Conservation of Angular Momentum

AI Thread Summary
The discussion centers on a physics problem involving the conservation of angular momentum when a cylindrical shell is dropped onto a spinning disk. The calculation for the final angular velocity after the shell is added is confirmed to be correct, yielding a speed of 1.5 revolutions per second. However, the method used to arrive at this result is criticized for employing incorrect units, despite ultimately leading to the right answer. In contrast, the assertion that kinetic energy is conserved is deemed incorrect, indicating a misunderstanding of energy conservation in this context. The conversation highlights the importance of proper unit usage and the distinction between angular momentum and kinetic energy conservation.
Jim011991
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Homework Statement


A uniform disk of mass 2kg and radius 10.0cm is spinning on a fixed frictionless bearing at 3 revolutions/second. A think cylindrical shell of mass 2kg and radius 10cm is carefully dropped onto the spinning disk so that they end up spinning together, with their edges coinciding.

a) How fast (in revolutions/second) are they spinning after the drop?
b) Is kinetic energy conserved is this process? If not, by what percentage has it changed?


Homework Equations


m*w*r^2


The Attempt at a Solution


a) 2*3*10^2 = 600
600 = 4*w*10^2
600 = 400w
w=1.5 rev/sec

b) Kinetic energy is conserved in this process.

Did I do this correctly?
 
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Point a)
In the end the result is correct, but it's wrong the way you get it.
You use wrong units, but in the end they cancel out each other, so you get the good result.

Point b)
No, it's wrong.
 
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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