[Boards: 3 / a / aco / adv / an / asp / b / bant / biz / c / can / cgl / ck / cm / co / cock / d / diy / e / fa / fap / fit / fitlit / g / gd / gif / h / hc / his / hm / hr / i / ic / int / jp / k / lgbt / lit / m / mlp / mlpol / mo / mtv / mu / n / news / o / out / outsoc / p / po / pol / qa / qst / r / r9k / s / s4s / sci / soc / sp / spa / t / tg / toy / trash / trv / tv / u / v / vg / vint / vip / vp / vr / w / wg / wsg / wsr / x / y ] [Search | Free Show | Home]

Well, an old puzzle my father and I were trying to solve. It

This is a blue board which means that it's for everybody (Safe For Work content only). If you see any adult content, please report it.

Thread replies: 26
Thread images: 4

File: snowball.png (7KB, 600x600px) Image search: [Google]
snowball.png
7KB, 600x600px
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
>>
File: doingsciencephotou1.jpg (68KB, 1024x577px) Image search: [Google]
doingsciencephotou1.jpg
68KB, 1024x577px
>>8194615
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.
>>
>>8194615
32 degrees Fahrenheit
>>
>>8194675
correct.
>>
>>8194615
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
>>
>>8194719
hey i forgot to tell you but the snowball is about the size of jupiter and has its own gravitational field
>>
>>8194733

Then there isn't an angle you lose contact at if you're on the surface of a planet?
>>
>>8194736
then how do planes manage to take off?
>>
>>8194736
what he forgot to tell you was that there is no spoon
>>
>>8194740

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.
>>
>>8194652
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.
>>
>>8194675
fpbp
>>
arccos 2/3 you beanie
>>
>>8194719
Not force, but velocity has to be tangent. Both can be tangent at the same time, but self-propelling or impeding can change this.
>>
>>8194875

If you're moving along the surface then your velocity is tangential by definition. Cheers!
>>
File: _20160710_140014.jpg (513KB, 2340x1476px) Image search: [Google]
_20160710_140014.jpg
513KB, 2340x1476px
>>8194615
>>8194756
The pic is how I solved it in "Fundamentals of mechanics".
Just to prove that the drunk guy got it right.
>>
File: 1436127030164.png (139KB, 436x438px) Image search: [Google]
1436127030164.png
139KB, 436x438px
>It's a /sci/ can't into highschool mecanics thread
>>
>>8195503
The only hard part of this bullshit is finding that the centripetal force equals mgcos(th). I can't into similar triangles.
>>
>>8194850
Did you just fart?
>>
>>8194756

I apologize for this mess.
>>
>>8195698
Messes are social constructs.
Thread posts: 26
Thread images: 4


[Boards: 3 / a / aco / adv / an / asp / b / bant / biz / c / can / cgl / ck / cm / co / cock / d / diy / e / fa / fap / fit / fitlit / g / gd / gif / h / hc / his / hm / hr / i / ic / int / jp / k / lgbt / lit / m / mlp / mlpol / mo / mtv / mu / n / news / o / out / outsoc / p / po / pol / qa / qst / r / r9k / s / s4s / sci / soc / sp / spa / t / tg / toy / trash / trv / tv / u / v / vg / vint / vip / vp / vr / w / wg / wsg / wsr / x / y] [Search | Top | Home]

I'm aware that Imgur.com will stop allowing adult images since 15th of May. I'm taking actions to backup as much data as possible.
Read more on this topic here - https://archived.moe/talk/thread/1694/


If you need a post removed click on it's [Report] button and follow the instruction.
DMCA Content Takedown via dmca.com
All images are hosted on imgur.com.
If you like this website please support us by donating with Bitcoins at 16mKtbZiwW52BLkibtCr8jUg2KVUMTxVQ5
All trademarks and copyrights on this page are owned by their respective parties.
Images uploaded are the responsibility of the Poster. Comments are owned by the Poster.
This is a 4chan archive - all of the content originated from that site.
This means that RandomArchive shows their content, archived.
If you need information for a Poster - contact them.