Rotating disk dropped onto another rotating disk

AI Thread Summary
The discussion focuses on a physics problem involving two rotating wheels and the application of conservation of angular momentum. The first wheel rotates at 500 rpm counterclockwise, while the second wheel, dropped onto the shaft, rotates at 125 rpm clockwise. The correct approach involves calculating the initial and final angular momentum, ensuring proper conversion between rpm and radians per second. Mistakes were noted in the initial calculations, particularly in the conversion of angular velocities and the application of the conservation principle. The final angular velocity of the combined system should be expressed in rpm along with its direction.
robertmatthew
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Homework Statement


A wheel, mounted on a vertical shaft of negligible rotational inertia, is rotating at 500 rpm (CCW from above).
Part a) asks to find the new angular velocity if an identical wheel is dropped onto the shaft. I got this part right.
Part b) is: Now suppose the dropped wheel starts with an angular velocity of 125 rpm is the opposite direction of the first wheel (CW from above). Determine the angular velocity for the resultant combination of the two wheels.


Homework Equations


Conservation of angular momentum: Li=Lf
L= Iω


The Attempt at a Solution


Li=Lf
I(ωCCW) + IωCW) = 2Iωf
I(26.180 rad/s) + I(-13.090 rads) = 2Iωf
13.09 = 2ωf
ωf = 6.545

Is that right? On my initial attempt (these are test corrections), I set the initial momentum of the first wheel equal to the sum of the final momentum and the initial momentum of wheel being dropped. My teacher wrote "same (stuck together)" on the paper, so I thought this might've been what he meant.
 
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robertmatthew said:

Homework Statement


A wheel, mounted on a vertical shaft of negligible rotational inertia, is rotating at 500 rpm (CCW from above).
Part a) asks to find the new angular velocity if an identical wheel is dropped onto the shaft. I got this part right.
Part b) is: Now suppose the dropped wheel starts with an angular velocity of 125 rpm is the opposite direction of the first wheel (CW from above). Determine the angular velocity for the resultant combination of the two wheels.

Homework Equations


Conservation of angular momentum: Li=Lf
L= Iω

The Attempt at a Solution


Li=Lf
I(ωCCW) + IωCW) = 2Iωf
I(26.180 rad/s) + I(-13.090 rads) = 2Iωf
13.09 = 2ωf
ωf = 6.545

Is that right? On my initial attempt (these are test corrections), I set the initial momentum of the first wheel equal to the sum of the final momentum and the initial momentum of wheel being dropped. My teacher wrote "same (stuck together)" on the paper, so I thought this might've been what he meant.
500 rpm is 4 times 125 rpm .

26.180 rad/s is only 2 times 13.09 rads/s .

You made a mistake converting one of these to rads/s .
 
I set the initial momentum of the first wheel equal to the sum of the final momentum and the initial momentum of wheel being dropped.

I can see how you got that but its not good practice to write it that way. If you read it literally it's actually wrong. Conservation of momentum would say...

Initial momentum of first wheel + Initial momentum of dropped = Final momentum of both.

If you rearrange that you get...

Initial momentum of first wheel = Final momentum of both - Initial momentum of dropped wheel

..which is not what you wrote. There is a minus sign on the right. That should only become a +ve when you substitute the actual data values for this problem.
 
SammyS said:
500 rpm is 4 times 125 rpm .

26.180 rad/s is only 2 times 13.09 rads/s .

You made a mistake converting one of these to rads/s .

Yeah, I mistakenly substituted the ωf of 26.180 rad/s from part a. Tried it with 52.36 rad/s instead and got 19.64 rad/s.
 
robertmatthew said:
Yeah, I mistakenly substituted the ωf of 26.180 rad/s from part a. Tried it with 52.36 rad/s instead and got 19.64 rad/s.
What is it ?

You used 52.36 rad/s for which wheel?

What did you use for the other wheel?
 
robertmatthew said:
Yeah, I mistakenly substituted the ωf of 26.180 rad/s from part a. Tried it with 52.36 rad/s instead and got 19.64 rad/s.

The question used rpm rather than rads/s so might want to give the answer in the same form and mention the direction.
 
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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