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Understanding centripetal force 
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#19
Sep1611, 05:28 AM

P: 614

R = mg  mv^2/r 


#21
Sep1611, 06:38 AM

P: 614

as speed increases, Reaction force decreases so apparent weight decreases



#22
Sep1611, 06:39 AM

P: 614

This matches the answer
could we now go through Q2 (with the change of "space habitat" to "satellite") 


#24
Sep1611, 08:11 AM

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 


#25
Sep1611, 08:14 AM

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P: 41,316

Note that the satellite is in free fall as it orbits. So what's the apparent weight of the people inside? No calculation needed. 


#26
Sep1611, 08:16 AM

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P: 2,316

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 V^{2} / 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. 


#27
Sep1611, 10:10 AM

P: 614

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 


#28
Sep1611, 10:15 AM

P: 614

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 


#29
Sep1611, 10:23 AM

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P: 41,316

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. 


#30
Sep1611, 11:25 AM

P: 614

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 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) 


#31
Sep1611, 12:03 PM

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#32
Sep1711, 02:03 PM

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 


#33
Sep1711, 02:30 PM

P: 614

(please also see last post for the previous question) 


#34
Sep1811, 12:47 AM

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P: 2,316

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. 


#35
Sep1811, 12:55 AM

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P: 2,316

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. 


#36
Sep1811, 03:09 AM

P: 614




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