Question 112: Centre of Mass and Rotational Motion

Question

Moment of inertia of a rod of mass ' M ' and length ' L ' about an axis passing through its center and normal to its length is ' $\alpha$ '. Now the rod is cut into two equal parts and these parts are joined symmetrically to form a cross shape. Moment of inertia of cross about an axis passing through its center and normal to plane containing cross is :

Accepted Answer

$$\alpha=\frac{\mathrm{M} \ell^2}{12}\quad	ext{..... (1)}$$
$$\begin{aligned}
&\begin{aligned}
& \alpha^{\prime}=2\left[\frac{\frac{\mathrm{M}}{2}\left(\frac{\ell}{2}ight)^2}{12}ight] \
& \alpha^{\prime}=\frac{\mathrm{M} \ell^2}{48}=\frac{\alpha}{4}
\end{aligned}\
&	ext { Correct option is (2) }
\end{aligned}$$

Answer

$\alpha / 4$

Answer

$\alpha / 8$

Answer

$\alpha$

Answer

$\alpha / 2$

Centre of Mass and Rotational Motion

Choose the correct answer: