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

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A cheerleader is practicing baton twirling with a baton that has two 0.6 kg masses on a 70 cm rod, twirling at 1.6 revolutions per second. The angular velocity was calculated to be 10.05 rad/s after converting revolutions to radians. The total kinetic energy was computed using the formula KE(total) = 1/2(I)(ω^2) for both masses, resulting in a total of 7.58 Joules. There was some uncertainty about the calculation method and whether the angular velocity was applied correctly, but the final formula used for kinetic energy was confirmed to be accurate. The discussion highlights the importance of understanding rotational dynamics in calculating kinetic energy.
Lo.Lee.Ta.
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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. :frown:

Thank you for helping me to figure this out! :)
 
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Lo.Lee.Ta. said:
KE(total) = 1/2(.6)(.30^2)(10.05^2) + 1/2(.6)(.40^2)(10.05^2)
looks right to me.
 
Oh, really? Yay! Thanks, Haruspex! :)
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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