Before I start, I want to say that this is not a duplicate of "Is it possible for information to be transmitted faster than light by using a rigid pole?", Since point A is not a real object, it is possible for A to exceed the speed of light. Notice that I have already taken the bending effect into account and decided that if the widths of the bars are small enough, then it's safe to treat the scissor as if it weren't bending. In other words, when they are thin, then the motion will obey the equation in a small region around point O. Please read to the end and have a look on my analysis in the last part before you explain.
Suppose we have a "scissor", which is composed by two bars with a same width of length "l". Bring them closer to each other, during this process, we have the equation(#):$$v=-\frac l4\frac{cos\frac{\theta}2}{sin^2\frac{\theta}2}\omega$$, where v is the velocity of A, \theta is the angle between BC and BA as shown in the picture.It's also an arbitrary function of time t, or $\theta=\theta(t)$. $\omega$ is the angular velocity of $\theta$, thus $\omega=\frac d{dt}\theta(t)$.Here is the derivation:
1. Draw a line vertical to L from A, intersecting with L at C.
- Since $sin\theta=\frac{AC}{AB}$, $$AB=\frac{AC}{sin{\theta}}$$.
- Since $cos\frac{\theta}2=\frac{AO}{AB}$, $$AO=AB*cos\frac{{{\theta}}}2$$
- from 2 and 3 we have: $$AO=\frac{cos\frac{\theta}2}{sin{\theta}}*AC$$
- Since $AC=l$, the equation in 4 becomes: $$AO=\frac{cos\frac{\theta}2}{sin\theta}*l$$or$$AO=\frac l{2sin\frac {\theta}2}$$
- Differentiate each side with respect to t, and denote $\frac d{dt}AO$ as "v", getting the equation(#):
$$v=-\frac l4\frac{cos\frac{\theta}2}{sin^2\frac{\theta}2}\omega$$
Again, we just look at a small area around point O where our equation is obeyed. Now, at any time $t_0$, $\theta$ and $\omega$ are independent of each other, thus $\omega$ is a free variable. At t= 0, let's set the value of $\omega$ really large and constant, so that the instantaneous speed of A is larger than even the speed of light. Later, since the factor $\frac{cos \frac{\theta}2}{sin^2 \frac {\theta}2}$ in equation(#) is quite large when $\theta$ is close to 0, and since large $\omega$ leads to small $\theta$ during a short period of time, the velocity of A will be increased by this factor. Thus overall, $\omega$ is set to be a large constant, l is a constant, the only changing factor is increasing. Therefore, the velocity of A will increase. Meanwhile, we emit a beam of light. A is in a superluminal motion. And A will arrive at a detector first, causing the bars to touch the detector, and sending the command to the machine. Then the light beam comes later, thus being detected later. Notice that here again, we place the detector close enough to point O, so that our equation is valid.
Remark: Someone may argue that we are not sure if A could attain a speed higher to that of light. So suppose at t=0, the force has already affected the area around point O, and is now traveling further.Meanwhile, ${\theta}_0=\frac{\pi}3$ A is speeded up to 0.999c, then since our formula is valid at t=0 and later period of time, let's apply it and set $l=0.001m, c=3*10^8m/s,{\theta}_0=\frac{\pi}3$. Then plug them in, we get $$\omega=-3.461*10^{11}rad/s $$. Now we want to know its velocity at $t=1*10^{-12}s$, and we get $v=6.89062*10^8m/s$, faster than light.
According to relativity, this shall violate casualty, and therefore could never happen. With which kind of mechanism is this paradox solved? I have taken the bending effect into account, and decided that as long as the widths of the bars are small enough, then it's safe to treat the scissor as if it weren't bending.
