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Q: Two identical uniform sticks are rotating about their stationary centers with equal angular speeds. The vertical stick is slowly raised until its top end collides with the center of the horizontal stick. The sticks join together to make a rigid object in the shape of a T. Assume that the collision takes place when the top stick lies in the plane of the paper. Immediately after the collision, one point (in addition to the CM) on the T will instantaneously be at rest. Where is this point?

I was thinking that when/where they collide, their rotations would be in opposite directions, and so would cancel out and make the instantaneously still point the point of connection. Is this not correct? How could one prove or disprove this analytically/mathematically? Besides just thinking about it or visualizing it, I'm not sure how to go about this problem. Could anyone offer some guidance?

diagram of the two sticks' motion

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After thinking about it some more, we need to calculate the velocity of the t-joint where the two pieces connect. For now, consider the rotation of the vertical rod. Its angular momentum is:

$L=\frac{m\ l^2}{12}\omega$

When they attach, angular momentum in the x direction is conserved. The center of mass of the combined pieces is $l/4$ below the t-joint, which will be the new center of x axis rotation. The moment of inertia of the combination about this axis is:

$inew=m \left(\frac{l}{4}\right)^2+\frac{m l^2}{12}+m \left(\frac{l}{4}\right)^2$

which is the moment of inertia of the original vertical bar about its original center of mass + the addition of the $l/4$ offset from the original center of mass + the contribution of the top bar. The equation for the new vertical angular velocity is:

$L=\ inew\ \omega new$

or

$\text{$\omega $new}=\frac{2}{5} \omega$

making the velocity at the t-joint:

$v=\frac{l}{4}\ \omega new=\frac{l\ \omega }{10}$

The top bar is still rotating with angular velocity $\omega$, so the point to the right of the t-joint on the top bar that has the same velocity in the opposite direction and is therefore instantaneously at rest is:

$\text{dist}=\frac{v}{\omega }=\frac{l}{10}$

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  • $\begingroup$ Thanks for responding! How would those cancel? I think I'm confused by your explanation - First you describe the right end of the top rod, and then the center of the top rod... These are two different points? Is there a way to prove this ? $\endgroup$ – Fin S.H. Jan 31 '19 at 3:42
  • $\begingroup$ We do need to assume an immediate change in velocity when the two collide. The top of the vertical rod is traveling with $v$ which it imparts to the top rod. The right end of the top rod moves with $-v$. The question is kind of unclear about timing. $\endgroup$ – Bill Watts Jan 31 '19 at 3:56
  • $\begingroup$ Thank you again, this makes a lot of sense. But in i(new), why is the contribution from the top/horizontal bar m(L/4)^2? $\endgroup$ – Fin S.H. Jan 31 '19 at 15:57
  • $\begingroup$ Also, how did you get that w(new) is equal to (2/5)w and why is v = (L/4)wnew? $\endgroup$ – Fin S.H. Jan 31 '19 at 16:03
  • $\begingroup$ The contribution is $m r^2$ with $r=l/4$ since all of the mass of the top rod is $l/4$ from the axis of rotation. Solve for $\omega new$ from the equation I use, using the value of $L$ I state near the beginning of my answer. $\endgroup$ – Bill Watts Jan 31 '19 at 18:11
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Assuming the sticks are in the shown position (but touching) when the lower stick fuses with the upper one, then the top end of the lower stick has been coming toward you, and imparts some of that motion to the center of the upper stick. But, the right end of the upper stick has been moving away from you and will continue to do so. The point where these two motions are instantaneously equal and opposite will be on the upper stick somewhat to the right of the center of the upper stick. I have not carefully checked Mr.Watts calculations, but they look reasonable.

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