# Circular Motion of a bullet

1. Oct 1, 2007

### sam.

1. The problem statement, all variables and given/known data

Derive a formula for the bullet speed v in terms of D, T, and a measured angle 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 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.

A diagram of this can be found here: http://ca.geocities.com/canbball/MRB_rr_8_a.jpg [Broken]

2. Relevant equations

Okay so I know that v=D/t
And that v = (2$$\pi$$r)/T

3. The attempt at a solution

I know that the disks rotate by 2$$\pi$$ in time T. What I don't understand is how to express this in terms of $$\theta$$.

Any help would be greatly appreciated!

Last edited by a moderator: May 3, 2017
2. Oct 2, 2007

### andrevdh

You say that the disc need to rotate at the same speed as what the bullet is travelling at. Why is this neccessary? All that is needed is that the size of the hole and the speed of rotation is such that the bullet could pass through it. The speed of the disc is anyway perpendicular to that of the path of the bullet, so theoretically it can travel at any speed one just need to adjust the distance, D, for the second disc, or the angle, in order for the bullet to pass through it. So your two formulas would not solve the problem.

What is required is that the time of travel for disc, to, to turn through the angle theta, and the time for the bullet to travel the distance D, need to be the same.