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Posted on 4/30/19 at 3:47 pm to tigerpawl
OT Physics Experts and/or Stoners
Assumptions:
1. A perfectly round and smooth globe with a circumference of 25,000 miles. (Image of earth for reference; no surface features; smooth)
2. A yellow string on the surface at the circumference is 25,000 miles long.
3. A large red metal hoop sits directly over the yellow string, but 6 foot above the surface on top of wooden support poles all the way around the globe.
4. Simultaneously each wooden support pole is kicked out from under the red metal hoop at exactly the same time.
Question: What happens to the red metal hoop???
Assumptions:
1. A perfectly round and smooth globe with a circumference of 25,000 miles. (Image of earth for reference; no surface features; smooth)
2. A yellow string on the surface at the circumference is 25,000 miles long.
3. A large red metal hoop sits directly over the yellow string, but 6 foot above the surface on top of wooden support poles all the way around the globe.
4. Simultaneously each wooden support pole is kicked out from under the red metal hoop at exactly the same time.
Question: What happens to the red metal hoop???
This post was edited on 4/30/19 at 3:48 pm
Posted on 4/30/19 at 5:21 pm to tigerpawl
25,006.283185
Posted on 4/30/19 at 5:26 pm to TigerstuckinMS
quote:
Are you fricking kidding me? The OT is stupid. It can't figure out PEMDAS.
L1 = 2 * pi * r1
L2 = 2 * pi * r2
r2 = r1 + 1 ft
L2 = 2 * pi * (r1 + 1 ft)
L2 - L1 = [2 * pi * (r1 + 1 ft)] - [2 * pi * r1]
L2 - L1 = (2 * pi) (r1 + 1 ft - r1)
L2 - L1 = (2 * pi) ft
L2 - L1 ~ 6.2 ft
I came up to that answer entirely different and rather easily. Plus I'm high as a kite right now off some cannabis.
Just find the diameter of earth, then add 2, then use that to find the circumference and then subtract the two, which I did forget to do but thought he wanted just the circumference and not the difference.
Posted on 4/30/19 at 5:28 pm to That's BS
quote:
Question: What happens to the red metal hoop???
It is inside the minimum stable orbit, so eventually it must contact the ground. Though it is in an unstable equilibrium initially, any perturbation, no matter how small, destroys the equilibrium and it will eventually contact the planet.
This post was edited on 4/30/19 at 5:33 pm
Posted on 4/30/19 at 5:30 pm to That's BS
quote:
OT Physics Experts and/or Stoners
Assumptions:
1. A perfectly round and smooth globe with a circumference of 25,000 miles. (Image of earth for reference; no surface features; smooth)
2. A yellow string on the surface at the circumference is 25,000 miles long.
3. A large red metal hoop sits directly over the yellow string, but 6 foot above the surface on top of wooden support poles all the way around the globe.
4. Simultaneously each wooden support pole is kicked out from under the red metal hoop at exactly the same time.
Question: What happens to the red metal hoop???
Posted on 4/30/19 at 5:31 pm to dawgsjw
quote:
Just find the diameter of earth, then add 2, then use that to find the circumference and then subtract the two
That's all I did. I just showed my work.
This post was edited on 4/30/19 at 5:33 pm
Posted on 4/30/19 at 5:36 pm to TigerstuckinMS
quote:meh. That's just way too much
Are you fricking kidding me? The OT is stupid. It can't figure out PEMDAS.
L1 = 2 * pi * r1
L2 = 2 * pi * r2
r2 = r1 + 1 ft
L2 = 2 * pi * (r1 + 1 ft)
L2 - L1 = [2 * pi * (r1 + 1 ft)] - [2 * pi * r1]
L2 - L1 = (2 * pi) (r1 + 1 ft - r1)
L2 - L1 = (2 * pi) ft
L2 - L1 ~ 6.2 ft
pi(D+2)-piD=pi(D+2-D)=2pi
Posted on 4/30/19 at 6:35 pm to PrivatePublic
quote:
Length of the first string +2pi, Bob
FIFY
I think you missed the reference
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