Rotating disk dropped onto another rotating disk

[robertmatthew]

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: L_i=L_f
L= Iω

The Attempt at a Solution

Li=Lf
I(ω_CCW) + Iω_CW) = 2Iω_f
I(26.180 rad/s) + I(-13.090 rads) = 2Iω_f
13.09 = 2ω_f
ω_f = 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.

 

[SammyS]

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: L_i=L_f
L= Iω

The Attempt at a Solution

Li=Lf
I(ω_CCW) + Iω_CW) = 2Iω_f
I(26.180 rad/s) + I(-13.090 rads) = 2Iω_f
13.09 = 2ω_f
ω_f = 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 .

 

[CWatters]

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.

 

[robertmatthew]

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 ω_f of 26.180 rad/s from part a. Tried it with 52.36 rad/s instead and got 19.64 rad/s.

 

[SammyS]

Yeah, I mistakenly substituted the ω_f 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?

 

[CWatters]

Yeah, I mistakenly substituted the ω_f 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.