HRK Energy/Rotational Motion Question

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The discussion revolves around a physics question regarding the simultaneous stopping of a hockey stick's rotation and its center of mass due to friction. Participants express confusion about the relationship between translational and rotational motion, particularly how they influence each other. The role of kinetic versus static friction is considered, with emphasis on the fact that kinetic friction acts until both motions cease. The idea that one end of the stick moves faster than the other during rotation is noted, leading to variations in frictional force. Ultimately, the conclusion is that both motions stop simultaneously due to the nature of friction acting on the stick.
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Homework Statement


A concept question from Halliday, Resnick, Krane:
"A disgruntled hockey player throws a hockey stick along the ice. It rotates about its center of mass as it slides along and is eventually brought to rest by the action of friction. Its motion of rotation stops at the same moment that its center of mass comes to rest, not before and not after. Explain why."

Homework Equations


None. I suppose the answer should be intuitive!

The Attempt at a Solution


I've tried but I have no idea how to go about doing this. It's extremely different from any other question they give...
 
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Would it have anything to do with the fact that the kinetic coefficient of friction is different from the static one?
 
I doubt it- it's always kinetic friction until both rotation and translation stop, right?
 
Yes. I guess it isn't a way one of the motions would affect the other directly.
Another thought - the spinning makes one end go faster, the other slower. At some point on the slower side, the speed with respect to the ice would be zero - higher friction force affecting the rotational motion. It might be worth thinking about how this changes as the translational speed gets smaller.
 
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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