Well, an old puzzle my father and I were trying to solve. It was a bonus question from my physics class last year but I can't remember the solution.
A skier of mass m is on a snowball of radius r. He begins to slide down the snowball, at what angle does he lose contact with the snowball?
45 degrees
360 degrees sempai
GRAVITY IS PULLING HIM DOWN NO MATTER WHERE HE IS ON THE SNOWBALL, HE HAS NO INWARD ACCELERATION, HE LOSES CONTACT AT 90 DEGREES IF NORTH IS 0.
32 degrees Fahrenheit
correct.
not enough information
The skier loses contact when the force acting on them is going tangent to the surface of the snowball. Gravity is gonna be mg straight down obviously, and the normal force is gonna be mg cos whatever where whatever is the angle they've traveled through starting from zero at the top. So drop the mg because we don't really care about the mass or the actual magnitude of the acceleration since the two forces we care about are proportional to one another.
The normal force is always gonna be pushing out radially, so what we really want to do is figure out how that downward force changes relative to the surface of the snowball as you move through the angle. At zero degrees it's directed straight inward, and at 90 degrees it's completely tangent to the surface of the ball. This also sounds like a cos function to me, except instead of the magnitude changing like it did with the normal force, what you see is the orientation relative to the surface of the ball changes with angle.
So here's the punchline: at what angle is the magnitude of the normal force equal to the proportion of the force due to gravity directed inward?
when the centrifugal force > normal force
hey i forgot to tell you but the snowball is about the size of jupiter and has its own gravitational field
Then there isn't an angle you lose contact at if you're on the surface of a planet?
then how do planes manage to take off?
what he forgot to tell you was that there is no spoon
With lift force, obviously. You go fast enough and angle your wings so you can make a negative pressure differential that pushes up on the bottom of the wings.
This is the answer my father and I got but no way to check our answer because I'm no longer in the class.
I'm a little drunk right now...so excuse me for a poro explanation.
The sum of the forces in the r direction is equal to centripetal force so:
-m(v^2)/r=N-m*g*cos(th) 1)
where th is the angle from the vertical (so he starts at th=0).
Now we use conservation of energy. At his final height, he has speed va, so
m*g*r = (1/2)*m*va^2 + m*g*r*cos(thf) 2)
where thf is the final angle. Now at the instant he slides off, the normal force is 0, so equation 1 becomes
-m(va^2)/r = -m*g*cos(thf) 3)
Note that the velocity is va now and the angle is thf. These are the velocity and angle where it loses touch respectively.
Oh God, I'm going to throw up. What am I doing? My cock will find a way. Equation 2 rearranges to:
2*m*g*r*(1-cos(thf)) = m*va^2
which can be substuted into 3) so that:
2*m*g*(1-cos(thf)) = m*g*cos(thf)
or
2 = 3*cos(thf)
such that
cos(thf) = 2/3.
He loses contact at a height of 2R/3. Fuck, what did you want?
He loses contact at an alge of arccos(2/3). Fuck, what is that? 48 ish degrees?
If that's right, I need another drink.
fpbp
arccos 2/3 you beanie
Not force, but velocity has to be tangent. Both can be tangent at the same time, but self-propelling or impeding can change this.
If you're moving along the surface then your velocity is tangential by definition. Cheers!
The pic is how I solved it in "Fundamentals of mechanics".
Just to prove that the drunk guy got it right.
>It's a Veeky Forums can't into highschool mecanics thread
The only hard part of this bullshit is finding that the centripetal force equals mgcos(th). I can't into similar triangles.
Did you just fart?
I apologize for this mess.
Messes are social constructs.