Question

Two thin rods (each of mass$\mathbf{\text{0.20\hspace{0.17em}kg}}$) are joined together to form a rigid body as shown in Fig.$\mathbf{\text{10-60}}$ . One of the rods has length$\mathbf{L}\mathbf{1}\mathbf{\text{=0.40\hspace{0.17em}m}}$ , and the other has length$\mathbf{L}\mathbf{\text{2=0.50m}}$. What is the rotational inertia of this rigid body about (a) an axis that is perpendicular to the plane of the paper and passes through the center of the shorter rod and (b) an axis that is perpendicular to the plane of the paper and passes through the center of the longer rod?

Short answer

a) The rotational inertia of the rigid body about an axis perpendicular to the plane and that passes through the center of the shorter rod is $0.019\text{kg}.{\text{m}}^2$.

b) 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 $0.019\text{kg}.{\text{m}}^2$.

Step-by-step solution

Step 1: Given

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

ii) Mass of each rod, $M=0.20\text{\hspace{0.17em}kg}$

Determining the concept

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.

Calculations(a)

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

$I=I_s+I_l$

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

$I=\frac{1}{12}(0.20\text{kg}){(0.4\text{\hspace{0.17em}m})}^2+\frac{1}{3}\left(0.20\text{\hspace{0.17em}kg}\right){\left(0.50\text{\hspace{0.17em}m}\right)}^2$

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

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$0.019\text{\hspace{0.17em}kg}.{\text{m}}^2$.

(b) Determining 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 

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,

$I=I_s+I_l$

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

$I=\frac{1}{12}\left(0.20kg\right){\left(0.50m\right)}^2+[\frac{1}{12}\left(0.20\text{\hspace{0.17em}kg}\right){\left(0.40\text{\hspace{0.17em}m}\right)}^2+\left(0.20\text{\hspace{0.17em}kg}\right){\left(\frac{0.50\text{\hspace{0.17em}m}}{2}\right)}^2]$

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

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