mersenneforum.org  

Go Back   mersenneforum.org > Fun Stuff > Puzzles

Reply
 
Thread Tools
Old 2003-08-06, 17:51   #1
eepiccolo
 
eepiccolo's Avatar
 
Dec 2002
Frederick County, MD

37010 Posts
Default 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
.
.
.
.
.
.
.
.
.
.
.
.
eepiccolo is offline   Reply With Quote
Old 2003-08-06, 18:20   #2
Maybeso
 
Maybeso's Avatar
 
Aug 2002
Portland, OR USA

2·137 Posts
Default

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 air-bags (no masses were harmed during this experiment.)

:D
Maybeso is offline   Reply With Quote
Old 2003-08-06, 18:25   #3
c00ler
 
Jul 2003

52 Posts
Default

My answer:
theta=arctg(1/2)
So, theta~= 26,5650511770779893515721937204533 deg

Solution will be posted later.
c00ler is offline   Reply With Quote
Old 2003-08-06, 18:27   #4
eepiccolo
 
eepiccolo's Avatar
 
Dec 2002
Frederick County, MD

2×5×37 Posts
Default

Quote:
Originally Posted by Maybeso
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 air-bags (no masses were harmed during this experiment.)

:)
All correct except for the last. The mass fell to its death. :mrgreen:

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.
eepiccolo is offline   Reply With Quote
Old 2003-08-06, 18:30   #5
eepiccolo
 
eepiccolo's Avatar
 
Dec 2002
Frederick County, MD

2×5×37 Posts
Default

Quote:
Originally Posted by c00ler
My answer:
theta=arctg(1/2)
So, theta~= 26,5650511770779893515721937204533 deg

Solution will be posted later.
You can post your solution anytime you want, because I'm afraid you have the wrong answer.
eepiccolo is offline   Reply With Quote
Old 2003-08-06, 18:46   #6
Pi Rho
 
Pi Rho's Avatar
 
Sep 2002

4410 Posts
Default

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. :(
Pi Rho is offline   Reply With Quote
Old 2003-08-06, 19:12   #7
c00ler
 
Jul 2003

110012 Posts
Default

v-speed 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

N-force 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:
c00ler is offline   Reply With Quote
Old 2003-08-06, 20:19   #8
patrik
 
patrik's Avatar
 
"Patrik Johansson"
Aug 2002
Uppsala, Sweden

52×17 Posts
Default

Quote:
Originally Posted by c00ler
1) m*v^2/2=m*g*R*sin(theta) (energy is not lost)
I think R*sin(theta) is the horisontal displacement, not the vertical.
patrik is offline   Reply With Quote
Old 2003-08-07, 06:37   #9
c00ler
 
Jul 2003

1916 Posts
Default

Patrik is right.
R*sin(theta) is horisontal.
Vertical is R*(1-cos(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.
c00ler is offline   Reply With Quote
Old 2003-08-16, 01:35   #10
Fusion_power
 
Fusion_power's Avatar
 
Aug 2003
Snicker, AL

11101111112 Posts
Default

did you ignore the gravity field of the snowball itself?

Fusion
Fusion_power is offline   Reply With Quote
Old 2003-08-18, 19:17   #11
c00ler
 
Jul 2003

1916 Posts
Default

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".
c00ler is offline   Reply With Quote
Reply

Thread Tools


Similar Threads
Thread Thread Starter Forum Replies Last Post
Irish physics davieddy Science & Technology 26 2011-01-23 17:03
Physics Problem jinydu Homework Help 22 2011-01-15 20:55
Physics problem Please Joshua2 Homework Help 9 2009-03-02 03:39
Physics Nobel. mfgoode Science & Technology 8 2006-10-13 16:18
A physics puzzle hyh1048576 Puzzles 0 2003-09-28 15:14

All times are UTC. The time now is 04:38.

Tue Mar 2 04:38:21 UTC 2021 up 89 days, 49 mins, 0 users, load averages: 1.39, 1.72, 1.83

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2021, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.