/sci/ - Science & Math

depends which way you shoot, if you point at yourself you can be stationary

>>5042289
beat me to it

If you are going at exactly the speed of the bullet, then you can fire "backwards", i.e. the opposite direction to what you are moving, and the bullet will be stationary after you shoot it. Then you will travel all the way around and it will it you in the face.

Now if you are moving slower, the calculation is more complicated so I don't know if it's still possible.

Transient velocity of bullet

Diameter of merry go round seat

Find time required for bullet to accelerate and travel diameter

Calculate 'w' angular velocity required to travel pi or 180º in the time it takes for the bullet to travel the diameter.

let's assume the radius is 1m, the bullet fires at 1000m/s

the half-circumference is pi

the bullet will travel the 2m diameter in 0.002 seconds, let's assume it'll suffice if you just arrive at that point at the same time, which means pi m in 0.002 seconds

(pi meters) per (0.002 seconds) = it needs to move at around 5 655 kilometer per hour rotational velocity

in short, you're more likely to get:

https://www.youtube.com/watch?v=hLRoXZkhW_8

Guns usually have a speed of about 1200m/s. The diameter of one of these is roughly 3m, thus the gun will cover that distance in 0,0025s, which is a very small amount of time. In order to be there in 0,0025s, V=(π*D)/t so V is roughly 3800m/s. Which very hard to achieve and probably one of these would be destroyed during the process of accelerating it, hence you should consider alternative ways of killing yourself.

let's suposse some stuff.
1)you'll shot aiming the center.
2)you'll die
3)Hard physics and coriolis efects are gonna be dismissed


Now, If you shoot to the center in a given point A, A' is the point opposed by the center, being at 2r from A. If bullet velocity is V, then t=2r/V is the time the bullet would need to go from A to A'. Now all you have to do is find an angular velocity that allows you to move pi·r in time t.

Now the guys in coats will correct and add more info, but all can be reduced to choose a firing point and a hitting point and trouhg the variable t making the equal, so teh bullet hit you. Since the bullet is very fast, teh angular velocity must be even higher, so if you ever desing an artifact that can acomplish such, you'll die from g forces messing your blood before the bullet.

Good luck, anon.

Let's say the velocity of the bullet in the ground frame makes an angle <span class="math">\theta[/spoiler] with your velocity as you ride on the merry-go-round.

You have to travel a distance <span class="math">2r\theta[/spoiler] whereas the bullet travels <span class="math">2r\sin\theta[/spoiler] where r is the merry-go-round radius. So the ground-frame speed of the bullet is <span class="math">\sin\theta/\theta[/spoiler] times the merry-go-round speed.

The components of the bullet velocity in your frame parallel and perpendicular to your velocity on the merry-go-round are then <span class="math">\sin\theta\cos\theta/\theta - 1[/spoiler] and <span class="math">\sin^2\theta[/spoiler] times the merry-go-round speed. Giving us a merry-go-round speed of the bullet speed times this:
<div class="math">\frac{1}{\sqrt{\displaystyle \left(\frac{\sin\theta\cos\theta}{\theta} - 1\right)^2 + \frac{\sin^4\theta}{\theta^2}}}</div>
Now we could simplify and apply calculus, but that ends up with us trying to solve <span class="math">\cot\theta - 1/\theta = \pm 1[/spoiler], so let's instead just ask Wolfram Alpha to find us the minimum numerically:
http://www.wolframalpha.com/input/?i=minimize+1%2FSqrt[%28Sin[x]*Cos[x]%2Fx+-+1%29^2+%2B+Sin[x]^4%2F
x^2]

So we get a minimum speed of approximately 0.793909 times the bullet speed.

>>5042517
Correction:
perpendicular component should say <span class="math">\sin^2\theta/\theta[/spoiler] times the merry-go-round speed.

OP made me lol with this funny yet legit question.
Have a bump.

Better question; how fast would it have to be going before all the kids hanging on by their fingertips flew off across the playground and knocked other kids off of the swings 20 meters away?

>>5042618
It depends on the size of the merry-go-round. Given a sufficiently large merry-go-round you could get up to any speed you want without flinging people off.

>ignoring coriolis effects at these speeds

ISHY and so on

>>5042649
>coriolis effects

But I didn't. Bullet travels in a straight line.

>>5042828
To an outside observer yes, but not for the person firing the gun

>>5042517
How are you writing theta?
>>5042618
The instantaneous velocity that they let go is tangent to the circle. They will leave at that speed at some height with some air resistance. It is actually very trivial to calculate.

>>5042842
>matlab

>>5042852
The equation is transcendental. You have to solve it graphically in its current form (or on a computer).

>>5042842
http://img823.imageshack.us/img823/6099/jsmath.png

>>5042872
No, I realize it made me sound like I didn't understand the LaTex. I meant what does theta look like if you were to draw the thing.

>>5042842
>How are you writing theta?
See below the post form:
>Use TeX/jsMath with the [ma th] (inline) and [eq n] (block) tags. Double-click equations to view the source.

>>5042838

It doesn't matter. The bullet travels from Point A to Point B in 't' seconds.

Person has to move from Point A to Point B in 't' seconds.

>>5042909
The question was at what speed it happens.

To know the required angular velocity of the merry go round you must know the speed of the bullet and the length of the path it takes from point A to point B.

If you shoot towards the centre it won't go straight over and meet you at the other side

>>5043005
>If you shoot towards the centre it won't go straight over and meet you at the other side

That's exactly what it will do if you calculate it correctly.

>>5043030
For it to go straight over the centre you can't spin at all

The bullet will have velocity straight towards the centre as a result of the shot, but will also have a velocity tangential to the circle as a result of the person firing it spinning around

>>5042281
FUCKING HELL GUYS YOU DONT EVEN NEED TO BE MOVING AT ALL TO HIT YOURSELF.

>point gun at face
>shoot

HOLYFUCK DID I JUST BLOW MY JAW OFF?

come on guys, critical thinking.

>>5043056 >For it to go straight over the centre you can't spin at all

Can you make a sketch of the bullets path on a fixed top down reference plane from t0

>>5043056 you
>>5044621 me

Nevermind I went to youtube to find what I wanted.

There is no tangential velocity.

The bullet goes straight, unless you count the time it takes to travel through and leave the barrel, with the possibility of the barrel creating a torque on the bullet because they briefly touch. We're ignoring the barrel in this situation.

Make sure you understand clearly that the bullet has a clearly defined velocity vector, there is no orbital acceleration to make it travel in a curved path.

The point on the circle where the bullet comes out is dependent on the velocity of rotation with respect to time.

>>5044655
If you are in free fall and push an object away from you. Assuming the air resistance is the same on the object as it is for you. Do you not agree that the object will relative to you just go in the direction you push it?

In the scenario you present, you are claiming that what would happen for the ball is the moment you let go, it is going to instantly fly upward relative to you because you don't think it shares your inertial frame.

When you shoot the bullet, it will share the velocity of the person. This velocity is tangential to the the radius of the merry go round. You will impart tangential velocity onto the bullet. If you turned opposite to the direction of the rotation of the merry go round and shot a bullet backwards while the merry go round was rotating with a tangential velocity equal to the bullet, the bullet would drop out of the barrel with no velocity!

>>5044719
You could put the gun in the center of the spinning go-round-thing.

That simplifies the equation a lot, since the bullet DOES travel in a straight line.

>>5044732
The tangential velocity is the product of the radius and the angular velocity. If you sit directly in the middle, the radius to you is zero. So, you have no tangential velocity and of course the only velocity component of the bullet is that which the gun imposes on it.

this is actually pretty complex to calculate.
The bullet leaves as a vector and that vector is influenced as well by the angular acceleration/velocity of the wheel as it's being fired. And since you'd have to be moving at incredibly high speed to achieve this in the first place the more that bullet is influenced at the moment it is fired.

>>5043093
>come on guys, critical thinking.

lol'd

This film was fictional, stop getting your science from films.

http://www.youtube.com/watch?v=lGZQi3ODB-U

>>5045198
i get mine from cereal boxes

>>5042517
>mfw glorious generalized solution

>>5042883
Not him, but it is the angle made to your velocity, ie between a tangent to the circle and the direction of the bullet

Also, the front of the bullet leaves the gun while the back of the bullet is still in the barrel, which would cause it to rotate during flight...messy.

>>5046848
Clearly, the barrel has a length of 0. Also there is no gravity (or alternatively, the bullet is shot high enough from the ground that it will not hit the ground). And no wind resistance. Also, v << c and lamda << h.