Question 607: Centre of Mass and Rotational Motion

Question

Three identical rods each of mass ' $M$ ' and length ' $L$ ' are joined to form a symbol ' $H$. The moment of inertia of the system about one of the sides of ' $H$ ' is

Accepted Answer

The given situation can be shown as
Let us take the moment of inertia of the system about rod $R_1$ then total moment of inertia is
$$I_T=I_1+I_2+I_3\quad	ext{.... (i)}$$
For $\operatorname{rod} R_1, I_1=0$
For rod $R_2$, using perpendicular axis theorem,
$$I_2=\frac{M L^2}{3}$$
For rod $R_3$, using parallel axis theorem,
$$I_3=I_{C M}+I_{(a t L)}=0+M L^2=M L^2$$
Now, putting the values of $I_1, I_2$ and $I_3$ in Eq. (i), we get
$$I_T=0+\frac{M L^2}{3}+M L^2 \Rightarrow I_T=\frac{4 M L^2}{3}$$

Answer

$\frac{2 M L^2}{3}$

Answer

$\frac{M L^2}{2}$

Answer

$\frac{M L^2}{6}$

Answer

$\frac{4 M L^2}{3}$

Centre of Mass and Rotational Motion

Three identical rods each of mass ' M ' and length ' L ' are joined to form a symbol ' H. The moment of inertia of the system about one of the sides of ' H ' is

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