Lost Kinetic Energy in Inelastic Collision of Putty and Pivoting Rod

[kingsmaug]

Homework Statement

A piece of putty of mass m = 0.75 kg and velocity v = 2.5 m/s moves on a horizontal frictionless surface. It collides with and sticks to a rod of mass M = 2 kg and length L = 0.9 m which pivots about a fixed vertical axis at the opposite end of the rod as shown. What fraction of the initial kinetic energy of the putty is lost in this collision?

Homework Equations

KE = 1/2mv^2
KE = 1/2Iw^2
L=mvr
L=Iw
I=mL^2/3
I=mr^2

r=L (I'm using the pivot as the point of origin)

The Attempt at a Solution

Based on the wording, it's an inelastic collision, and the putty sticks to the rod.

So, momentum is conserved:

mvL = (mL^2/3+mL^2)w

Using numbers, I found that w = 1.176 rad/s

The initial KE is 1/2mv^2= 2.34

The final KE is 1/2Iw^2= 0.79

KElost/KEi=(KEi-KEf)/KEi = 0.66

but the answer is wrong.

 

[AlephNumbers]

Did the problem statement give any information regarding where the putty sticks to the rod? I see you assumed that it stuck to the end of the rod. If you guessed "wrong", that could be your source of error. Under this assumption I am also calculating a different value for the angular velocity post-collision.

 

[haruspex]

I found that w = 1.176 rad/s

If the mass is striking the end of the rod, that's too low. Are you confusing yourself by using 'm' for both masses in the equation?

 

[kingsmaug]

I don't believe I'm using the wrong ones.

So working it over using the proper masses:

Homework Equations

KE = 1/2mv^2
KE = 1/2Iw^2
L=mvr
L=Iw
I=ML^2/3
I=mr^2

r=L (I'm using the pivot as the point of origin)

mvL=(ML^2/3+mL^2)w

w=mvL/(ML^2/3+mL^2) = 1.47059 rad/sec (I don't understand why it isn't the same, but whatever...)

KEi = 1/2mv^2 = 2.34375 J

KEf = 1/2(ML^2/3+mL^2)w^2 = 1.24081 J

(KEi-KEf)/KEi = 0.470588 Which is correct.

And looking through my calculator's log, I found exactly where I went wrong the first time. I accidentally used 2 as the velocity instead of 2.5. :| The stupid feelings.