Question: In Figure, three particles of mass m = 23 gm are fastened to three rods of length d = 12 cmand negligible mass. The rigid assembly rotates around point Oat the angular speed v = 0.85 rad/s. About O, (a) What are the rotational inertia of the assembly? (b) What are the magnitude of the angular momentum of the middle particle? (c) What are the magnitude of the angular momentum of the assembly?

1. The mass of the particle is$m=23g=23\times {10}^{-3}kg$
2. The length of the rod is$d=12cm=12\times {10}^{-2}m$
3. The magnitude of angular momentum of the assembly is$\omega =0.85rad/s$

Use the expression of angular momentum in terms of rotational inertia and angular velocity.

Formula:

$\mathit{L}\mathbf{=}\mathit{I}\mathit{\omega }$

(a)

The rigid assembly is rotating about the origin. Hence each particle is rotating about the same origin as shown inthegiven figure.

The moment of inertia of the rotating particle about the origin O is the product of mass of the particle and square of distance r from the origin O.

I = mr²

Use this relation for each particle and find the rotational inertia of the assembly.

$$\begin{gathered} I=m{\left(d\right)}^2+m{\left(2d\right)}^2+m{\left(3d\right)}^2 \\ \Rightarrow I=14m{\left(d\right)}^2 \\ \Rightarrow I=14\times 23\times {10}^{-3}kg\times {\left(12\times {10}^{-2}m\right)}^2 \\ \Rightarrow I=4.6\times {10}^{-3}kg{m}^2 \end{gathered}$$

(b)

The expression of the angular momentum of middle particle is the relation between its rotational inertia and its distance 2 d² from the origin O

$$\begin{gathered} L_m=I_m\omega \\ \Rightarrow L_m=m\left(2d^2\right)\omega \\ \Rightarrow L_m=23\times {10}^{-3}kg\times {\left(2\times 12\times {10}^{-2}m\right)}^2\times 0.85rad/s \\ \Rightarrow L_m=1.1\times {10}^{-3}kg.\frac{m^2}{s} \end{gathered}$$

(c)

The expression for the angular momentum of the assembly is

$$\begin{gathered} L=I\omega \\ \Rightarrow L=4.6\times {10}^{-3}kg{m}^2\times 0.85rad/s \\ \Rightarrow L=3.9\times {10}^{-3}kg\frac{m^2}{s} \end{gathered}$$