# Homework Help: Sum of Torque vs. Conservation of Angular Momentum! Quick help needed!

1. Dec 2, 2012

### riseofphoenix

Number 11)

This is what I did:

Ʃτ = F2r2 + F1r1 = 0
(195)(7) + F1(0.7) = 0
F1(0.7) = -(195)(7)
F1 = -1365/0.7
F1 = -1950 N
F1 = 1950 N

Number 12)

This is what I did:

Since this is a massless rod and the location of the axis is through the end, I = ML2

Linitial = Lfinal (Conservation of Angular Momentum)
(I1ω1 + I2ω2)initial = (I1ω1 + I2ω2)final
[ M1L12ω1 ] + [ M2L22ω2 ] = [ M1L12ω1 ] + [ M2L22ω2 ]
[ (14 kg)(4)2(6) ] + [ (2 kg)(10)2(6) ] = [(14 kg)(10)22) ] + [ (2 kg)(4)22) ]
1344 + 1200 = 1400(ω2) + 32(ω2)
2544 = 1432(ω2)
2544/1432= ω2
1.776 = ω2

Is that right?

Number 13)

I think this one is right. I googled it...

Number 14)

fobserver = fsource [ (v + vobserver)/(v + vsource) ]
fobserver = 1550 [ (340 + 9 m/s)/(340 + 24 m/s) ]
fobserver = 1550(349/364)
fobserver = 1550(0.958791209)
fobserver = 1486 Hz

Is that right?

Last edited: Dec 2, 2012
2. Dec 2, 2012

### SammyS

Staff Emeritus
You're familiar enough with this site, that you should know by now that you shouldn't be posting 4 questions in one thread.

Your solution to problem appears to be correct.

3. Dec 2, 2012

### SammyS

Staff Emeritus
I suspect that the answer is correct, but you used the wrong moments of inertia throughout.

Why do you have the 1/3 in all of them (moments of inertia) ? These are mass-less rods.

4. Dec 2, 2012

### riseofphoenix

Wait but...what should they be then?
I've been looking online for "moment of inertia for massless rod" and I can't find anything other than (1/3)ML2

So it would just be ML2?

Last edited: Dec 2, 2012
5. Dec 2, 2012

### riseofphoenix

And sorry about posting 4 questions all in one post! I just needed someone to verify my work real quick

Is it ok if I ask you to check one very last thing for me? I promise it'll be quick

6. Dec 2, 2012

### SammyS

Staff Emeritus
What is the mass of a massless rod?

The only objects with mass in problem 12 are the two weights.