Angular Velocity and Inertia in a Collision

[corey2014]

Homework Statement

A uniform thin rod of length L=2.2 m and mass 5 kg can rotate in a horizontal plane about a vertical axis through its center. The rod is at rest when a 0.2 kg ball of putty, moving in the horizontal plane of the rod, hits and sticks to one end. As viewed from above, the ball's velocity vector makes an angle of θ =70° with the rod. If the ball's speed just before impact is 20 m/s, what is the angular velocity of the rod immediately after the collision?

Homework Equations

KE=.5mv^2
KE=1/12*M*L^2*w^2

The Attempt at a Solution

I thought this was pretty straightforward, unless I have my formulas wrong, but we just plug the numbers in... and M = 5+.2... right? because energy is conserved...

 

[ehild]

Energy is not conserved here. The putty sticks to the rod, it is kind of inelastic collision.

ehild

 

[corey2014]

so then am I supposed to use m1*u1+m2*u2=(m1+m2)v2?

 

[ehild]

No, the momentum does not conserve either, as there is a force (at the axis) during the collision.
The conserving quantity is : angular momentum

ehild

 

[corey2014]

ok so angular momentum where we have L=I(omega), but then the momentum coming in would be 20(.02)? also does I=mr^2/12

 

[ehild]

You need the angular momentum of the putty, which is m v r sin(θ), r is the distance from the centre of the point where the putty strikes the rod.

The moment of inertia of the rod is OK. Bit you need to take into account also the contribution of the putty to the final moment of inertia.

ehild

 

[corey2014]

ok so what I tried to do is I took mvrsin(theta)=I(omega)
then substituting in and solving for omega i get (omega)=mvrsin(theta)/((1/12)MD^2+mr^2)

where m=.2kg
M=5kg
v=20m/s
r=2.2/2=1.1m
D=2.2m
theta=70deg

I thought this was right but I am not getting the correct answer
I get (omega)=1.83rad/sec, but that doesn't work

There could be an error in there program also...

then I tried with r replacing the D value and it still didn't work

 

[ehild]

It must be correct. Do you have the solution in the book?

ehild

 

[corey2014]

no its an online problem... and the first case would be correct yes?

also, thanks for the help

 

[ehild]

Yes, 1.83 rad/s should be correct.

ehild