Question

A drum major twirls a 96 -cm-long, 400g baton about its center of mass at 100 rpm. What is the baton's rotational kinetic energy?

Short answer

The rotational kinetic energy of the baton is$1.68\text{J}$

Step-by-step solution

Step 1. Given information

The baton mass is 400 g.

The moment of inertia of a thin rod rotating about its center is $I=\frac{1}{12}M{L}^2$.

For the baton,

$I=\frac{1}{12}(0.400\text{kg}){(0.96\text{m})}^2=0.031\text{kg}{\text{m}}^2$

Step 2. Calculation

The rotational kinetic energy of the baton is

$\begin{array}{l}{K}_{\text{rot}}=\frac{1}{2}I{\omega }^2\\ =\frac{1}{2}\left(0.031\text{kg}{\text{m}}^2\right){\left((100\text{rev}/\text{min})\left(\frac{2\pi \text{rad}}{\text{rev}}\right)\left(\frac{\text{min}}{60\text{s}}\right)\right)}^2\\ =1.68\text{J}\end{array}$