# Understanding centripetal force

by jsmith613
Tags: centripetal, force
P: 614
 Quote by Doc Al This is correct. This is not correct. Take the correct equation and solve for R.
sorry
R = mg - mv^2/r
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P: 40,276
 Quote by jsmith613 sorry R = mg - mv^2/r
Good. Now how would you answer the question?
 P: 614 as speed increases, Reaction force decreases so apparent weight decreases
 P: 614 This matches the answer could we now go through Q2 (with the change of "space habitat" to "satellite")
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P: 40,276
 Quote by jsmith613 as speed increases, Reaction force decreases so apparent weight decreases
Good.
 Quote by jsmith613 could we now go through Q2 (with the change of "space habitat" to "satellite")
 P: 614 Q:As the rotational speed of a satellite increases what happens to the apparent weight of the people inside The force, mg, of the space station acts towards the earth F = ma F = mv2/r BUT v = omega (w) * r SO F = m*w2*r F = (GMm)/r2 (GMm)/r2 = m*w2*r If w (omega) increases then Force increases this seems wrong because GMm/r2 should always be constant and therefore not affected by speed :S this is where i get stuck
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P: 40,276
 Quote by jsmith613 Q:As the rotational speed of a satellite increases what happens to the apparent weight of the people inside
I assume they mean orbital speed, not that the satellite is rotating about it's center of mass.

Note that the satellite is in free fall as it orbits. So what's the apparent weight of the people inside? No calculation needed.
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P: 2,315
 Quote by jsmith613 http://www.youtube.com/watch?v=1CD8Q...eature=related could someone please explain what is the forces are here (especially Fn)
I don't like the presenters explanation very much. I look at the problem this way:

When the bucket is overhead, the water will tend to accelerate down with acceleration g, just as it would if there was no bucket there at all.

In order to not have the water coming out of the bucket, you must accelerate the bucket down at g or higher.

You could pull the bucket down with great acceleration, but it will soon hit the ground.

If you rotate the bucket fast enough, the bucket will have acceleration [centripetal acceleration towards the centre] given by V2 / R. Provided that is at least equal to g, we have the bucket accelerating down with a sufficiently large acceleration for the water not to fall out of the bucket.
P: 614
 Quote by Doc Al I assume they mean orbital speed, not that the satellite is rotating about it's center of mass. Note that the satellite is in free fall as it orbits. So what's the apparent weight of the people inside? No calculation needed.
I copied the question exactly from a website. The website has the answer as an increase.
The only force acting on the satellite is mg (towards the earth) and we would use the F = GmM/r^2 to find this force (I presume)

I don't see why no calculation is needed for your interpretation of the Q
P: 614
 Quote by PeterO I don't like the presenters explanation very much. I look at the problem this way: When the bucket is overhead, the water will tend to accelerate down with acceleration g, just as it would if there was no bucket there at all. In order to not have the water coming out of the bucket, you must accelerate the bucket down at g or higher. You could pull the bucket down with great acceleration, but it will soon hit the ground. If you rotate the bucket fast enough, the bucket will have acceleration [centripetal acceleration towards the centre] given by V2 / R. Provided that is at least equal to g, we have the bucket accelerating down with a sufficiently large acceleration for the water not to fall out of the bucket.
i would presume the logic is
if the bucket accelerates fast than the water downwards then the bucket will move before the water has a chance to get out.
My next question is why won't the bucket and the water accelerate at the same speed downward as they are effectively one object
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P: 40,276
 Quote by jsmith613 I copied the question exactly from a website. The website has the answer as an increase.
So they are talking about the rotation of the satellite about its center of mass, not its orbiting around the earth.
 The only force acting on the satellite is mg (towards the earth) and we would use the F = GmM/r^2 to find this force (I presume)
What you need to consider is the force exerted by the rotating satellite (a space 'habitat') on the passenger. That's his apparent weight. The force of the earth on the satellite is not relevant to this question.

Imagine the 'satellite' as a rotating cylinder and a person is standing on the inner surface of that cylinder. As the rotational speed increases, what happens to the force with which he is pressed against the 'floor'?

FYI: 'mg' is only valid near the earth's surface.

 I don't see why no calculation is needed for your interpretation of the Q
Think of astronauts in the space shuttle once they are in orbit. What's their apparent weight?
P: 614
 Quote by Doc Al What you need to consider is the force exerted by the rotating satellite (a space 'habitat') on the passenger. That's his apparent weight. The force of the earth on the satellite is not relevant to this question. Imagine the 'satellite' as a rotating cylinder and a person is standing on the inner surface of that cylinder. As the rotational speed increases, what happens to the force with which he is pressed against the 'floor'? FYI: 'mg' is only valid near the earth's surface.
ok so with a free body diagram,
mg = down
Reaction force = towards centre
he is accelerating towards the centre of the satellite (i presume this is up)
F = ma
ma = R - mg
mv^2/r = R - mg
mg = R + mv^2/r

so as the satellite rotates faster, his weight increases :)

I am confused though as to why on earth the effect of a faster rotation is completley different than on a satellite

 Quote by Doc Al Think of astronauts in the space shuttle once they are in orbit. What's their apparent weight?
well weight = mass * acceleration
I would presume their weight is dependant on how fast the rocket is accelerating
Once in orbit, they must have a net acceleration towards the earth (v^2/r) so the guys weight = mass * (v^2/r)
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P: 40,276
 Quote by jsmith613 ok so with a free body diagram, mg = down Reaction force = towards centre he is accelerating towards the centre of the satellite (i presume this is up) F = ma ma = R - mg mv^2/r = R - mg mg = R + mv^2/r so as the satellite rotates faster, his weight increases :)
The only force on the person that you need to consider is the 'reaction force' from the 'floor' of the satellite.

 I am confused though as to why on earth the effect of a faster rotation is completley different than on a satellite
On the earth you are standing on the outside of the rotating body; in the satellite, you're standing on the inside. (Look up 'artificial gravity'.)

 well weight = mass * acceleration I would presume their weight is dependant on how fast the rocket is accelerating Once in orbit, they must have a net acceleration towards the earth (v^2/r) so the guys weight = mass * (v^2/r)
So I guess you've never see pictures of the astronauts in orbit floating around in their ship?
 P: 614 of course, but they must weight something Although, if your weight is given by the reaction force then floating astronouts have no weight
P: 614
 Quote by Doc Al The only force on the person that you need to consider is the 'reaction force' from the 'floor' of the satellite.

(please also see last post for the previous question)
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P: 2,315
 Quote by jsmith613 My next question is why won't the bucket and the water accelerate at the same speed downward as they are effectively one object
I don't like the way you used speed [I underlined it]!!!!
They do accelerate at the same rate [speed is not a term that relates well to acceleration]. The fact is that acceleration is greater than g, so the only way the water can accelerate at a rate greater than g, is if something applies an extra force to it [gravity is not enough] The thing that applies the extra force is the bucket - but that can only happen if the water remains in contact with the bucket - it doesn't come out.
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P: 2,315
 Quote by jsmith613 of course, but they must weight something Although, if your weight is given by the reaction force then floating astronouts have no weight
Common usage refers to people's weight, and their apparent weight.

Weight is mg - and for an orbiting satellite g might be only 8.7 if the satellite is high enough.

Apparent weight refers to how heavy you feel.

If you ride a roller coaster, you feel light as you go over the top of a hump, you feel heavy as you go through a dip, and who knows how you feel as you go through the inside of an inverted loop, or a spiral.

At all times you mass is the same, and g is 9.8, so your weight is constant, but your apparent weight varies wildly.

When people say an astronaut is weightless, they mean her apparent weight is zero.
P: 614
 Quote by PeterO I don't like the way you used speed [I underlined it]!!!! They do accelerate at the same rate [speed is not a term that relates well to acceleration]. The fact is that acceleration is greater than g, so the only way the water can accelerate at a rate greater than g, is if something applies an extra force to it [gravity is not enough] The thing that applies the extra force is the bucket - but that can only happen if the water remains in contact with the bucket - it doesn't come out.
What I still don't get is that the water is accelerating downwards at the top of the flight, no?

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