The speed of a bullet may be determined by allowing the bullet to pass throu

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SUMMARY

The speed of a bullet can be accurately determined using two rotating paper disks positioned 80 cm apart. By measuring the angular displacement of 31 degrees between the bullet holes and knowing the disks' rotational speed of 900 revolutions per minute, the bullet's speed can be calculated. The final result, derived through the appropriate conversion of units and application of rotational kinematics, yields a bullet speed of 139 m/s. This method effectively utilizes angular velocity and time to derive linear speed.

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[SOLVED] The speed of a bullet may be determined by allowing the bullet to pass throu

Hello! I am new here and this is my first time posting...I am working on a physics problem and I have been stumped on it for a while now. Any hints would be greatly appreciated.

Homework Statement



The speed of a moving bullet can be determined by allowing the bullet to pass through two rotating paper disks mounted a distance d apart on the same axle. From the angular displacement Δθ of the two bullet holes in the disks and the rotational speed of the disks, we can determine the speed v of the bullet. Find the bullet speed for the following data:

d=80 cm, ω=900 rev/min, and Δθ=31º.

Homework Equations



My initial thought was that I would need to use the rotational kinematics formulas in some way or another:
1. ω = ω(initial) + αt
2. θ = θ(initial) + 1/2(ω + ω(initial))t
3. θ = θ(initial) + ω(initial)t + 1/2αt^2
4. ω^2 = ω(initial)^2 + 2α(θ - θ(initial))

to somehow get t, and then use the horizontal displacement over t to find v, the velocity of the bullet.


The Attempt at a Solution


Some of my thoughts: I do not know what alpha, the angular acceleration is, so I cannot use it in order to find t. This rules out equations 1, 3, and four, leaving me with equation 2 to work with. Only one angular velocity was given in the problem, so the omegas in the formula confuse me, and I'm not sure if I need to use the same number for omega initial and omega final, or if I am supposed to assume constant angular velocity at all. If so, I think I can use 0 as theta initial and the 31 degrees as theta final to find t.

If this is in fact the correct way to approach this problem (I have searched the chapter in the book and can't seem to find another approach), the last thing on my mind is the issue of radians and degrees. Should I change every aspect of the problem to radians, or to degrees? My initial impression was radians, but from the way I'm looking at it, I'm not sure if it would make a difference, or if there is a correct way to do this. I am again assuming that the velocity remains constant, so that is how I can find it, by dividing d by time.

ANY hint or small tip in the right direction would be greatly appreciated. Thanks ahead for any help.
 
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Some of my thoughts: I do not know what alpha, the angular acceleration is
well, from the problem statement, I will conclude that the discs are rotating with constant angular velocity, so alpha is zero.
 
Try to understand this: (my english is poor, sorry for that)

The first disc rotated by 31 degrees in the same time as the bullet took to travel 80 cm.

omega of disc(which will be constant) =900x(2xpi)/60 rad/sec. (pi=3.14)

convert 31 degrees into rad by multiplying by (pi/180). = 31x (3.14/180)

time taken by disc to rotate 31 degree = (31x(3.14/180) / omega) seconds

this time =0.80/v.

____________________________________
I may be wrong...
 
Last edited:
Thanks so much for the help! Here's what I did and how I got it:

First converted d into meters, 80 cm = 0.80m
Converted omega into rad/s, 900 rev/min(1 min/60 s)(2pi/1 rev) = 94.2 rad/s
Converted degrees into radians (like you said), 31 degrees(pi/180 degrrees) = 0.541 rad


Then I used equation 3 from above...theta initial and alpha are both zero, so I just took 0.541 rad / 94.2rad/s to get 5.74x10^-3 s for t (0.00571 s)

Then I took 0.80m/0.00571s = 139 m/s

How does this look, and thanks so much for the help!
 
Again, I'm pretty sure that it looks right.
 
Bump...sorry if we're not allowed to bump.
 
do an order of magnitude estimate
disks 1m apart, rotate in 1/15 seconds hits are 1/6 of a turn apart = 0.01s
so answer is roughly 100m/s
You have made things slightly complicated for yourself with radians etc.

That isn't actually realistic for anything other than a musket ball - but you never know how accurately the question was set!
 
Thanks...but is it still suitable, the way I did it in radians? So far, everything we've learned in class has been in radians for objects rotating about a fixed axis.
 
A bit unnecessary but the right answer.
 
  • #10
OK...thank you so much for the assistance, I really appreciate it.
 

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