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I have this question assigned, but I really am stuck on how to do it:

A bullet is shot through two cardboard disks attached a distance $D$ apart to a shaft turning with a rotational period $T$, as shown.

bullet

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, $\theta=0$ means that the holes are in a line and $\theta=\pi$ 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.

So far I've compiled some information that I thought might help me, but I don't know how to apply any of it:

$$\theta=2\pi$$ $$rotationalperiod=T$$ so the angular velocity is: $$w={2\pi\over T}$$

I know that I need $D$ on its own, and the only way I could think of that was to do the distance formula $$v={D\over t}$$ $$D={vt}$$

I'm not sure if this information is enough to derive an equation, if it is I just don't know how to apply it all. Is it valid to use the $t$ in this formula in place of the $T$ from the angular velocity one? What other information do I need?

Community
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matryoshka
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1 Answers1

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Is it valid to use the $t$ in this formula in place of the $T$ from the angular velocity one?

No, $T$ is the time for one revolution. The time $t$ that you calculate is the time between the bullet hitting the first and the second plate. Two very different things.

$\theta=2\pi$

This is not correct. $\theta$ is defined as the angle between the holes, not a whole round $2\pi$.

What other information do I need?

Let's consider the angle $\theta$ in the same way as you considered $D$. $D$ is the distance passed in time $t$ at the speed $v$. Similarly, $\theta$ is angular distance passed in time $t$ at the angular speed $\omega$. You can therefor set up a similar expression for the angular motion:

$$\theta=\omega t$$

Use this with the other expressions you already have,

$$\omega=\frac{2\pi}{T}\quad\text{and}\quad D=vt\:,$$

and you are done.

Steeven
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  • In this problem wouldn't the first disk make one revolution when the bullet enters in the second hole? – Oussama Boussif Sep 03 '15 at 20:42
  • @OussamaBoussif Why would it do that? The discs are spun around, and then a high-speed bullet is shot through them. The spinning speed could be anything and is not dependent on the bullet speed. – Steeven Sep 03 '15 at 20:49
  • Ah I think I misunderstood things again. Anyways the answer is it: $$\frac{D\omega}{2\pi-\theta}$$? – Oussama Boussif Sep 03 '15 at 20:54
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    @OussamaBoussif The question states: Assume that the bullet must travel through the set of disks within a single revolution. It is of course possible that the first disc has spun exactly one whole revolution, and it could also have spun less. This does not give any problems, since the solution will be general for any angle up to one full revolution. – Steeven Sep 03 '15 at 20:55
  • @OussamaBoussif I have just given a method. Let us let the asker solve it :)

    (But no, that can't be the answer - the answer must be $v$ expressed with only $D$, $T$ and $\theta$ as the question states.)

     – Steeven Sep 03 '15 at 20:56
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    Ah okay thanks because when I saw that one revolution constraint I thought that the first disk shouldnt spin one revolution.For my answer just plug $\omega=2\pi/T$ and you'll get it in terms of the given quantities. Thanks again! – Oussama Boussif Sep 03 '15 at 20:58
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    @OussamaBoussif I see. There is surely a big difference between within a single revolution. and in a single revolution. – Steeven Sep 03 '15 at 21:00
  • Ah okay. Thanks for your time and for your explaining. – Oussama Boussif Sep 03 '15 at 21:01