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Homework Statement
A bullet is shot through two cardboard disks attached a distance D apart to a shaft turning with a rotational period T, as shown.
Derive a formula for the bullet speed v in terms of D, T, and a measured angle theta between the position of the hole in the first disk and that of the hole in the second. If required, use pi, not its numeric equivalent. Both of the holes lie at the same radial distance from the shaft. theta measures the angular displacement between the two holes; for instance, [itex]\theta = 0[/itex] means that the holes are in a line and [itex]\theta=\pi[/itex] means that when one hole is up, the other is down. Assume that the bullet must travel through the set of disks within a single revolution.
Homework Equations
[itex]\theta(radians) = \frac{s}{r}[/itex]
[itex]v = \frac{2{\pi}r}{T}[/itex]
where s is the arc length, r is the radius, and T is the period.
The Attempt at a Solution
I really do not know how to approach this problem. Nothing on circular motion has been covered in lecture and all I know about it is what I have read about it, which isn't much. Ideally, I assume the bullet will travel successfully through the disks in one revolution if both holes have the largest displacement (which would be if one is at its highest point and the other at its lowest, implying [itex]\theta = \pi[/itex].
Then:
[itex]\theta(radians) = \frac{s}{r}[/itex]
[itex]\theta = \pi[/itex]
[itex]\pi = \frac{s}{r} => r = \frac{\pi}{s}[/itex]
Now, to substitute this into equation 2:
[itex] v = \frac{2{\pi}r}{T} => \frac{2{\pi}(\frac{s}{{\pi}})}{T} => \frac{2s}{T}[/itex]
But I can't go any further than that, and in fact, I am quite doubtful that I am even on the correct path to solving this problem thus far. Any help? Thanks in advance.