>high school physics class 2009
>learning about kinetic energy
>bored, make up a physics problem
>it stumps entire class, including teacher
>i never solved it
>2016
>i found the problem looking through old notebook
>let's fucking do this
>takes one day to complete
>solve it today
>feel like a champion
50/50 you either let go or you don't
>>8005537
45 degrees celsuis
I dont know but I can give a bound of 0 < theta < 180
>>8005537
your life is sad
>>8005546
>not using radians
it's gonna be between 0 and pi / 2
>>8005537
>One day to complete
>>8005537
45deg
>>8005537
I'm pretty sure it doesn't matter
90 is fastest speed, 45 is optimal angle. I'd say halfway between.
>>8005551
nah, speed has slowed down by that time.
>>8005564
I don;t mean exactly halfway. My intuition tells me it's closer to 90 than 45. Though I don't know how to actually solve the problem.
- 45 degrees.
>>8005569
yep, closer to 90 than 45
what angle will give you an initial escape angle of 45 degrees?
Trivial problem I would expect any highshool kid who has passed a AP physics test to be a able to solve.
>took a day to solve
>look at all this extra useless information
Are you even trying?
Can't be answered unless we are given the distance the person is from the ground and the length of the rope.
>>8005605
woops didn't see that at the bottom
>>8005597
only useless info is the 'this is on earth' part
the ratio of the rope length to the ceiling height matters
also, eat my asshole
>>8005607
>Bohr model of the atom
>>8005607
http://datagenetics.com/blog/september42014/index.html
~77 degrees
Is there an actually efficient way of solving this?
>>8005610
awww shit, nice. my curve (with derivative) looks pretty similar to that dudes
i don't think there's an efficient way to solve it... i think you just gotta work it out by hand. hardest part is just attention to detail - my final equation ended up being long as fuck
>>8005634
oh and with that graph the x axis, theta, is (pi/4 - the theta in the original pic)
>>8005644
its a pretty neat problem. Much better than rinse repeating the same generic problems over and over again.
>>8005537
Isn't it just a common understanding that you want something to go the farthest you throw it at a 45 degree angle? Didn't they figure that out in the 1500's? I guess we're not assuming it's a spherical object though so I dunno.
>>8005676
no man, the easiest guess is at 90 degrees because that way gravity has no effect over the total spped, however if you go past 90 you will be higher so you will most probably go furhter thanks to that
I haven't tried to solve this (yet), but are you guys making sure you don't hit the ceiling?
The velocity of a pendulum as a function of theta is given by the equation v(theta)=sqrt(2gLm(1-sin(theta)). The y component of this velocity is cos(theta)*v(theta), and the x component is sin(theta)*v(theta). The distance the person will travel is the x component of velocity times the time it's in the air: v(theta)*sin(theta)*t. The time in the air is given by the kinematic equations involving the y component and gravity: t*v(theta)*cos(theta)=t^2g/2. Solving for t gives us t=2v(theta)cos(theta)/g. Plugging this into the equation for the distance we get: 2v(theta)^2cos(theta)sin(theta)/g. If we want the maximum distance we take the derivative of this equation with respect to theta and set it equal to 0.
4Lm(1-sin(theta))cos(theta)sin(theta)
4Lm(1-sin(theta))*[cos(theta)^2-sin(theta)^2]-4Lm*cos(theta)^2sin(theta)=0
(1-sin(theta))[cos(theta)^2-sin(theta)^2]=cos(theta)^2sin(theta)
cos(theta)^2-sin(theta)^2-sin(theta)cos(theta)^2+sin(theta)^3=cos(theta)^2sin(theta)
csc(theta)-2sin(theta)-cos(theta)^2+sin(theta)^2=cos(theta)^2
csc(theta)-2sin(theta)+sin(theta)^2=2cos(theta)^2
csc(theta)-2sin(theta)+1-cos(theta)^2=2cos(theta)^2
csc(theta)-2sin(theta)+1=3cos(theta)^2
1-2sin(theta)^2+sin(theta)=3cos(theta)^2sin(theta)
1-2sin(theta)^2+sin(theta)=3sin(theta)-3sin(theta)^3
Let sin(theta)=x and we have the cubic
3x^3-2x^2-2x+1=0. We can use the cubic equation and get solutions:
x=1, -0.768, and 0.434
So now we solve for theta and get answers:
theta=90, -50, and 25
theta=90 is a min, theta=-50 is not possible in this set up, and so the answer is theta=25. I just realized I didn't take into account the person slamming into the ceiling and ending his fall early, but Idk if you wanted that to be considered.
>>8005549
This is bait right
This is pretty tough. I reached theta = 50.1°, but donno if it's correct
>>8005711
When I did it I got -50, but I discounted that solution because it was negative. Maybe the sign doesn't matter, and now that I think about it 25 degrees would probably be a minimum as well because it's obviously going faster at 50 degrees and it's closer to the optimal angle of 45 so it should be that.
This was the formula I got with 58.3 degrees as an answer given that g = 9.8 and L = 1m
I don't have paper, but some thoughts on how this should be solved.
1. Find v(y) where y is the guys position, y initial is zero. This is easy, mgy = (1/2)mv^2, and y = l * sin(theta).
2. Take the y component of the velocity you just found. If you do the trig it should end up being v_y = v * cos(theta).
3. Using that velocity, use basic kinematic equation vf = vi + at, (I hope these subscripts are self-explanatory). You'll do this in two parts to find the total time, when vf = 0 and then when he hits the ground.
4. You now have the total time he's in the air as a function of theta. Plug this into the x kinematic equation along with the x component of the velocity and differentiate x with respect to theta and solve the min/max problem.
Pretty sure if you follow these steps you'll get the right answer, the rest is just the math.
>>8005676
only if the initial speed is constant
not so in this case
I got 57.06 degrees
>>8005537
-2arctan((1+(2/(17+3sqrt(33))^(1/3)-(17+3sqrt(33))^(1/3)))/3) ~ 57.06 degrees
>>8005610
Good link.
>>8005691
Lrn2TeX nigga
>>8005609
bohr model is more aesthetically pleasing than anything else
>>8005537
Separate the problem into two parts - first is the velocity of the person when letting go of the rope at [math]\theta[/math], second is how far he's going to go after he's let go.
For the first part, just use conservation of energy. Potential energy [math](V)[/math] is given by:
[math]\displaystyle V=-mgL\sin\theta[/math]
The kinetic energy [math](T)[/math] is given by:
[math]\displaystyle T=\frac{mv^2}{2}[/math]
Total energy [math](E)[/math] is given by:
[math]\displaystyle E=T+V=\frac{mv^2}{2}-mgL\sin\theta[/math]
Initially [math]T=0, V=0[/math] so [math]E=0[/math]
By conservation of energy:
[math]\displaystyle 0=T+V\\ \frac{mv^2}{2}=mgL\sin\theta \\ v=\sqrt{2gL\sin\theta}[/math]
Ok, this is the velocity the little tarzan is travelling at when he leaves the rope.
Split [math]v[/math] up into [math]x, y[/math] components:
[math]\displaystyle v_x=v\sin\theta\\v_y=v\cos\theta[/math]
Now the velocities when he leaves the rope are solved. For the range he travels:
Solve for the time before he hits the ground first.
Initial [math]y_0=L(2-\sin\theta)[/math]
Inital [math]v_{y0}=v\cos\theta[/math]
Therefore:
[math]\displaystyle y(t)=L(2-\sin\theta)+v\cos\theta-\frac{gt^2}{2}\\
\displaystyle t=\frac{v\cos\theta}{g}+\frac{\sqrt{v^2\cos^2\theta+2gL(2-\sin\theta)}}{g}[/math]
>>8006030
>>8005537
Now you have the time they'll travel in terms of [math]\theta[/math], you want the range they'll travel. This is simple.
[math]\displaystyle x=x_0+v_xt\\
\displaystyle x=L\cos\theta+v\sin\theta t[/math]
The total time he gets to travel is given by [math]T[/math] in my post above, so:
[math]\displaystyle X=L\cos\theta+v\sin\theta T[/math]
Now, to find the farthest [math]X[/math] for some [math]\theta[/math], differentiate with respect to [math]\theta[/math] and find the maximum.
[math]\displaystyle \frac{dX}{d\theta}=-L\sin\theta+\frac{dv}{d\theta}\sin\theta T + v\cos\theta T + v\sin\theta \frac{dT}{d\theta}\\ \displaystyle 0=\frac{dv}{d\theta}\sin\theta T+v\cos\theta T+v\sin\theta\frac{dT}{d\theta}-L\sin\theta[/math]
Now, just put in [math]v,\frac{dv}{dt},T,\frac{dT}{d\theta}[/math] and you have the solution. This is probably best done on the computer, as it's a horrible thing to solve analytically.
Hello friends, brainlet here
Can someone explain me why the length of the rope and the floor to ceiling distance matter?
>>8005537
That looks like an interesting question, I will save this picture
>>8006311
Length of the rope affects the total kinetic energy the person can have. The distance to the bottom affects how far away they can land
>>8006334
But you only need to find the best angle? Sure a longer rope makes them go further but why wouldn't it make them proportionally go further at each angle?
>>8006311
And now you know!
>>8005564
>pointless math is fun, though...
No, I meant the angle doesn't matter
Nice problem! I spent most of my HS years mastrubating to anime (come to think of it, I still spend most of my time mastrubating to anime).
>>8005541
it's 30 degree kelvin you fucking knuckledragger
>>8005690
Assuming all of the persons body and mass is concentrated at the end of the rope, if the person lets go it is impossible for them to hit the roof.
If they don't let go then they will just hit the ceiling as that will bring them back to 0 velocity at the same gravitational potential energy
[math]\frac{1}{2}m v_0^2=m g l \sin(\theta)[/math]
[math] v_0=\sqrt{ 2 mg l \sin(\theta)}[/math]
[math] v_{0x}= v_0 \sin(\theta)[/math]
[math] v_{0y}= v_0 \cos(\theta)[/math]
[math] x=v_{0x}t[/math]
[math] y= 2l-\sin(\theta) +v_{0y}t-\frac{1}{2}g t^2[/math]
Time to reach floor:
[math]t: 0= 2l-\sin(\theta) +v_{0y}t-\frac{1}{2}g t^2[/math]
Solve for time. Solve for [math]x[/math], which is function of [math] \theta[/math].
Maximum is found when derivative is zero.
>>8006373
So if the person has enough space, the length of the rope doesn't matter at all? That makes sense to me intuitively, but I was under the impression that it wasn't correct
>>8006449
it's [math]2l-l\sin\theta[/math], sauce-pot
>>8006488
Yeah - the length of the rope doesn't matter if all you're wanting is to know the [math]\theta[/math] they should let go at. It does matter if you're wanting to find out how far away they'll land.
Guise
Guise
GUISE
What about air resistance?
>>8006512
Spherical cows in a fucking vacuum mate
>>8006512
how do you know it's air that slows things, and not that things consciously move slower when they see air coming?
>>8006436
>30 ̶d̶e̶g̶r̶e̶e̶ kelvin
ftfy
remember it's K and not °K
>>8006087
What the fuck is L? The only thing that matters is the height at the start and the length of the rope, and these are given. 57.06 degrees is correct.
>>8005537
>high school student 2016
Ftfy
>>8006500
Hmm
What about the height of the person then? Don't we need that to make sure they don't hit the floor?
>>8005537
I see a solution was already posted, and just taking a glance at this, I'd say you're probably best off using Lagrange or Hamilton formalism for this.
>>8005541
That was good Anon...
>>8005597
>Are you even trying?
So solve it
>>8006737
The problem is piecewise.
There's the time before you let go and the time after you let go and the time when you hit the ground.
Each of these impose different constraints on the E.O.M. How could Lagrangian/Hamiltonian mechanics possibly make this easier?
>>8006737
I don't think you understand what they are
>>8005686
>spped
what? its basically an angled throw problem
and if I remember correctly you throw the furthest with a 45° angle.
figuring out why its 45° isnt hard
>>8005537
The angle that will allow you to travel the furthest distance without having to leap from a height where you would hit the ceiling.
>>8007195
>figuring out why its 45° isnt hard
actually this is not true. I initially thought the same but the difference to the angled throw problem is that the initial velocity changes as well.
Therefore you can solve it by using the same way as for the angled throw if you replace v0 with v0(theta), as the initial velocity now depends on the angle you "throw" at.
>>8005537
I'm not in the mood for dynamics
So the answer is 60
>>8005537
Letting go anywhere between 0 and 90 will land you at the same distance
>>8005537
For a given initial velocity V, the largest distance covered will be achieved when the angle with the ground is 45 degrees.
However in this problem, the initial velocity is a function of the angle to the ground, so that's making it more complex. I guess you will have to determine an expression for the distance traveled as a function of the angle theta, differentiate and set to zero.
We actually measured this (in high school physics AP, so maybe no completely accurate). We found ~43 degrees was optimal.
the position of his legs is also important
You guys are ignoring two important things.
The length of the rope is actually important.
If he swings past 90 degrees then he has an upward component to his velocity.
The dynamic equation for s (The distance he travels from his start position) is given by.
[math]l+sin(90-\theta)+\frac{2cos(90-\theta)\sqrt{2glsin(\theta)} (sin(90-\theta)+\sqrt{2glsin^2(90-\theta)sin(\theta)-2gl(sin(\theta)-2)})}{g}[/math]
>>8008998
dumb nerd this is not science
>>8008998
If we then find the maximum for theta assuming l is constant we find that the maximum is about 61.5 degrees
isn't this just basic highschool physics?
kinematics - projectile
it's 45 deg
>>8009683
Why are you assuming the initial velocity will be the same, no matter what the degree is?
im getting 47 degrees using theta in reference to the horizontal or 43 degrees in reference to the vertical
>>8009907
Welp, I just noticed a mistake in my solution because I had y defined from the top down but then from the bottom up during the energy balance but it just means the (1 - sin(theta)) is replaced with sin(theta)
>>8005537
Solve for x in terms of theta and it's just a simple optimization problem.
>>8010314
What is x?
eh?
>>8010436
I can't read that shit
Have you considered a pen and paper?
Or LaTeX?
To everyone worrying about hitting the ceiling it can't happen cuz conservation of energy
>>8005537
Relevant:
http://arxiv.org/abs/1208.4355
>>How far can Tarzan jump?
>>The tree-based rope swing is a popular recreation facility, often installed in outdoor areas, giving pleasure to thrill-seekers. In the setting, one drops down from a high platform, hanging from a rope, then swings at a great speed like "Tarzan", and finally jumps ahead to land on the ground. The question now arises: How far can Tarzan jump by the swing? In this article, I present an introductory analysis of the Tarzan swing mechanics, a big pendulum-like swing with Tarzan himself attached as weight. The analysis enables determination of how farther forward Tarzan can jump using a given swing apparatus. The discussion is based on elementary mechanics and, therefore, expected to provide rich opportunities for investigations using analytic and numerical methods.
>>8005541
i keked
>>8007232
dont have any paper... its probably diffeq then?
59.663 Degrees
http://www.real-world-physics-problems.com/physics-of-skateboarding.html
This is something similar that happened to me. I remembered about asking my physics professor how do skaters lift their skateboards WITH them when doing a Ollie. He couldn't give me a solid answer. Now i know why.
Easy shit
https://www.desmos.com/calculator/0he0llzxih
