Question 427: Centre of Mass and Rotational Motion

Question

Moment of inertia of a disc of mass '$$M$$' and radius '$$R$$' about any of its diameter is $$\frac{M R^{2}}{4}$$. The moment of inertia of this disc about an axis normal to the disc and passing through a point on its edge will be, $$\frac{x}{2}$$ MR$$^{2}$$. The value of $$x$$ is ___________.

Accepted Answer

$\begin{aligned} & \mathrm{I}=\mathrm{I}_{\mathrm{cm}}+\mathrm{Md}^2 \\ & =\frac{\mathrm{MR}^2}{2}+\mathrm{MR}^2 \\ & =\frac{3}{2} \mathrm{MR}^2 \\ & \mathrm{x}=3\end{aligned}$

Centre of Mass and Rotational Motion

Moment of inertia of a disc of mass '$M' and radius 'R' about any of its diameter is \frac{M R^{2}}{4}. The moment of inertia of this disc about an axis normal to the disc and passing through a point on its edge will be, \frac{x}{2} MR^{2}. The value of x$ is ___________.

Choose the correct answer: