Speed of a Bullet: Deriving Formula with D, T & Theta

  • Thread starter Thread starter tigerseye
  • Start date Start date
  • Tags Tags
    Bullet Speed
AI Thread Summary
To derive the bullet speed formula in terms of distance D, rotational period T, and angle theta, the time taken for the bullet to cross the gap must be calculated. The angle theta indicates the fraction of the full revolution completed by the disks during the bullet's travel. If the bullet penetrates the second disk 45 degrees later, it implies the disks rotated 1/4 turn in that time. Using this relationship, the time can be determined as T/4, allowing for the calculation of bullet speed by dividing the gap length D by the time taken. This approach provides a clear method to find the bullet's speed without needing the radius.
tigerseye
Messages
16
Reaction score
0
I've been stuck on this problem forever and I don't know how to do it if you can't use r as the radius.
A bullet is shot through two cardboard disks attached a distance D apart to a shaft turning with a rotational period T (see the attached picture). 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.
 

Attachments

  • untitled.JPG
    untitled.JPG
    7.5 KB · Views: 456
Physics news on Phys.org
You know the time required to make a full revolution and from the angle you can determine the time it takes the bullet to traverse the gap. Find the speed of the bullet by just dividing the length of the gap by the time it takes to cross it.
 
I no this sounds stupid but how do you find how long it takes to cross the gap !
 
If the bullet penetrated the second disk 45^\circ[/tex] later than where it penetrated the first disk, then the disks traveled 1/4 turn in the time it took the bullet to penetrate both disks, so...
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top