- #1
KaiserBrandon
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Homework Statement
A stick of mass m and length L is pivoted at an end. It is held horizontal and then released. It swings down, and when it is vertical the free end elastically collides with a ball. (assume that the ball is initially held at rest and then released a split second before the stick strikes it.) If the stick loses half of its angular velocity during the collision, what is the mass of the ball? What is the speed of the ball right after the collision?
Homework Equations
moment of inertia for the stick: [tex]I=\frac{1}{3}*mL^{2}[/tex].
[tex]L=I\omega[/tex].
[tex]\omega =\frac{v}{r}[/tex]
[tex]K_{rotational}=\frac{1}{2}I\omega^{2}[/tex]
The Attempt at a Solution
Alright so first I found the angular velocity the instant before it hit the ball by using conservation of energy:
[tex]mgL = \frac{1}{2}(\frac{1}{3}mL^{2})\omega^{2}[/tex]
and found that [tex]\omega=\sqrt{6g/L}[/tex].
then I used E-conservation for before and after the collision:
[tex]\frac{1}{2}I\omega^{2}=\frac{1}{2}I(\omega/2)^{2}+\frac{1}{2}m_{ball}v_{ball}^{2}[/tex]
and substituting I and [tex]\omega[/tex], I reduced this for the mass and velocity of the ball:
[tex]m_{ball}v_{ball}^{2}=\frac{3}{2}mLg[/tex]
Then I used linear momentum conservation here. My reasoning here is that the momentum of the centre of mass of the stick before is equal to the momentum of the centre of mass of the stick afterwards plus the momentum of the ball afterwards.
so before
[tex]v_{cm}=\frac{L}{2}\omega[/tex]
and just divide this by 2 for the velocity after the collision. So using
[tex]mv=mv/2+m_{ball}v_{ball}[/tex]
I found an expression for mv of the ball, which is
[tex]\frac{m}{2}\sqrt{3/2gL}=m_{b}v_{b}[/tex]
isolating v, and substituting it into the [tex]m_{b}v_{b}^{2}[/tex] expression, I got
[tex]m_{b}=\frac{1}{4}m[/tex]
and then putting this into the previous v expression, I get
[tex]v_{b}=2\sqrt{3/2Lg}[/tex]
So does what I did make sense? The only thing I can think of is that by using my method, the ball would have the same end velocity no matter where it struck the stick, as this does not matter in my p-conservation equations. I can't think of a way to use angular momentum, since the ball doesn't really have angular momentum.