#### Answer

The rotational kinetic energy is 1.7 J

#### Work Step by Step

We can convert the angular velocity to units of rad/s.
$\omega = (100~rpm)(2\pi~rad/rev)(\frac{1~min}{60~s})$
$\omega = 10.5~rad/s$
We can find the rotational inertia of the baton.
$I = \frac{1}{12}ML^2$
$I = \frac{1}{12}(0.40~kg)(0.96~m)^2$
$I = 0.0307~kg~m^2$
We can find the rotational kinetic energy.
$KE_{rot} = \frac{1}{2}I~\omega^2$
$KE_{rot} = \frac{1}{2}(0.0307~kg~m^2)(10.5~rad/s)^2$
$KE_{rot} = 1.7~J$
The rotational kinetic energy is 1.7 J