Sum of Torque vs. Conservation of Angular Momentum Quick help needed

[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

Is that answer right?

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

L_initial = L_final (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)²(ω₂) ] + [ (2 kg)(4)²(ω₂) ]
1344 + 1200 = 1400(ω₂) + 32(ω₂)
2544 = 1432(ω₂)
2544/1432= ω₂
1.776 = ω₂
1.78 rad/s = ω₂

Is that right?

Number 13)

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

Number 14)

f_observer = f_source [ (v + v_observer)/(v + v_source) ]
f_observer = 1550 [ (340 + 9 m/s)/(340 + 24 m/s) ]
f_observer = 1550(349/364)
f_observer = 1550(0.958791209)
f_observer = 1486 Hz

Is that right?

 

[SammyS]

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

Is that answer right?

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.

 

[SammyS]

Number 12)

This is what I did:

Since this is a long uniform rod and the location of the axis is through the end, I = (1/3)ML²

L_initial = L_final (Conservation of Angular Momentum)
(I₁ω₁ + I₂ω₂)_initial = (I₁ω₁ + I₂ω₂)_final
[ (1/3)M₁L₁²ω₁ ] + [ (1/3)M₂L₂²ω₂ ] = [ (1/3)M₁L₁²ω₁ ] + [ (1/3)M₂L₂²ω₂ ]
[ (1/3)(14 kg)(4)²(6) ] + [ (1/3)(2 kg)(10)²(6) ] = [ (1/3)(14 kg)(10)²(ω₂) ] + [ (1/3)(2 kg)(4)²(ω₂) ]
448 + 400 = 466.67(ω₂) + 10.67(ω₂)
848 = 477.34(ω₂)
848/477.34 = ω₂
1.776 = ω₂
1.78 rad/s = ω₂

Is that right?

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.

 

[riseofphoenix]

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.

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?

 

[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

 

[SammyS]

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?

What is the mass of a massless rod?

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