A bullet is shot through two cardboard disks attached a distance D apart to a sh

Question

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 between the position of the hole in the first disk and that of the hole in the second. If required, use, not its numeric equivalent. Both of the holes lie at the same radial distance from the shaft. measures the angular displacement between the two holes; for instance, = 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.

Explanation / Answer

The distance between the 2 disks (is D) is traveled by the bullit in t seconds, so D = v.t (we assume the bullets speed to be constant)

The disk turn T times a second, so every time t the disk turns 2.pi.t/T times. You can see that by realizing that if the time equal the rotational period T there is a 2.pi rotation, and that is a full circle.
So at a time t smaller than the rotational period, the angle that both disks rotate is theta = 2.pi.t/T.
Take at the time the bullit hits the first disk theta = 0 than at the time the bullit hits the second (both) disks have rotated through theta = 2.pi.t/T

The time is above also expressed in D and v as D = v.t so t = D/v. Fill this in the other equation and you get:
theta = 2.pi.D/(v.T)

So the speed is: v = 2.pi.D/(theta.T)

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