Is there a REASONABLE explanation for this (physics) phenomenon?

Factual Questions
21

If a jar is full of water, most of the mass won’t rotate; it’s basically mass sliding down a frinctionless slope. If the jar is empty, it’s a solid object rolling down a ramp. Clearly there’s some energy that’s going into rotating the object which, if it weren’t rotating, would go into linear motion.

If that’s not intuitive, try this: imagine you have a cart, and on the cart is a turntable (Lazy Susan) that turns freely. On the turntable is a big tractor tire, so big that the edge of the tire is protruding out the side of the cart. If the edge of that tire is rubbing against the wall (so the tire rotates as you push the cart), does it slow down the cart? Is it easier to push the cart if it isn’t rubbing?

22

I employed the formulas to figure out what the correct numbers are, but it can be thought of without the specific formulas. The one thing you can’t get away from, though, are the basic laws. Conservation of energy is the way I decided to explain it, if you dismiss allowing that concept to be part of the explanation, I don’t think you can explain it. Much of what we think is intuitive is just flat out wrong.

Non-formulaically I can say that rolling objects with lots of mass on the outer rim have more energy than objects with mass near the axis, when both objects are going the same speed and have the same total mass. (the different speeds at different points along the spinning stick correspond to different energies) Therefore, in order for the perimiter weighted jar to be going the same speed as the center weighted jar, it has to have more energy. Conservation of energy says it can’t have more energy after dropping the same distance, because energy isn’t created or destroyed.

23

If most of the water doesn’t rotate as scr4 said, and I overlooked that, then the answer is simple. The torque has increased and the moment of inertia is virtually the same. Hence, increased acceleration.

24

Ok, then let me try this. Say I have two objects. One is a weighted rim connected to a hub with almost invisible and weightless spokes. The other is a disc, same diameter, and it weighs the same. If I try to spin those two objects by twirling them at the center, the solid one will be a little easier to spin, because all of its weight wont be at the outer rim. The rim, being weighted and having its weight all distributed at the outer edge, will be a bit more difficult to get started. I can ALMOST feel that in my inner hand and mind. The disc would probably be a bit easier to twirl. A bit. Am I on the right track? Because that makes a certain amount of intuitive sense, and that’s all I’m looking for. I surely do agree with Cheesesteak that a lot of what’s intuitive is wrong, however. And, of course, that’s the danger I’m running by looking for something like this.

25

If the two have the same mass, the rim will have more inertia and will be more difficult to sping.

Bt the way, this isn’t the same question as was asked in the OP.

26

Imagine this, you’re going to do some cutting.

You cut the disk into 10 concentric rings, each one 10% the mass of the whole, numbered 1 to 10 (inner to outer). Connect each ring to the hub with massless spokes and set them next to each other. Spinning the whole set 1-10 is exactly like spinning the disk. Note that spinning #1 will be easier than spinning #10 because it’s so much closer to the hub.

You cut the Big Ring into 10 identical slices, perpendicular to the hub, so you have 10 rings of the exact same diameter each with 10% the mass of the whole. Connect these rings to the hub with massless spokes and set them next to each other. Spinning the whole set is exactly like spinning the big ring. Note, however, that each of these rings is now identical to ring #10 above, it’s a ring with the maximum diameter and 10% the weight of the whole.

Since the rings #1-9 from the first set are easier to spin than the remaining 9 (#10 sized) rings of the second set, the disk is easier to spin than the big ring.

27

The same torque per unit mass is correct, that it cancels out of the angular acceleration equation is not. Lets look at the equations for simple rotation:

T=Ialpha*

Here is our first problem. Since the jar is both rotating and translating we need to relate translation and rotation (assuming no slippage). Saving the intermediate steps we get:

mgRsin(theta)=(I+m*R[sup]2[/sup])alpha

This is not what makes your cancellation incorrect, as by the way your are solving the problem you would still get mass dropping out.

In the above equation, m is the mass, g is the local acceleration due to gravity, R is the outer radius of the glass, and theta is the angle of the ramp. For this example we are ignoring the ends of the jar and just taking it as a thin walled cylinder. For just the glass we get:

*I=m[sub]g[/sub]R[sup]2[/sup]

m[sub]g[/sub] is the mass of the glass. Entering this into the earlier equations gives us:

gsin(theta)=2Ralpha*

or

alpha=(gsin(theta))/(2R)

In this case the mass cancels. Lets look at what happens when we add the water though.

(m[sub]g[/sub]+m[sub]w[/sub])gRsin(theta)=(I+(m[sub]g[/sub]+m[sub]w[/sub])*R[sup]2[/sup])alpha

where m[sub]w[/sub] is the mass of the water.

Now for the moment of inertia:

*I=m[sub]g[/sub]R[sup]2[/sup] + 0.5m[sub]w[/sub]R[sup]2[/sup]

Note the 0.5 coefficient in front of the water term due to it being a filled cylinder, not just thin-walled like the glass.

This gives us:

I=(m[sub]g[/sub]+0.5m[sub]w[/sub])R[sup]2[/sup]

Thus,

(m[sub]g[/sub]+m[sub]w[/sub])gRsin(theta)=((m[sub]g[/sub]+0.5*m[sub]w[/sub])*R[sup]2[/sup]+(m[sub]g[/sub]+m[sub]w[/sub])*R[sup]2[/sup])alpha

or

alpha=(gsin(theta))/(2R)(m[sub]g[/sub]+m[sub]w[/sub])/(m[sub]g[/sub]+0.75m[sub]w[/sub])

Note how the masses do not cancel out.

Your lead jar comparison is inapt as too many variables change to illustrate the problem well. The masses only cancel when the material is homogenous. Also, you are taking a simplified view of mass by completely divorcing it from geometry. The full equation for finding the moment of inertia, from which the above equations were derived, is an integral of the density over the volume. Both mass and geometric terms are inside the integral, so you see that mass only pops out nicely if the material is of constant density, which a combination of two materials is not.

So we see that the mass itself is not a particularly useful term when trying to explain this phenomenon as we cannot always say if divides out as you aver. It is more understandable, and in fact more correct, to talk about changes in the torque and the inertia relative one another, a relation that is affected both by mass and geometry. Plus, it is a lot easier to understand this way when you can’t use equations.

This has me baffled. Aren’t we already rotating about the center of mass? We are able to use these nice simple equations becuase we are on a principal axis of a simple body. Are you talking about a condition with no translational motion? If so how does the parallel axis theorum come into play?

28

CC: Try this one on for size …
We agree that if we dropped the full & the empty jar side by side from alongside the top of your ramp they’d hit the floor at the same time (same “end-t”) with the same impact velocity (same “end-v”), even though they have different weights/masses. F=ma , d=1/2 at^2 and all that.

Intuitively, each jar “wants” to descend due to gravities’ pull, and does so. One’s heavier, one’s lighter, but gravity’s pull is proportional to weight / mass & so the different masses’ differing inertia (i.e. resistance to acceleration) cancel out & they fall identically. The only acceleration on each jar is dead vertical & down they go.
Now imagine we have a perfectly frictionless ramp. We set the two jars on their BASES at the top & let 'em go. What happens?

They slide like tall hockey pucks down the ramp & arrive at the bottom at the exact same time at the exact same velocity. Same end-t, same end-v. Similar mechanics & formulas as before, BUT …

Each jar still “wants” to descend due to gravity, but the ramp gets in the way. The jars undergo TWO accelerations; one vertical and one horizontal. With the same force available, it takes longer to descend the same vertical distance & the vertical component of the end-velocity is less. The horizontal component of the end-velocity is more & it all balances out nicely as a vector sum; there ain’t no free lunch, even with a frictionless ramp.
If we add friction to our ramp or our jar bases things start to get more complex, but idealized friction being proportional to normal force they’ll both still arrive at the bottom together (i.e. same end-t, same end-v). Agreed?
Now, back to our still-frictionless magic ramp. We lay the jars (assume they’re perfect cylinders) on their sides at the top of the ramp & we let go. What happens?

They SLIDE on their sides to the bottom at the same end-t & same end-v. In fact, the result is identical to the previous experiment where they slid on their bases.

They don’t start rotating, because it takes FRICTION between the jar & ramp to convert the horizontal component of motion into rotation. In fact, on a frictionless ramp, you could spin one of them up to 100,000 RPM in the “uphill” direction, leave the other not spinning at all, and they’d still slide downhill identically side by side. Agreed?
Now for the payoff…

We’re going to lay the jars on their sides on a real friction-equipped ramp & let them ROLL down instead of SLIDE down. In fact, we’re going to posit a ramp with enough friction that there will be ZERO sliding, as if it was a pinion gear on an inclined gear-rack. The only way the jar can move is to rotate such that it’s linear velocity exactly equals the rotational velocity of its surface. (That’s not really a very exotic condition; any wheel that’s not skidding is rotating at a rate that moves the circumference at exactly it’s axle’s linear speed.)

When we do that, there are now THREE accelerations that each jar must perform. Each “wants” to descend due to gravity like always, but to do that it must move horizontally along the ramp. And to do THAT, it now must rotate. And that’s the magic difference.

The empty & full jars have different rotational inertia. In other words, one resists rotational acceleration more than the other does. The resistance is a function of the radial distribution of mass. You’ve bought that point already & it’s pretty easy to explain intuitively in terms of lever arms. For any given value for total weight/mass, a compact mass (steel rod) spins up easily, a flywheel-like mass is harder to spin up, and something that looked like a racing bicycle wheel with a solid lead-filled tire would be very hard to spin up.

Back to our jars.

Each jar has an “engine”, gravity, acting vertically. Each jar’s engine power is proportional to the jar’s mass. So if one masses 5Kg & the other masses 10Kg, the second one has twice as much engine power to push twice as much mass.

But ounce for ounce, which jar has the greater rotational inertia, the greater resistance to rotating, which is a prerequisite for rolling which is a prerequisite for moving along the ramp face which is a prerequisite for moving both horizontally & vertically? That vertical motion is what finally “satisfies” gravity’s “desire”.

Answer: the empty jar has its mass concentrated at the outer edge. The full jar is a compact mass. Ounce-for-ounce, the empty jar is harder to spin up. And ounce-for-ounce is the relevant metric for “engine power” available.

Since the empty jar is harder to spin up, it takes more “power” to do so. Relatively speaking, it’s engine is less powerful than the full jar’s engine is, at least for the task at hand. So it accelerates more slowly, and ends up with a lower end-v after a longer end-t

QED.

Now let’s talk about a different case & see the situation from the other end of the telescope so to speak. That may cement the idea for you.

Imagine a magic world where rotational inertia didn’t exist.

Just like our magic frictionless ramp from a few paragraphs ago, imagine we have magic objects that have no resistance to changing RPM. For magic objects like those, our two jars would roll down the rack-and-pinion ramp just as fast as they could slide down the frictionless ramp. No engine power would be consumed spinning them up, so all of it could be spent accelerating them along the ramp face. They’d have identical end-t, end-v and end-RPM

But given that we don’t live in that world, that we DO have different rotational inertia for different configurations of mass, we’re going to get different results for our two real world jars on real world ramps.

29

There were a few problems with your last post, but most of the conclusions hold.

Actually not, though it will be close. This only holds in a dragless environment (no atmosphere). With atmosphere you have a drag term that is not proportional to mass and therefore mass does not drop out of the equation. With two identically shaped objects the heavier one will hit the ground first.

Again, you need aerodynamic drag. It is correct if you say you are not only on a frictionless ramp, but in dragless air. The rest of the post is a very nice explanation though.

30

Indeed. Very clear and I, for one, appreciate the effort. Except that the piece that I was looking for some assistance with is taken for granted in this section - the issue of greater rotational inertia. I’m not fine with that assertion and I was looking for a way to conceive of it - really, a way to explain it to youngsters - without having to restate it in terms of fomulas or simply to declare it by fiat, which is what happens here when LSLGuy says the empty jar has greater resistiance to motion. That is precisely what is not intuitive and exactly what I’ve been seeking in trying to find a reasonable or familiar analog or example to help get that point across.

By this time, there’s no arguement. It’s just that describing their weight distribution as an explanation for the difference in rotational inertia doesn’t quite do it. It doesn’t seem natural that that difference would automatically be translated into the empty jar having greater inertia. We know it does - all the formulas tell us that. That particular observation supports that, if we know anything about rotational inertia. But what can we experience, or what have we observed that will convince us that’s what is at work in this case? Nome sane?

31

Sorry, I should indeed have specified that I was assuming a homogenous distribution of mass. The OP’s problem didn’t seem to be so specifically bound up with why it happens in this particular case (water in a jar) as with why it happens at all, so I was addressing the most general

My reasoning was that at any given instant, the jar is rotating around the line where it contacts the ramp, so the parallel axes theorem is needed to determine the correct moment of inertia. I was led to that conclusion by the observation that only the point of contact is instantaneously fixed, and thus that must be the instantaneous axis of rotation. I would welcome an explanation of why this is incorrect, though - as there were a couple of good explanations of the specific situation already I was focusing my attention on the OP’s query about why mass distribution affects rotational inertia, so my analysis may not have been as rigorous as it ought to have been.

32

That is a reasonable way to look at it. You will get exactly the same answer I did. I said that you have a combination of translational and rotational motion, with rotation being about the center of mass. Solving these two equations simultaneously leave you with an equation with (I+extra term)alpha instead of Ialpha. That extra term is exactly what you get by using the parallel axis theorum. In fact, I think that is the better way to do it, so kudos.

Let me try to help you with that. When you want to accelerate something you need to apply a force to it. The bigger the mass the bigger the force you need for the same acceleration, or the bigger the mass the less the acceleration for a given force. Our good old F=ma. This much should be intuitive. A given force might accelerate a baseball a great deal while the same force will only make a bowling ball creep along. Mass can be said to be an object’s resistance to acceleration, the bigger the mass, the harder it is to accelerate.

The same thing applies to objects that are spinning. Let us look at a ball tied to the end of a string. The other end is tied to a fixed post. If you move the ball fast it will spin around the post on the end of the string. At any given instant it’s motion is tangential to the circle it spins around (It is not moving towards or away from the center post). If you push on it in this direction of motion it will accelerate faster around the post. This is a simple case of F=ma. The harder you push the more it accelerates. The more it accelerates, the faster it spins around the post.

OK, now lets take a quick sidelight to look at levers. If you have a heavy weight you want to lift you can put it on one end of a seesaw close to the center of rotation and push on the other side of the seesaw. The farther away from the center of rotation that you push, the easier it is to lift the weight. This is because of torque. Torque is force times distance. The same force applied farther from the center of rotation gives more torque (you are increasing the distance term). The same works in reverse. Take that torque and it will impart a force on the weight sitting on the seesaw. The farther the weight is from the center, the less force you put on it for a given torque. This should be pretty intuitive as we have all done something like it before. It is much easier to lift a kid on the other end of the seesaw when he is sitting close to the center than when he is sitting on the very end.

This torque that you are applying can come from you pushing down on the other end, or just twisting the seesaw where it rotates, it will work out the same. So you combine this and see that for a given torque you apply it can impart a small force to a weight far out on the seesaw, or a large force on a weight close in to the center of the seesaw.

Now lets go back to our ball on a string, but make the string that is holding the ball a massless rod. It still spins the same, but this will allow us to put a torque at the center of rotation to make it spin rather than having to push on the ball itself. As we saw with the seesaw, if the ball is far away, then we are putting much less force on the ball than if the ball was on a very short rod. Thus, for a given torque the ball on the shorter rod will accelerate more even though both balls have the same mass. We can say that the ball farther out has a greater resistance spinning, or rotational inertia, than the ball close in.*

Now we expand this to any object. You can consider any object, a jar, the water, whatever, to be an large number of little balls on rods. We can see that if the total mass is the same, than when the balls are farther away from the center of rotation (the jar) they will have a greater resistance to rotation than if they are close to the center (the water).

Even if one object is heavier or denser than another, we can just say that it is composed of more of our little balls. The jar has a whole bunch of balls of mass m located at a distance R from the center of the jar. Each one of these little balls, having the same mass, will impart the same torque (due to the effect of gravity) to try to make the jar roll down the ramp. So long as the jar is symmetric, then it doesn’t matter whether these balls are close to the center of the jar or far away. We add them all up and get a grand total torque. As they are also at the same distance from the center, then they have the same resistance to rolling, or moment of inertia. We can add all of these up to get a grand total moment of inertia as well.

Now we add the water.** The water also consists of a bunch of balls all of the same mass (they are just a little more spread out since the water is less dense). Each of the water balls adds exactly the same amount of torque as the glass balls, since they weigh the same, but as the water balls are closer to the center they each add less rotational inertia than the glass balls did. In fact, the ones right in the center add nothing at all to inertia.***

So if the jar alone accelerated down the ramp at one speed, we see that the jar and water (which now has a lot more torque but only some more rotational inertia) will accelerate more rapidly and reach the bottom sooner.

*I am ignoring the conversion from acceleration to radial acceleration. It does the same thing as the force to torque conversion (adding another factor of R) and would make understanding the example more difficult without equations.

**I am assuming the water all rotates with the jar. In reality it would not at the water-filled jar would move even more quickly.

***Ignoring the parallel axis theorum. The results still work out the same qualitatively.

33

An empty jar has more of its mass moving faster, for the same angular speed, than a full jar of the same total mass. That means it takes more energy to make it spin at the same speed.

For an example that’s easy to imagine or do, take some long, fairly heavy bar (like a crowbar or quarterstaff). Imagine rotating it around its long axis at some angular speed – pretty easy to do, right? It takes very little energy to get it rotating. Now imagine holding it at its center and spinning it around a line perpendicular to its axis, at the same angular speed. This is much harder, right? It’s harder because for the same angular speed you’re making parts of the bar move much faster than in the first case.

So, since it takes more energy to spin the empty jar at a given speed, if you give both empty and full jars the same total energy (e.g., by letting them roll a given distance down a ramp) the empty one has to be rolling more slowly.