The problem is as follows:
The figure from below shows a cylinder spinning. Find the angular speed of the cylinder so that the perforations made in a minimum time by the bullet carrying a linear velocity $v$ make an angle with $O$ a central angle of $120^{\circ}$.
The alternatives given in my book are as follows:
$\begin{array}{ll} 1.&\frac{v}{\pi R}\\ 2.&\frac{3v}{R}\\ 3.&\frac{3\pi v}{R}\\ 4.&\frac{\pi v}{R^2}\\ 5.&\frac{\pi v}{6R}\\ \end{array}$
This problem is kind of confusing. I don't know exactly how to understand what's the central angle they are referring to. What would it be?.
The diagram is a bit confusing too, if the bullet is hitting the center of the cylinder, how can it make an angle?. What central angle is it referring?.
The only thing which I think it intended to say was that in a tiny fraction of time the cylinder and the bullet both have the same tangential speed. But would that idea help in the solution of this problem?.
Even if this is the case, how to use this logic?.
The only thing which I could do was to transform the given angle in rads as the angular displacement.
$\theta=120^{\circ}\times\frac{\pi}{180^{\circ}}=\frac{2\pi}{3}$
But again I don't know what is the $O$ they are referring, could it be that $O$ is a point situated in the center or is it located at a certain distance from the center?. Would that matter?.
All and all can comeone help me with this question?. I'm very confused. Help please?.
I think the question is a little bit confusing, too. But here's what I understand. Suppose the bullet strikes the cylinder at time $t=0$ and let us suppose that $x_{0} = 0$. Here, $x_{0}$ is the initial position of the bullet. The cylinder will keep rotating and I'm assuming you are measuring angles in the canonical way, that is, as you do in usual trigonometry. So, if the perforation makes an angle of $2\pi/3$ with $O$ in the minimum time, it means that this perforation goes from $\theta = \pi$ (where the bullet initally stoke) to $\theta = 2\pi/3$. In other words, the perforation rotates and angle of $\pi/3$ in the same period of time in which the bullet goes from the first to the second perforation. But what is this period of time? Because we set $t=0$ when the bullet strikes the cylinder first, the period of time we're looking for is: $$ 2R = vT \Rightarrow T = \frac{2R}{v}$$ Thus, once the cylinder rotates with constant angular speed, we have $\theta(t) = \omega t$, and it follows that: $$\theta\bigg{(}\frac{2R}{v}\bigg{)} = \frac{\pi}{3} = \omega \frac{2R}{v} \Rightarrow \omega = \frac{\pi v}{6R}$$
Note: To be fully rigorous, I should have considered $\theta(t) = \omega t + \pi$, since at $t = 0$ we have $\theta(0) = \pi$. However, I'm using the simplified equation for uniform rotation: $$\omega = \frac{\Delta \theta}{\Delta t}$$