20030806, 17:51  #1 
Dec 2002
Frederick County, MD
370_{10} Posts 
Physics Sledding Problem
OK, here is another physics puzzle. Actually, I'd call this more of a problem than a puzzle.
A sled with mass m starts at the top of a very large frictionless snowball of radius R, with an infinitesimal initial speed, and slides straight down the side (see the figure). At what point does the sled lose contact with the snowball and fly off at a tangent? In other words, at the instant the sled loses contact with the snowball, what angle theta does a radial line from the center of the snowball to the sled make with the vertical? http://home.earthlink.net/~eepiccolo...%20problem.gif . . . . . . . . . . . . 
20030806, 18:20  #2 
Aug 2002
Portland, OR USA
2·137 Posts 
To satisfy those concerned with details, I'm going to presume:
 I can ignore any effects from the air.  I'm conducting this experiment on earth, so g = 9.8 m/s/s. (reasonable, given the space budget.)  The sled is equipped with airbags (no masses were harmed during this experiment.) :D 
20030806, 18:25  #3 
Jul 2003
5^{2} Posts 
My answer:
theta=arctg(1/2) So, theta~= 26,5650511770779893515721937204533 deg Solution will be posted later. 
20030806, 18:27  #4  
Dec 2002
Frederick County, MD
2×5×37 Posts 
Quote:
Oh, and since this is a physics problem, significant digits matter, so if you use a value of g with two significant digits (9.8 m/s^2), the answer should have two significant digits. 

20030806, 18:30  #5  
Dec 2002
Frederick County, MD
2×5×37 Posts 
Quote:


20030806, 18:46  #6 
Sep 2002
44_{10} Posts 
After a bit of crunching, I came up with 48.19 degrees as the value of theta. I may be mistaken, but I believe the value of gravity is irrelevant as long as it is greater than zero and the snowball isn't attached to a ceiling somewhere, which in either case, theta = 0.
I will grieve for the mass m and massless sled. :( 
20030806, 19:12  #7 
Jul 2003
11001_{2} Posts 
vspeed of sled
1) m*v^2/2=m*g*R*sin(theta) (energy is not lost) so 2*g*R*sin(theta)=v^2 and v^2/R=2*g*sin(theta) we know that tangencial velocity (a_t) is equal to v^2/R when sled is on the snowball Nforce of normal reaction of snowball F_t tangencial projection of summ of forces F_t=m*g*cos(theta)N but F_t=m*a_t so m*a_t=m*g*cos(theta)N m*2*g*sin(theta)=m*g*cos(theta)N N=m*g*(cos(theta)2*sin(theta)) When the sled is on the snowball N can be calculated as shown but N>=0 so cos(theta)2*sin(theta)>=0 or tg(theta)<=1/2 N=0 iff the sled loses contact (there is no glue on the snowball) when it loses contact it is still on the snowball so the formula for N is correct so N=0 <=> tg(theta)=1/2 Am I right now? P.S. Sorry for my english. My physics's english is too bad. ops: 
20030806, 20:19  #8  
"Patrik Johansson"
Aug 2002
Uppsala, Sweden
5^{2}×17 Posts 
Quote:


20030807, 06:37  #9 
Jul 2003
19_{16} Posts 
Patrik is right.
R*sin(theta) is horisontal. Vertical is R*(1cos(theta)). So N=m*g*(3*cos(theta)2). And N=0 <=> cos(theta)=2/3 So theta=arccos(2/3)~=48,1896851042214019341420832694217 deg (Pi Rho's answer is close to that value) Now I'm sure my answer is correct... if there are no more mistakes. 
20030816, 01:35  #10 
Aug 2003
Snicker, AL
1110111111_{2} Posts 
did you ignore the gravity field of the snowball itself?
Fusion 
20030818, 19:17  #11 
Jul 2003
19_{16} Posts 
Yes!
If the snowball is so big that it's gravity field is important than it's radius is very very big (some 100's or even 1000's of kilometers) and g<>9.8m/s/s because it depends on the distance between object and the center of the Earth, and don't forget about rotation of the Earth, the geographical coordinats of the snowball, and the date experiment takes place, ... I'm unable to solve such "puzzle". 
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