A rigid body can be hinged about any point on | vedclass

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Question

Uing paralell axis theorm, $I_{y}=I_{C M}+M x_{1}^{2} \quad$ where $x_{1}$ is the co-ordinate of $\mathrm{CM}$.

Now, as $I=2 x^{2}-12 x+27 \quad k

Vedclass · Accepted Answer

$x = 3$

Vedclass · Answer

Uing paralell axis theorm, $I_{y}=I_{C M}+M x_{1}^{2} \quad$ where $x_{1}$ is the co-ordinate of $\mathrm{CM}$.

Now, as $I=2 x^{2}-12 x+27 \quad k g m^{2}$

For $I_{y}=2\left(0^{2}\right)-12(0)+27=27 \mathrm{kgm}^{2}$

$I_{C M}=2 x_{1}^{2}-12 x_{1}+27 k g m^{2}$

Thus, $27=2 x_{1}^{2}-12 x_{1}+27+2 x_{1}^{2} \quad \Longrightarrow \quad x_{1}=3 m$

A rigid body can be hinged about any point on the $x$ -axis. When it is hinged such that the hinge is at $x$, the moment of inertia is given by $I = 2x^2 - 12x + 27$ The $x$ -coordinate of centre of mass is

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