View Full Version : importance of inertia for rotation in a gravity field?

Nicolas

2009-Jan-29, 03:05 PM

A bit more general than space and astronomy, but Newtonean physics and gravity nonetheless:

Imagine a horizontal structure, supported at both ends. We kick away one of the supports. Now it is supported at one end only only and starts to drop by rotating around that point.

Does the inertia of that structure influence how fast it rotates/its rotation accelerates when dropping due to a uniform gravity field? Or is the only thing determining the rotational speed (acceleration) the moment of unbalance around the pivot point?

I know that the moment of inertia determines how large the force is you need to apply at the end of the structure to change its rotational velocity, but somewhere in the back of my mind something tells me that there was something tricky about the influence of uniform gravitational fields...

hhEb09'1

2009-Jan-29, 03:13 PM

Does the inertia of that structure influence how fast it rotates/its rotation accelerates when dropping due to a uniform gravity field? Or is the only thing determining the rotational speed (acceleration) the moment of unbalance around the pivot point?It sounds like those two alternatives are the same thing!

But, yes, something that is heavily weighted towards the pivot point will fall/turn differently than something that is weighted towards the opposite end. If the weight is concentrated towards the pivot end enough (so that the center of mass is at the pivot), the object won't move at all.

Smoke Ring

2009-Jan-29, 03:22 PM

The boards center of gravity is the only determining factor and of course the strength of G.

hhEb09'1

2009-Jan-29, 03:35 PM

The boards center of gravity is the only determining factor and of course the strength of G.Hmmm, lets put 2 kg masses at each of the ends of a lightweight bar one meter long, with the pivot at one end.

Compare that to the same bar, with the same pivot, with a 4 kg mass in the center.

I haven't worked this through yet.

Nicolas

2009-Jan-29, 03:51 PM

The pivot point doesn't have to be at the outer edge of the structure, it may also be somewhere close to halfway.

So, does it matter if you've got 1kg at 1m to the right of the pivot and 2kg at 1m to the left, versus 1000kg at 1m to the right versus 1001kg at 1m to the left?

hhEb09'1

2009-Jan-29, 05:39 PM

The pivot point doesn't have to be at the outer edge of the structure, it may also be somewhere close to halfway.But, in the OP, you did say the two supports were at each end, and one gets kicked away. :)

But, it does look like you're asking a general question about moment of inertia.

So, does it matter if you've got 1kg at 1m to the right of the pivot and 2kg at 1m to the left, versus 1000kg at 1m to the right versus 1001kg at 1m to the left?The center of gravity is going to be a lot closer to the pivot in the second example, that makes some difference.

Nicolas

2009-Jan-29, 06:24 PM

I know my addition is against the description in the OP. I thought it wasn't confusing enough as it was. ;)

Good point about the CG.

OK, if anyone can come up with a situation where only the inertia is different (if such a situation exists), can (s)he tell me whether it would matter for the rotational acceleration due to a uniform gravity field only?

I'm sorry, my mind is completely filled with largely unrelated things I've been working on for 2 weeks straight now, but I've got this issue in between. I can't seem to focus on it. I just have this thing in the back of my mind about uniform gravity fields and exceptions...

grav

2009-Jan-30, 05:07 AM

The center of gravity is going to be a lot closer to the pivot in the second example, that makes some difference.Yes, but there is also much more force acting at that distance from the pivot due to the greater weight, so it applies the same overall torque and turns in the same way. The two examples you gave earlier in post #4 are also the same.

Jeff Root

2009-Jan-30, 06:04 AM

I haven't looked at it, but I expect that the Wikipedia article on pendulums

(pendula?) tells everything about what you're asking.

-- Jeff, in Minneapolis

grav

2009-Jan-30, 09:32 AM

Yes, but there is also much more force acting at that distance from the pivot due to the greater weight, so it applies the same overall torque and turns in the same way. The two examples you gave earlier in post #4 are also the same.Oops. Something was bugging me about this, and it looks like I was wrong, way wrong. The torque is the same in each of the examples, but the moment of inertia definitely makes a difference in the resulting angular acceleration. The angular acceleration is equal to the torque divided by the moment of inertia, or (sum of m g r) / (sum of m r^2). Multiplying that result by some distance from the pivot gives the acceleration at that point. Of course, that's just the initial acceleration, since as the lever bar begins to turn, the acceleration changes according to the angle between the lever bar and the direction of gravity as well. It might be a bit easier to think about a single pulley geared for different widths (or cone-shaped) with weights falling from ropes constantly without changing angles, rather than that of a lever or pendulum.

In the example Hh gave, a 4 kg mass placed at .5 meter from the pivot will still accelerate at 1g at that point. So at a distance of 1 meter, the acceleration is 2g. The 2 kg mass placed at the pivot adds nothing in terms of torque or inertia, and the other 2kg mass accelerates at 1g at the point it is placed at 1 meter, therefore with half of the angular acceleration of the 4 kg mass placed at half the distance (or any amount of mass placed there and none elsewhere). In Nicolas's example, the distances are the same and so is the torque, so the angular acceleration only depends upon the inertial values with a ratio of 3 to 2001, and the angular acceleration of the 1 and 2 kg masses, then, is 667 times greater than that of the 1000 and 1001 kg masses. The acceleration of the 1 and 2 kg masses would be the same as that of 500 and 1000 kg masses, though.

hhEb09'1

2009-Jan-30, 11:04 AM

OK, if anyone can come up with a situation where only the inertia is different (if such a situation exists), I think I did in post #4: :)

Hmmm, lets put 2 kg masses at each of the ends of a lightweight bar one meter long, with the pivot at one end.

Compare that to the same bar, with the same pivot, with a 4 kg mass in the center.It looks like grav has a start on the analysis. Does that answer your concerns in the OP?

Nicolas

2009-Jan-30, 12:28 PM

I think it does. Summarized: the moment of inertia will have an influence on the angular acceleration, the rest being equal (as far as possible if you want a different moment of inertia).

Thanks!

grav

2009-Jan-30, 06:56 PM

Summarized: the moment of inertia will have an influence on the angular acceleration, the rest being equal (as far as possible if you want a different moment of inertia).I'm quite not sure what you mean by "the rest being equal", but equally distributed weights will produce a different moment of inertia also, as demonstrated with your example. Even the moment of inertia of the lever bar must be taken into account, so a more accurate formula, considering the moment of inertia of the lever bar separately from the weights, would be (a / r) = (sum of m g r) / ( Ibar + sum of m r^2). For a lever bar (very thin and wire-like) with a pivot at its center, Ibar-center = mbar L^2 / 12, and pivoted at the end is Ibar-end = mbar L^2 / 3, found here (http://en.wikipedia.org/wiki/List_of_moments_of_inertia). Of course, with a lever bar pivoted at the end (or anywhere but its center of mass), the torque of the bar must be added in as well. For a bar of some width that is pivoted at the center, it looks like it would become I = m (L^2 + w^2) / 12, using the cuboid example in the link, and a cylindrical bar pivoted at the center would be I = m (L^2 + 3 radius^2) / 12.

mugaliens

2009-Jan-30, 08:46 PM

It sounds like those two alternatives are the same thing!

But, yes, something that is heavily weighted towards the pivot point will fall/turn differently than something that is weighted towards the opposite end. If the weight is concentrated towards the pivot end enough (so that the center of mass is at the pivot), the object won't move at all.

The beam can also be equally weighted at the ends, and infinitely light along the middle. That gives it the greatest moment of inertia, and thus, the slowest angular acceleration. In short, the end will fall with an initial acceleration equal to that of an independantly dropped object.

On the other hand, if the weight is concentrated in the middle, and infinitely light from the middle outward, it's moment of inertia is much less, and it's angular acceleration will be greater. The end will fall with an initial acceleration equal to twice that of an independantly dropped object.

grav

2009-Jan-31, 08:38 PM

I need to make a couple more quick corrections on my part.

The 2 kg mass placed at the pivot adds nothing in terms of torque or inertia, ...This is only true if all of the mass is placed precisely at the pivot point. If the mass's distribution extends a little bit to each side of the pivot point, then it will add inertia, and slow the overall angular acceleration of masses placed elsewhere. And the further out two equal masses are placed at equal distances from the pivot, the slower the angular acceleration becomes. However, if the two masses are slightly different but placed at equal distances, as with Nicolas's example, the overall angular acceleration is less the further out they placed, but the acceleration at the point they are placed remains the same. So if we replaced the lever bar with a pulley, and a 101 kg mass is placed on one side and a 100 kg man on the other, the man will accelerate upward at the same rate regardless of the radius of the pulley, which works out to a rate of (mmass - mman) / (mmass + mman) * g = g / 201. The rate of acceleration can be adjusted by changing the difference between the masses. For example, with a 100.001 kg mass and 100 kg man, the man will ascend at a rate of acceleration of g / 200001. This is all neglecting friction and the moment of inertia of the pulley itself. For a pulley, the moment of inertia is I = mpulley r^2 / 2.

For a bar of some width that is pivoted at the center, it looks like it would become I = m (L^2 + w^2) / 12, ...This is the other correction. The first few pictures in the link had the height of the figures going from top to bottom of the page, so I assumed it should be the same with the cuboid, making it the width of the lever bar in this context. But I'm thinking it doesn't make sense that the moment of inertia should depend upon the width of the bar if the pivot contact is to span that width. All moments should be taken as the distance from the pivot axis in that case, I would think, so the formula for that becomes I = m (L^2 + D^2) / 12, where D is the depth of the lever bar. To help verify this, the next picture in the link shows the rotation of a flat plane, with reference to the height in the formula for the moment of inertia, but it would have no height if that were to run from top to bottom of the page, so height definitely becomes depth in this case.

Also, if the moment of inertia depends upon the depth of the lever arm, then so does that of the masses, and by how much higher they are placed from the actual axis of rotation. If the axis of rotation is placed at the top of the lever bar, then "flat" masses acting along the same horizontal line will be the same as we have been doing, although the moment of inertia of the lever bar will also change from that placed through the center which is what the earlier formula is based upon. If the masses also have some height, then, that changes the moment of inertia accordingly for them as well, then. I'm unsure about whether it would change the torque too, but probably so.

grav

2009-Jan-31, 08:50 PM

Sorry, guys. I didn't mean to get all 3D on you. :)

Nicolas

2009-Feb-02, 09:29 AM

Someone on another board still claims that it pivots at the same speed because "ignoring air resistance, all objects fall at the acceleration of gravity". He refuses to take the rotational inertia into account. Luckily nature doesn't refuse whenever it doesn't like something :).

hhEb09'1

2009-Feb-02, 11:10 AM

Someone on another board still claims that it pivots at the same speed because "ignoring air resistance, all objects fall at the acceleration of gravity". He refuses to take the rotational inertia into account. Luckily nature doesn't refuse whenever it doesn't like something :).Maybe have them work through this example, I like Mugs' response to it.

Hmmm, lets put 2 kg masses at each of the ends of a lightweight bar one meter long, with the pivot at one end.

Compare that to the same bar, with the same pivot, with a 4 kg mass in the center.

I haven't worked this through yet.

The beam can also be equally weighted at the ends, and infinitely light along the middle. That gives it the greatest moment of inertia, and thus, the slowest angular acceleration. In short, the end will fall with an initial acceleration equal to that of an independantly dropped object.

On the other hand, if the weight is concentrated in the middle, and infinitely light from the middle outward, it's moment of inertia is much less, and it's angular acceleration will be greater. The end will fall with an initial acceleration equal to twice that of an independantly dropped object.In other words, in one situation, you have a weight at the middle of the bar falling, and in the other it is at the end of the bar falling. They both fall under the influence of gravity (to start) but the ends of the bars will be falling at different rates.

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