is the co-ordinate of CM.

Now, as $I=2x^2-12x+27kg{m}^2$

For $I_y=2\left(0^2\right)-12\left(0\right)+27=27kg{m}^2$

$I_{CM}=2x_1^2-12x_1+27kg{m}^2$

Thus, $27=2x_1^2-12x_1+27+2x_1^2⟹x_1=3m$