Say there is a solid bar in space. Imagine it vertically, with the top point labeled $A$, and the bottom point labeled $B$ (these aren't points on the bar, but points in space between which the bar originally is situated).

We smash a rocket into the side labeled $B$, giving it a small amount of momentum. We know that because the mass was hit off center, it will rotate, and because the entire bar has momentum, it will move forward.

However, will the top part of the bar, the part that starts off at the point labeled $A$, ever be horizontally behind $A$ as the bar starts moving forwards and spinning, or will it allways stay in front of it, meaning that the bar moves forwards before it has a chance to spin its topmost point behind $A$, and why?

I've drawn a picture to better explain the two situations; the top situation is the top part never goes behind $A$, and the bottom situation is the top part does go behind $A$.

enter image description here

Edit after closed for not enough effort (sorry, I thought I had put my reasoning in before, I'll put it in now!):

My thoughts are these. When something is hit off-center, the reason it spins is because one side of it accelerates before the other does, since it received the momentum before that momentum spread to the other side.

If an object is hit in the center of mass, it doesn't spin because the center of mass has equal mass on EVERY side, meaning that every piece of mass around the center of mass accelerates at the same rate.

Now, if we were to hit the bar in the center of mass (and let's say its a uniform bar so smack in the middle), obviously the top part of the bar wouldn't move behind A. That wouldn't make any sense, since both sides would accelerate at the same rate.

Now, if we hit it a bit off the center, although the bottom side would accelerate before the top, momentum would still spread to the center of mass before it gets to the top, since its closer. I don't think the top part can accelerate before the center of mass does simply because the center of mass is closer to the point of impact.

Therefore I think its the top situation.

However, I'm not sure, and would appreciate some comments and responses!

  • 1
    $\begingroup$ Think that it all depends on the distribution of mass of the object and its moment of inertia. If you imagine all of the mass concentrated at the center and just two relatively light wires extending from opposite ends, then it seems like you'll have one answer. If, on the other hand, you imagine an object with mass concentrated at the ends (like a barbell) then you'll likely have the opposite answer. Don't think that there is a simple way to tell for a bar of uniformly distributed mass, as you seem to have drawn, without going through detailed calculations. Have you tried doing a calculation? $\endgroup$ – user93237 Oct 13 '18 at 22:29

First, the bar has mass $M$ and length $l$. The rocket has mass $m$ and initial speed $v_i$. We need to conserve linear and angular momentum. If we use the center of the bar as the reference point, we have at the moment before collision,

$$\begin{aligned} p_i &= mv_i \\ L_i &= \frac{l}{2}mv_i \end{aligned}$$

The rocket also has a kinetic energy of

$$K_i = \frac{1}{2} m v_i^2$$

Now, "smash the rocket" is open to interpretation. If we assume the rocket bounces off elastically, we have conservation of kinetic energy, so,

$$K_i = \frac{1}{2} m v_i^2 = K_f = \frac{1}{2} m v_f^2 + \frac{1}{2} M V_f^2 + \frac{1}{24}Ml^2\omega^2$$

And we also must conserve the two momenta, so,

$$\begin{aligned} p_i &= mv_i = p_f = mv_f + MV_f\\ L_i &= \frac{l}{2}mv_i = L_f = \frac{l}{2}mv_f + \frac{1}{12}Ml^2\omega \end{aligned}$$

In total, we have

$$\begin{align} \tag{1} mv_i &= mv_f + MV_f\\[0.75em] \tag{2} \frac{l}{2}mv_i &= \frac{l}{2}mv_f + \frac{1}{12}Ml^2\omega\\[0.75em] \tag{3} \frac{1}{2} m v_i^2 &= \frac{1}{2} m v_f^2 + \frac{1}{2} M V_f^2 + \frac{1}{24}Ml^2\omega^2 \end{align}$$

Now, in order to not cross point A ever, the end of the bar has to be moving with a positive horizontal velocity after the collision. That is,

$$V_f \geq \frac{l\omega}{2}$$

However, we can quickly see by rearranging eqs. (1) and (2) that in this case,

$$V_f = \frac{l\omega}{6}$$

and so the condition you seek cannot be met with an elastic collision.

For a perfectly inelastic collision, the rocket sticks to the rod after collision. You can solve by calculating the new moment of inertia of the system and again using conservation of linear and angular momentum. Maybe it is then possible to meet this condition...

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