A cheerleader is practicing her baton twirling. The baton has two .6kg masses

[Lo.Lee.Ta.]

Hi, everyone! Thank you for using your time to try to help me! You are kind! :)

1. A cheerleader is practicing her baton twirling. The baton has two .6kg masses attached to opposite ends of a massless rod 70cm long. If the baton is twirled about a point 30cm from one end at 1.6 revolutions per second, what is the baton's kinetic energy?

2. KE(total) = KE(1) + KE(2)
v=ωr
# of revolutions = # of radians / 2pi
I for a baton with a massless rod in between: I = mr^2

3. Okay, so I don't know if I'm doing this right...

*Changing revolutions/s to radians/s to know what the angular velocity (ω) is:
1.6 revolutions/s x #rad/2pi = 10.05rad/s

So now we know the angular velocity (ω) = 10.05rad/s

*KE(total) = KE(1) + KE(2)
KE(total) = 1/2(I)(ω^2) + 1/2(I)(ω^2)
KE(total) = 1/2(.6)(.30^2)(10.05^2) + 1/2(.6)(.40^2)(10.05^2)
KE(total) = 7.58J

...I have no idea if that's right... I'm not sure if I use the same thing for the angular velocity... I think both .6kg masses move at the same speed... not sure though.

Thank you for helping me to figure this out! :)

 

[haruspex]

KE(total) = 1/2(.6)(.30^2)(10.05^2) + 1/2(.6)(.40^2)(10.05^2)

looks right to me.

 

[Lo.Lee.Ta.]

Oh, really? Yay! Thanks, Haruspex! :)