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A 1m long, 2kg stick is nailed to the wall with a single nail, allowing it to pivot and freely rotate at the end. A 1kg ball, with speed 3m/s makes contact with the stick at some distance x (unknown) below the pivot point. The ball collides elastically with the stick, and stops dead after collision.

  • Find the stick's resulting initial angular velocity.
  • Find the distance x. (Hint: conserve L)

I'm having a lot of difficulty with this problem and I don't know what I'm missing. I know that kinetic energy is conserved so $\ \Delta KE=\frac{1}{2}mv^2-\frac{1}{2}I\omega^2=0$ and that $\ I_{end}=\frac{1}{3}Ml^2$ which in this case is $\ \frac{1}{3}*2kg*1m^2=\frac{2}{3}$, but yet i'm still missing something crucial to figure out how to caculate $\omega$ and $x$.

Any help would be greatly appreciated.

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closed as off-topic by ACuriousMind, John Rennie, user10851, JamalS, Kyle Kanos Mar 15 '15 at 15:10

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I know that kinetic energy is conserved so $\ \Delta KE=\frac{1}{2}mv^2-\frac{1}{2}I\omega^2=0$ and that $\ I_{end}=\frac{1}{3}Ml^2$ which in this case is $\ \frac{1}{3}*2kg*1m^2=\frac{2}{3}$

You are looking for $\omega (= y)$

from conservation of $KE = (1*3^2/2)$you know that $\frac{1}{2}mv^2-\frac{1}{2}I\omega^2=0\rightarrow 1*3^2 = y^2 2/3\rightarrow y^2 =3*3^2/2$,

therefore you already know the value of the angular velocity $\omega= y$ enter image description here

Now, if you know the rules of elastic collision you know that the cue ball stops dead when the object ball has same mass (in this case = 1Kg)

You can find the distance x considering that the point of the rod hit by the ball will rotate with angular velocity $\omega = 3.67423$ r/s but with velocity of $v=\omega*r (=x) $ m/s and that it will have the same Ke*(2) = 9 , therefore $x^2= 2mv_0^2/\omega^2$ $$x =\sqrt{2Ke/y^2}$$

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Use the hint in your second bullet point. The ball has angular momentum about the pivot point before it strikes the stick.

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Your conservation of kinetic energy equation should help you solve the for the stick's initial angular velocity.

Think of it this way: the tennis ball has initial momentum since it is moving, right? And the stick is not moving, so it has no momentum. At the end of the collision, the tennis ball stops completely, so it has no momentum, but the stick is now moving, so it has angular momentum.

Thus, your equation solves it perfectly. Energy of ball at beginning = energy of rod at end.

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  • $\begingroup$ Thank you! I think I didn't realize that in order for the ball to transfer all of its energy into the stick and thus to stop moving, it must collide with the stick at the end of the stick x=1m. Bit of a trick question in my opinion. $\endgroup$ – Tyler Rudig Mar 15 '15 at 4:13

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