Angular speed of rod after projectile collides into it

[StrawHat]

[SOVLED] Angular speed of rod after projectile collides into it

Homework Statement

Homework Equations

L = r x p
I_parallel = I_CM + md²

The Attempt at a Solution

L = r x p = mv_ir = mv_id/2
mv_i(d/2) = (I_parallel + md²)ω
mv_i(d/2) = (1/12M(d/2)^2 + m(d/2)^2)ω

Plugging and solving eventually leads me to...
ω = (mv_i)/(1/24Md+1/2md)
...Which is apparently incorrect.

 

[Ibix]

What's the moment of inertia of a thin rod about its center?

 

[StrawHat]

1/12mr²

 

[Ibix]

And what should r be in this problem? Hint: I don't think it's d/2.

As a matter of interest, how good is the submission system at algebra? Will it understand that a/(1/24b+1/2c)=24a/(b+12c), or will it only recognise expressions that your prof entered? The only one I've ever used could only do exact string comparison, so if the answer was 0.3 and you typed .3, that was marked wrong...

 

[StrawHat]

Why wouldn't it be d/2? If the diameter of the circular motion is d, then wouldn't the radius be d/2?

 

[Ibix]

You have the correct functional form for the second moment of inertia of the rod, but are misinterpreting the thing you called r. It's the length of the rod (d, in this case). You can easily look it up, or derive it from first principles by integrating \mu r^2dr (where \mu is the mass per unit length of the rod) over the range -d/2 to d/2.

 

[StrawHat]

Oh okay, I thought r was supposed to be the distance between the rotation point and the end of the rod.

I got the right answer of ω = (mvi)/(1/6Md+1/2md)

Thanks for your help!

 

[StrawHat]

Okay, I've tried part B of the question and this is what I've done so far.

K_final = 1/2Iω²
K_initial = 1/2mv²

I divided both of those to get (Iω²)/(mv_initial)

Calculating that further, and I get this:

I'm guessing that it's not K_final/K_initial, but (K_initial - K_final)/K_final?

EDIT: Okay, I got the answer. Thank you all!

 

[TSny]

I'm guessing that it's not K_final/K_initial, but (K_initial - K_final)/K_final?

Almost. After subtracting, divide by K_initial rather than K_final. (It's like one of those percent decrease problems in algebra.)