Question

Mounted on a low-mass rod of length $0.3m$are four balls. Two balls (shown in red on the diagram), each of mass$0.82kg$, are mounted at opposite ends of the rod. Two other balls, each of the mass$0.29kg$(shown in blue on the diagram), are each mounted a distance $0.08m$from the center of the rod. The rod rotates on the axle through the center of the rod (indicated by the "X" in the diagram), perpendicular to the rod, and it takes $0.9s$to make full rotation.

(a) What is the moment of inertia of the device about its center?

(b) What is the angular speed of the rotating device?

(c) What is the magnitude of the angular momentum of the rotating device?

Short answer

The moment of inertia of the device is$0.0457kg·m^2$.

The angular speed of the rotating device is $6.98rad/s$.

The magnitude of the angular momentum of the rotating device is $0.3189kg·m^2/s$.

Step-by-step solution

Definition of Angular Momentum.

The rotating analogue of linear momentum is angular momentum (rarely referred to as moment of momentum or rotational momentum). Because the total angular momentum of a closed system remains constant, it is an essential number in physics.

The give data is-

Length of a low-mass rod,$L=0.32m$

Mass of each blue ball,$m_{blue}=0.29kg$

Mass of each red ball,$m_{red}=0.82kg$

Distance of each blue ball from the center of the rod,$L_{blue}=0.08m$

Distance of each red ball from the center of the rod, $L_{red}=\frac{L}{2}$

$=\frac{0.32m}{2}$

$=0.16m$

Period of rotation,$T=0.9s$

Figure shows a low mass rod consisting of four balls.

The figure shows a low-mass rod consisting of four balls

Find the moment of inertia, angular speed and angular momentum of the rotating device.

(a) The moment of inertia of an object of mass $m$rotating about an axis at a distance $L$from the axis of rotation is given by

$I=M{L}^2.....\left(1\right)$

The rod is a low-mass rod, so we can neglect the contribution of the rod’s moment of inertia to the total moment of inertia of the object. The rod has two red balls each at its end at a distance $\frac{L}{2}$from the center of the rod, and two balls each at a distance$0.08m$from the center of the rod. So, the moment of inertia of the object is sum of the moment of inertias of each ball about the center of rod. Hence, the moment of inertia of the device is

$$\begin{gathered} I_{total}=2I_{red}+2I_{blue} \\ {I}_{total}=2m{L^2}_{red}+2m_{blue}{L}_{blue}^2......\left(2\right) \end{gathered}$$

On substituting the known values in the equation (2), the moment of inertia of the device can be calculated as

$I_{total}=2m{L}_{red}+2m_{blue}{L}_{blue}^2$

$=2\left(0.82kg\right){\left(0.16m\right)}^2+2\left(0.29kg\right){\left(0.08m\right)}^2$

$=0.0457kg·m^2$

(b) Angular speed of the rod, $\omega =\frac{2\pi rad}{T}$

$$\begin{gathered} =\frac{2\pi rad}{0.9s} \\ =6.98rad/s \end{gathered}$$

(c)The rotational angular momentum of an object rotating with angular speed$\left(\omega \right)$in terms of its moment of inertia is

$L_{rot}=I\omega ....\left(3\right)$

Here,$I\to$moment of inertia of the object.

On substituting the given values in the equation (3), the magnitude of the rotational angular momentum of the rotating device can be calculated as

$L_{rot}=I_{total}\omega$

$=\left(0.0457kg·m^2\right)\left(6.98rad/s\right)$

$=0.3189kg·m^2/s$