91Ó°ÊÓ

i) Length of rods, $L_1=0.40\text{ }\text{m}$ and $L_2=0.50\text{m}$

ii) Mass of each rod, $M=0.20\text{ k²µ}$

Using the perpendicular axis theorem for rotational inertia about an axis of rotation for a rod, find the rotational inertia of the rigid body about an axis perpendicular to the plane and intersects the shorter rod at its center. According to the perpendicular axis theorem, the moment of inertia of a body about an axis perpendicular to the plane is the sum of the moment of inertia about the two perpendicular axes along the plane and that they are intersecting with the axis perpendicular to the plane. The moment of inertia of longer rod

$\left({\mathbf{I}}_l\right)\mathbf{=}\frac{1}{3}{\mathbf{ML}}^2$

Inertia of shorter rod

$\mathbf{\left(}{\mathbf{I}}_s\mathbf{\right)}\mathbf{=}\frac{1}{12}{\mathbf{ML}}^2$

The moment of inertial about the axis passing through the center of the shorter rod and perpendicular to the plane of paper is-

$\mathbf{I}\mathbf{=}{\mathbf{I}}_l\mathbf{+}{\mathbf{I}}_s$

where, I is the moment of inertia, M is mass, L is length and d is distance.

For rotational inertia of the rigid body about an axis perpendicular to the plane and passing through the center of the shorter rod,

$I=I_s+I_l$

$I=\frac{1}{12}{m}_1{L}_1^2+\frac{1}{3}{m}_2{L}_2^2$

$I=\frac{1}{12}(0.20\text{ }\text{kg}){(0.4\text{″¾})}^2+\frac{1}{3}\left(0.20\text{ k²µ}\right){\left(0.50\text{″¾}\right)}^2$

$I=0.019\text{ }\text{kg}.{\text{m}}^2$

Hence, the rotational inertia of the rigid body about an axis perpendicular to the and that passes through the center of the shorter rod is$0.019\text{ k²µ}.{\text{m}}^2$.

Rotational inertia of the rigid body about an axis perpendicular to the plane of the paper and passes through the center of the longer rod,

$I=I_s+I_l$

$I=\frac{1}{12}{m}_2{L}_2^2+[\frac{1}{12}{m}_1{L}_1^2+m_2{\left(\frac{L_2}{2}\right)}^2]$

$I=\frac{1}{12}\left(0.20kg\right){\left(0.50m\right)}^2+[\frac{1}{12}\left(0.20\text{ k²µ}\right){\left(0.40\text{″¾}\right)}^2+\left(0.20\text{ k²µ}\right){\left(\frac{0.50\text{″¾}}{2}\right)}^2]$

$I=0.019\text{ }\text{kg}.{\text{m}}^2$

Hence, the rotational inertia of the rigid body about an axis perpendicular to the plane of the paper and passes through the center of the longer rod is$0.019\text{ }\text{kg}.{\text{m}}^2$.