Question

A low-mass rod of length $0.30m$has a metal ball of mass $1.4kg$at each end. The center of the rod is located at the origin, and the rod rotates in the $yz$plane about its center. The rod rotates clockwise around its axis when viewed form a point on the $+x$axis, looking towards the origin. The rod makes one complete rotation every$0.5s$ (a) what is the moment of inertia of the object (rod plus two balls)? (b) what is the rotational angular momentum of the object? (c) What is the rotational kinetic energy of the object?

Short answer

The moment of inertia of the object is-$0.0765kg·m^2$

The rotational angular momentum of the object is-$0.961kg·m^2/s$

The rotational kinetic energy of the object is-$6.043J$

Step-by-step solution

Definition of Inertia and Angular momentum.

The moment of inertia is a quantitative measure of a body's rotational inertia, or its resistance to having its speed of rotation about an axis changed by the application of a torque.

The rotating analogue of linear momentum is angular momentum (also known as moment of momentum or rotational momentum). Because it is a conserved quantity—the total angular momentum of a closed system remains constant—it is a significant quantity in physics.

The given data is-

Length of a low-mass rod,$L=0.30m$

Mass of each metal ball,$m=1.7kg$

Period of rotation,$T=0.5s$

Angular speed of the rod, $\omega =1rev/0.5s$

$=\left(1rev/0.5s\right)\left(\frac{2\pi rad}{1rev}\right)$

$=12.57rad/s$

Find the moment of inertia, rotational angular momentum and rotational kinetic energy of the object.

(a) The moment of inertia of an object of mass $m$rotating about an axis at a distance $L$from the axis is by$I=M{L}^2....\left(1\right)$

Because the rod has a low mass, the contribution of the rod's moment of inertia to the overall moment of inertia of the object can be ignored. There are two metal balls on each end of the rod.

its end at a distance $\frac{L}{2}$from the center of the rod. So, the moment of inertia of the object is sum of the moment of inertias of each object about the center of rod. Hence, the moment of inertia of the object is

$$\begin{gathered} I_{total}=I_{ball}+I_{ball} \\ =m{\left(\frac{L}{2}\right)}^2+m{\left(\frac{L}{2}\right)}^2 \\ =2m\left(\frac{L^2}{4}\right) \end{gathered}$$

$I_{total}=\frac{1}{2}m{L}^2....\left(2\right)$

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

$$\begin{gathered} I_{total}=\frac{1}{2}\left(1.7kg\right){\left(0.30m\right)}^2 \\ =0.0765kg·m^2 \end{gathered}$$

(b) 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 can be calculated as

$L_{rot}=\left(0.0765kg·m^2\right)\left(12.57rad/s\right)$

$=0.961kg·m^2/s$

(c) The rotational kinetic energy of an object rotating with angular speed$\left(\omega \right)$is

$K_{rot}=\frac{1}{2}I{\omega }^2......\left(4\right)$

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

On substituting the numerical values in the equation (4) , the rotational kinetic energy of the disk can be calculated as

$K_{rot}=\frac{1}{2}\left(0.0765kg·m^2\right){\left(12.57rad/s\right)}^2$

$=6.043J$