View Full Version : downward acceleration greater than g
mmfoley
Jan21-04, 10:29 AM
Hello,
I am taking a Natural and Physical Sciences course. As math and sciences aren't my strongest subjects, I am really struggling, reading things over and over again, and they are just not sinking in. For example, I understand the acceleration of gravity - g. I think that something can have a downward acceleration greater than g, - at least I think this is correct based on my notes, but I do not understand exactly how.
Would this be an example? – When an object first falls it is unbalanced, until the force of air resistance is large enough to balance the acceleration of gravity. So until it balances out, it is falling faster g. Since the object is falling down, this would mean that while unbalanced it has a downward acceleration greater than g.
Help!!
Thank you!
mmfoley
No, that is an example of something falling slower than g.
The most common example of something that is 'falling' faster than g is a rolloer coaster in a loop. With sufficient speed, the roller coaster will be accelerating at faster than g at the top of the loop so the coaster is pushing down on the passengers even though they are upside down.
mmfoley
Jan21-04, 10:40 AM
Thank you! I was at least on the right track, I just got it backwards.
Actually the objeect isn't falling faster than g. It's speed is increasing but g (the accelaration) stays the same. It's no problem to fall faster than g, for example if you throw yourself out of an airplane you will be accelarating at a constant 9.8 m/s^2, one g (If there was no air to provide resistance).
If you have a rocket strapped to your back, point it towards the sky (That is, point the exhaust at the sky) and fire it you'll then be accelarating at a rate of one g plus whatever accelaration the rocket provides.
Likewise, if you point it at the ground and fire it you'll accelarate at a rate of one g minus the accelaration the rocket provides. And if the rocket can provide enough accelaration, you'll stop dead in mid air and possibly start flying up again if the thrust from the rocket is enough.
What you were talking about in your question is terminal velocity, the biggest velocity an object can have before it can't go any faster because of the resistance from the atmosphere on Earth. I don't know how the physics courses are where you're from, but my guess is you don't have to know anything about that, basic classical physics doesn't take into account things like air resistance and stuff like that.
Hope that helped some,
Gunnar.
Edit: Too late. Oh well. :)
russ_watters
Jan21-04, 03:11 PM
Anyone know how to achieve an acceleration due to gravity greater than G on a single, rigid object, with gravity being the only acting force?
Originally posted by russ_watters
Anyone know how to achieve an acceleration due to gravity greater than G on a single, rigid object, with gravity being the only acting force? Rotation?
That would be my guess. If an object were rotating as it fell, the center would fall at g and the acceleration at all the other points would vary as the object rotated.
Cool question Russ.
pallidin
Jan21-04, 05:29 PM
How would rotation affect this? Isn't the net internal effect zero?
Perhaps I am missing something here.
ophecleide
Jan21-04, 08:02 PM
The acceleration of the center of mass would be equal to g, but other parts (such as the tip of a rod) could certainly experience acceleration greater than g. Even if there was not net displacement and the rod was just oscillating back and forth (like a pendulum), parts of the stick would have an accelerationg greater than g at certain points of the oscillation. Does that make sense?
pallidin
Jan21-04, 08:49 PM
Ah, indeed. Thanks.
russ_watters
Jan22-04, 03:30 AM
Originally posted by ophecleide
The acceleration of the center of mass would be equal to g, but other parts (such as the tip of a rod) could certainly experience acceleration greater than g. Even if there was not net displacement and the rod was just oscillating back and forth (like a pendulum), parts of the stick would have an accelerationg greater than g at certain points of the oscillation. Does that make sense? That's the one I was looking for. You guys ever see a large tower fall over? Because of the uneven acceleration, they generally break in half.
A good site with pics: http://aci.mta.ca/TheUmbrella/Physics/P3401/Investigations/ChimneyISM.html
vBulletin® v3.8.7, Copyright ©2000-2012, vBulletin Solutions, Inc.