Physics Sledding Problem

eepiccolo · 2003-08-06, 17:51

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
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Maybeso · 2003-08-06, 18:20

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

c00ler · 2003-08-06, 18:25

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

Solution will be posted later.

eepiccolo · 2003-08-06, 18:27

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 · 2003-08-06, 18:30

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.

Pi Rho · 2003-08-06, 18:46

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. :(

c00ler · 2003-08-06, 19:12

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:

patrik · 2003-08-06, 20:19

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.

c00ler · 2003-08-07, 06:37

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.

Fusion_power · 2003-08-16, 01:35

did you ignore the gravity field of the snowball itself?

Fusion

c00ler · 2003-08-18, 19:17

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".

2003-08-06, 17:51	#1
eepiccolo Dec 2002 Frederick County, MD 2·5·37 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 . . . . . . . . . . . .

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2003-08-06, 18:20	#2
Maybeso 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 air-bags (no masses were harmed during this experiment.) :D

2003-08-06, 18:25	#3
c00ler Jul 2003 5² Posts	My answer: theta=arctg(1/2) So, theta~= 26,5650511770779893515721937204533 deg Solution will be posted later.

2003-08-06, 18:46	#6
Pi Rho Sep 2002 2²×11 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. :(

2003-08-06, 19:12	#7
c00ler Jul 2003 5² Posts	v-speed of sled 1) mv^2/2=mgRsin(theta) (energy is not lost) so 2gRsin(theta)=v^2 and v^2/R=2gsin(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=mgcos(theta)-N but F_t=ma_t so ma_t=mgcos(theta)-N m2gsin(theta)=mgcos(theta)-N N=mg(cos(theta)-2sin(theta)) When the sled is on the snowball N can be calculated as shown but N>=0 so cos(theta)-2sin(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:

2003-08-07, 06:37	#9
c00ler Jul 2003 5² Posts	Patrik is right. Rsin(theta) is horisontal. Vertical is R(1-cos(theta)). So N=mg(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.

2003-08-16, 01:35	#10
Fusion_power Aug 2003 Snicker, AL 3BF₁₆ Posts	did you ignore the gravity field of the snowball itself? Fusion

2003-08-18, 19:17	#11
c00ler Jul 2003 5² 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".