Question

A $300g$ball and a $600g$ball are connected by a $40cm$-long massless, rigid rod. The structure rotates about its center of mass at $100rpm$. What is its rotational kinetic energy?

Short answer

Therefore the rotational kinetic energy of the object is$1.65\mathrm{J}$.

Step-by-step solution

Step : 1 Introduction 

The length of the rod is $40cm$convert unit of the length cm to m

$I=\left(40\mathrm{cm}\right)\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)$

$0.4m$

The mass of the ball is $300g$convert unit mass from g to kg

$M_1=\left(300\mathrm{g}\right)\left(\frac{1\mathrm{kg}}{1000\mathrm{g}}\right)$

$0.3kg$

The mass of bass $2is600g$convert units os mass from g to kg

$M_2=600\mathrm{g}\cdot \frac{1\mathrm{kg}}{1000\mathrm{g}}$

$0.6kg$

The angular velocity of rigid rod is $100rpm$. convert units of angular velocity from rpm to rads.

$\begin{array}{r}\omega =100\frac{\mathrm{rev}}{\mathrm{m}}\cdot \frac{2\pi \mathrm{rad}}{1\mathrm{rev}}\cdot \frac{1\mathrm{m}}{60\mathrm{s}}\\ =10.5\frac{\mathrm{rad}}{\mathrm{s}}\end{array}$

Step :2 Length of the rod 

Setting the point with the $300g$ball as $x=0$with the positive x axis along the length of the rod.

$\begin{array}{r}{x}_{en}=\frac{1}{M}\sum _i m_i{x}_i\\ =\frac{M_1{x}_1+M_2{x}_2}{M_1+M_2}\end{array}$

Subistitute $0.3\mathrm{kg}\text{for}{M}_1,0.6\mathrm{kg}\text{for}{M}_2,0\text{for}{x}_1\text{and}0.4\mathrm{m}\text{for}{x}_2$in above equation

$\begin{array}{r}{x}_{\mathrm{em}}=\frac{\left(0.3\mathrm{kg}\right)\left(0\mathrm{m}\right)+\left(0.6\mathrm{kg}\right)\left(0.4\mathrm{m}\right)}{\left(0.3\mathrm{kg}\right)+\left(0.6\mathrm{kg}\right)}\\ \end{array}$

$0.267m$

This gives the distance between ball $1$and the center of the mass$r_1=0.267\mathrm{m}$.It also gives distance between the center of mass and ball$2,r_2=0.4\mathrm{m}-0.267\mathrm{m}=0.133\mathrm{m}$

Step :3 Momentum of inertia 

The momentum of inertia about the center of mass

$\begin{array}{r}{I}_{\mathrm{ca}}=\sum _i m_i{r}_1^2\\ =M_1{r}_1^2+M_2{r}_2^2\end{array}$

Substitute $0.3\mathrm{kg}\text{for}{M}_1,0.6\mathrm{kg}\text{for}{M}_2,0.267\text{for}{r}_1\text{and}0.133\text{for}{r}_2$

$\begin{array}{r}{I}_{\mathrm{cm}}=\left(0.3\mathrm{kg}\right)(0.267\mathrm{m}{)}^2+(0.6\mathrm{kg}\left)\right(0.133\mathrm{m}{)}^2\\ \end{array}$

$=0.03\mathrm{kg}\cdot {\mathrm{m}}^2$

The rotational kinetic energy of an object is

$K_{nx}=\frac{1}{2}{I}_{cm}{\omega }^2$

Substitute $0.03\mathrm{kg}\cdot {\mathrm{m}}^2\text{for}{I}_{\mathrm{em}}\text{and}10.5\frac{\mathrm{rad}}{\mathrm{s}}\text{for}\omega$in the above equation to solve the rotational kinetic energy of the object.

$\begin{array}{r}{K}_{me}=\frac{1}{2}\left(0.03\mathrm{kg}\cdot {\mathrm{m}}^2\right){\left(10.5\frac{\mathrm{rad}}{\mathrm{s}}\right)}^2\\ \end{array}$

$=1.65j$