Rate of fall depends on rotation?

[rzyn]

Homework Statement

A conveyer belt is rolling in a factory on the equator. The top is going east at double the earth's linear velocity and the bottom is stationary relative to the center of the earth. What will be the acceleration due to gravity?

Relevant Equations

F=m×a=mv^2/R => a=v^2/R

v1=460m/s×2, v2=0
m1=m2=5kg
R=6378km
a1=460m×460m/s/s/6378/km/1000/m×km=0.13m/s/s
a2=0
F1=m1×a1=0.66kg×m/s/s
F2=m2×a2=0
G=9.81m/s/s
F3=(m1+m2)×G=98.1kg×m/s/s
(F3-F1)/10/kg = 9.744m/s/s

Where the heck did I go wrong?

 

[berkeman]

Can you attach a sketch of this configuration? I'm having trouble visualizing such a conveyor belt. Thanks.

 

[rzyn]

Can you attach a sketch of this configuration? I'm having trouble visualizing such a conveyor belt. Thanks.

Overall shape is like a long horizontal board, say 10m long, along the equator. Top surface, 5kg is going east with 920m/s linear velocity. Bottom surface, 5kg with 0 linear velocity. Apparent velocities from the ground are 460 m/s east and 460 m/s west, so that the belt is basically just undergoing east-west oscillation in the horizontal plane (and in the equatorial plane). The rollers have negligible radius and weight so that basically the only substantial inertia is contained in the top and bottom surfaces of the belt.

Rate of acceleration if dropped?

 

[haruspex]

What will be the acceleration due to gravity?

Start by defining "acceleration due to gravity". (Or does the question actually ask for the apparent acceleration due to gravity?)

F3=(m1+m2)×G=98.1kg×m/s/s
(F3-F1)/10/kg = 9.744m/s/s

You've lost me. I see that your F1 and F2 are centripetal (or centrifugal) forces, but what is F3?

 

[DaveC426913]

920 m/s?? That's not a conveyor belt, that's a rail gun!

 

[jbriggs444]

As I understand the setup, the total mass of the conveyor belt is 10 kg. That includes 5 kg for the upper surface (eastward-moving) and 5 kg for the lower surface (westward-moving). We neglect the mass of the rollers, tensioners, motors, gears and ironwork holding the assembly together.

The question is: at what rate would the assembly accelerate downward if the floor were removed.

@rzyn, does that match your understanding of the question?

Does it matter if the 10 meter belt is curved to match the surface of the earth?
Does it matter if the 10 meter belt is flat instead?

How accurately would you need to measure the height of the middle of the belt against a line between the ends to distinguish between the two cases?

 

[haruspex]

The question is: at what rate would the assembly accelerate downward if the floor were removed.

In which frame? I.e. how is acceleration due to gravity defined?

 

[rzyn]

In which frame? I.e. how is acceleration due to gravity defined?

The same frame of reference used to arrive at the G (the earth) because the G is used in the problem.

 

[rzyn]

As I understand the setup, the total mass of the conveyor belt is 10 kg. That includes 5 kg for the upper surface (eastward-moving) and 5 kg for the lower surface (westward-moving). We neglect the mass of the rollers, tensioners, motors, gears and ironwork holding the assembly together.

The question is: at what rate would the assembly accelerate downward if the floor were removed.

@rzyn, does that match your understanding of the question?

Does it matter if the 10 meter belt is curved to match the surface of the earth?
Does it matter if the 10 meter belt is flat instead?

How accurately would you need to measure the height of the middle of the belt against a line between the ends to distinguish between the two cases?

The geometry of the masses themselves in relation to the Earth given that R is large and v's are high, is negligible. Only 3 significant digits.

 

[rzyn]

Start by defining "acceleration due to gravity". (Or does the question actually ask for the apparent acceleration due to gravity?)

You've lost me. I see that your F1 and F2 are centripetal (or centrifugal) forces, but what is F3?

Force due to effective gravity in its non-rotating state.

 

[rzyn]

920 m/s?? That's not a conveyor belt, that's a rail gun!

Yeah but only looks half that from earth

 

[haruspex]

Force due to effective gravity in its non-rotating state.

Ok, that is how you are defining F3, but you still have not defined "acceleration due to gravity". Think carefully what that expression means.

 

[rzyn]

Ok, that is how you are defining F3, but you still have not defined "acceleration due to gravity". Think carefully what that expression means.

*effective gravity

 

[BvU]

Where the heck did I go wrong?

Says who ?
Did you render the full problem statement ? Is the 9.81 a given ? 'My' value for G is

 

[rzyn]

Ok, that is how you are defining F3, but you still have not defined "acceleration due to gravity". Think carefully what that expression means.

Does the object fall at the same rate of acceleration as an identical object not containing oscillation?

 

[jbriggs444]

Does the object fall at the same rate of acceleration as an identical object not containing oscillation?

What does Newton's second law say?

 

[rzyn]

Says who ?
Did you render the full problem statement ? Is the 9.81 a given ? 'My' value for G is
View attachment 255081

Ah, I mean little g.

The g has to be adjusted for the eastward mass and westward mass.

But when you do that, because of v^2 the eastward mass will lose more acceleration than the westward mass will gain.

It has to fall at the same g though whether it contains oscillation or not, right?

 

[jbriggs444]

The g has to be adjusted for the eastward mass and westward mass.

Think carefully about that.

There are four places where accelerations matter. Not just two. The accelerations at the ends may not be parallel.

There is a simple observation that can be made about the total change in momentum of a piece of belt around a complete cycle.

 

[BvU]

Ah, I mean little g.

That is obvious from the value you use. Am I correct in assuming that 9.81 m/s² is a given ?
Either way, the other questions remain :

Says who ?
Did you render the full problem statement ? Is the 9.81 a given ?

And I happily add a new one: in #1 the exercise asks

What will be the acceleration due to gravity?

and that could be interpreted as 'for someone riding the belt' (at mach 3 never mind).
In #3 you write

Rate of acceleration if dropped?

as if you want to let the whole thing fall into a hole. A different interpretation altogether. Which is it ?

 

[rzyn]

Think carefully about that.

There are four places where accelerations matter. Not just two. The accelerations at the ends may not be parallel.

There is a simple observation that can be made about the total change in momentum of a piece of belt around a complete cycle.

Hmm. I've thought it over. It appears a craft could not lose weight just by oscillating east-west within.

Now I'm wondering, suppose there were a Ferris wheel instead.

If it were stationary relative to inertial space, it would be heavier?

As it started rotating with the Earth it would be its normal weight.

Then if it started rotating faster than the Earth's rotational rate it would lose weight?

 

[jbriggs444]

Hmm. I've thought it over. It appears a craft could not lose weight just by oscillating east-west within.

Yes. This is correct.

Now I'm wondering, suppose there were a Ferris wheel instead.

If it were stationary relative to inertial space, it would be heavier?

As it started rotating with the Earth it would be its normal weight.

Then if it started rotating faster than the Earth's rotational rate it would lose weight?

Please carefully explain your reasoning. It is not correct.

 

[rzyn]

Think carefully about that.

There are four places where accelerations matter. Not just two. The accelerations at the ends may not be parallel.

There is a simple observation that can be made about the total change in momentum of a piece of belt around a complete cycle.

Yes. This is correct.Please carefully explain your reasoning. It is not correct.

F=mv^2/R
As R increases on a rotating reference frame v increases linearly however v^2 increases parabolically therefore F also increases parabolically for a corresponding point-mass.

A stationary Ferris wheel, simply by being vertical rather than horizontal would have the effective gravity at the top reduced by more than could be explained by altitude alone.

Were the wheel to start rotating top east the effective gravity at the top would be decreasing parabolically with velocity (rate of change in effective gravity with v is 2v/R) while effective gravity at the bottom would be increasing parabolically but at a much lower rate due lower v.

Therefore with effective gravity decreasing at the top but increasing less at the bottom, an increase in velocity would result in an acceleration up against gravity.

But F = m×a, so surely increasing oscillation alone cannot cause acceleration against gravity. Centripetal acceleration about the axel have to be zero-sum. So what gives?

 

[jbriggs444]

F=mv^2/R

Sure. But for what v and for what R?

It is not enough to blindly invoke a formula. You have to make sure it applies to the physical situation. That formula applies to uniform circular motion. You do not have uniform circular motion here.

 

[rzyn]

Sure. But for what v and for what R?

It is not enough to blindly invoke a formula. You have to make sure it applies to the physical situation. That formula applies to uniform circular motion. You do not have uniform circular motion here.

Hmm I thought you could do a calculation for whatever particle happened to be at the very top of the wheel. Sum up the Earth's linear velocity with the wheel's linear velocity to get v. Then add acceleration due to gravity. Then subtract acceleration due to rotation around the earth. The add centripetal acceleration toward the axel.

Then calculate similarly for the particle at the very bottom of the wheel. Centripetal forces about the wheel will sum-zero but the you've got a top point with less effective gravity than the bottom point, essentially stretching the wheel up from the top. Then as the wheel's rotational rate increased, there'd be an upward acceleration against gravity.

That's an errant approach?

 

[jbriggs444]

Hmm I thought you could do a calculation for whatever particle happened to be at the very top of the wheel. Sum up the Earth's linear velocity with the wheel's linear velocity to get v. Then add acceleration due to gravity. Then subtract acceleration due to rotation around the earth. The add centripetal acceleration toward the axel.

Then calculate similarly for the particle at the very bottom of the wheel. Centripetal forces about the wheel will sum-zero but the you've got a top point with less effective gravity than the bottom point, essentially stretching the wheel up from the top. Then as the wheel's rotational rate increased, there'd be an upward acceleration against gravity.

That's an errant approach?

You could take such an approach. But you would need to integrate around the entire circumference. You would need to be careful. The $v$ in $a=\frac{v^2}{r}$ is for uniform circular motion. Again, you do not have uniform circular motion (in the inertial frame). You have cycloidal motion. The obvious way to proceed is to decompose that motion into uniform circular motion of the wheel about the Earth and uniform circular motion of the rim of the wheel about its center.

Then there is an easy observation to make. For every particle on the wheel rim with a [wheel-relative] inward acceleration toward the center of the wheel, there is a similar particle opposite to it with an inward acceleration that is in exactly the opposite direction and with exactly the same magnitude. So the circular motion of the wheel rim about the wheel center is irrelevant.

You would find that there is no unbalanced vertical "stretching" force due to the wheel's rotation. The outward centrifugal force due to the rotation of the wheel is utterly uniform, applying equally in the horizontal direction. The only unbalanced vertical stretching is due to tidal gravity, slightly weaker at the top and higher at the bottom. Slightly rightward on the left and slightly leftward on the right. Tidal centrifugal force from the Earth's rotation contributes uniformly outward in all directions. Of course, tidal gravity is negligible in practice.

Reactionless drives like flying saucers are impossible. You cannot generate net thrust by cleverly spinning wheels, waving hands or gyrating wildly.

 

[haruspex]

*effective gravity

Is that what the question says or just your interpretation?

I asked in post #4 whether it says just "acceleration due to gravity" or whether it is qualified as 'apparent' or 'effective', but you did not answer that.

Then in post #8 you wrote that the it is in "the same frame as used to arrive at g". Well, that frame does not rotate with the earth, so does not include a correction to arrive at apparent gravity.

What I am trying to get you to see is that "acceleration due to gravity", not qualified as 'apparent' or 'effective', would just be g. The spin of the Earth or of the belt would be irrelevant.

That said, @jbriggs444 has shown you that even if it says effective gravity the belt's movement makes no overall difference: it only creates a tension in the belt.

 

[rzyn]

You could take such an approach. But you would need to integrate around the entire circumference. You would need to be careful. The $v$ in $a=\frac{v^2}{r}$ is for uniform circular motion. Again, you do not have uniform circular motion (in the inertial frame). You have cycloidal motion. The obvious way to proceed is to decompose that motion into uniform circular motion of the wheel about the Earth and uniform circular motion of the rim of the wheel about its center.

Then there is an easy observation to make. For every particle on the wheel rim with a [wheel-relative] inward acceleration toward the center of the wheel, there is a similar particle opposite to it with an inward acceleration that is in exactly the opposite direction and with exactly the same magnitude. So the circular motion of the wheel rim about the wheel center is irrelevant.

You would find that there is no unbalanced vertical "stretching" force due to the wheel's rotation. The outward centrifugal force due to the rotation of the wheel is utterly uniform, applying equally in the horizontal direction. The only unbalanced vertical stretching is due to tidal gravity, slightly weaker at the top and higher at the bottom. Slightly rightward on the left and slightly leftward on the right. Tidal centrifugal force from the Earth's rotation contributes uniformly outward in all directions. Of course, tidal gravity is negligible in practice.

Reactionless drives like flying saucers are impossible. You cannot generate net thrust by cleverly spinning wheels, waving hands or gyrating wildly.

Yes I understand that centripetal forces about the axel are zero-sum and need only be considered for determining tensile forces developed in the spokes.

The part I'm having trouble visualizing is how topside east rotation of the Ferris wheel could not exert upward tension through a spoke which would not be balanced by a downward tension in opposite spoke.

If you put a free-to-rotate horizontal bicycle wheel on the floor of a merry-go-round and then start up the merry-go-round, the full mass of the wheel cannot be incorporated into the angular momentum of the merry-go-round because it's not rotating with the ride. If you start braking the wheel to lock in its rotation with that of the merry-go-round, then you can incorporate the mass of the wheel as though it were fused into the floor. And you'll slow down the ride as you brake the wheel against the floor, as angular momentum is being transferred from the ride into the wheel. As angular momentum is transferred from the ride into the wheel, centripetal force toward the center of the ride will also be put into the wheel.

In theory if you spun up the bicycle wheel to tremendous rotational rates, couldn't you extract all the angular momentum of the ride into the wheel, and in the process all the centripetal force of the ride as well? So you'd end up with a stopped ride and a wheel that, were it allowed to slide across a radial path from center, would be ejected from the ride with all the centrifugal acceleration that was once distributed across the ride now concentrated into the wheel? In theory, with a powerful enough motor, one could stop a merry-go-round and eject a wheel, using only torque?

Surely it has to be so according to my visuals. Do the maths disagree?

 

[rzyn]

Is that what the question says or just your interpretation?

I asked in post #4 whether it says just "acceleration due to gravity" or whether it is qualified as 'apparent' or 'effective', but you did not answer that.

Then in post #8 you wrote that the it is in "the same frame as used to arrive at g". Well, that frame does not rotate with the earth, so does not include a correction to arrive at apparent gravity.

What I am trying to get you to see is that "acceleration due to gravity", not qualified as 'apparent' or 'effective', would just be g. The spin of the Earth or of the belt would be irrelevant.

That said, @jbriggs444 has shown you that even if it says effective gravity the belt's movement makes no overall difference: it only creates a tension in the belt.

Hmm maybe the teacher's wording is sloppy.

Judging from follow-up question it seems like he's trying to get us to determine whether its angular momentum in a wheel that could cause weight loss or east-west oscillation alone, or neither.

According to the account holder you referred to, it's neither.

But I'm having a really hard time visualizing how angular momentum with the same rotational orientation as the Earth would not cause weight loss.

As I expressed to that account holder, wouldn't the angular momentum in a wheel cause the Earth's rotation to slow slightly, and thus transfer the centripetal force that used to be distributed across the earth, right into the wheel?

Wouldn't that sort of eject the wheel away from the earth?

 

[jbriggs444]

As I expressed to that account holder, wouldn't the angular momentum in a wheel cause the Earth's rotation to slow slightly, and thus transfer the centripetal force that used to be distributed across the earth, right into the wheel?

This seems to be a different theory than the one you'd adopted earlier. As I understand it, we have a wheel rotating in the same direction as the Earth. By conservation of angular momentum, the rotation of the Earth itself slows significantly in response. That much of the new theory is entirely correct. The rotation of the Earth does slow as the Ferris wheel spins up. Not by much.

Let us take a moment to figure out how big that "not by much" is. We want to compare the angular momentum of the Ferris wheel with that of the Earth.

We have a Ferris wheel which weighs, let us say, 100 metric tons (100,000 kg) and has a radius of 10 meters and which is spun up to something atrocious like 1000 radians per second (10,000 rpm or so). Assume for the sake of argument that the mass is all concentrated at the rim. The angular momentum, $L$, of the Ferris wheel is given by $L=I \omega$. The moment of inertia, $I$, of the Ferris wheel is given by $I=mr^2$. Put it all together and you have: $$L_{\text{wheel}}=I\omega=mr^2\omega=100,000 \times 100^2 \times 1000 = 1 \times 10^{12} \text{ kg m}^2\text{/s}$$
We also have an Earth which weighs about 6 x 10²⁴ kilograms and has a radius of 4000 kilometers. It is spun up to a rotation rate of only about 7 x 10^-5 radians per second. Assume for the sake of argument that its density is uniform. The angular momentum, $L$, of the Earth is given by $L=I\omega$. The moment of inertia of the Earth is given by $I=\frac{2}{5}mr^2$. Put it all together and you have: $$L_{\text{Earth}} = I\omega=\frac{2}{5}mr^2\omega = \frac{2}{5} \times 6 \times 10^{24} \times 4,000,000^2 \times 7 \times 10^{-5}$$ $$= 4.4 \times 10^{33} \text{ kg m}^2\text{/s}$$

The ratio is a factor of 4 x 10²¹ or so. Even if I messed up by a few powers of ten, the Earth still won't slow down by very much. [Edit: Googled it. I was off by only a factor of two]

If the Earth did slow down by much, the effect would be to reduce the centrifugal force associated with the rotating Earth and, therefore, increase the apparent force of local gravity on the Ferris wheel.

This effect makes the spinning wheel heavier than the stationary wheel.

[But not by enough to measure]

thus transfer the centripetal force that used to be distributed across the earth, right into the wheel?

There is no law of conservation of centripetal force. The centripetal force associated with the rotation of the Earth does not have to be redistributed when the Earth slows down. It simply disappears.

 

[rzyn]

This seems to be a different theory than the one you'd adopted earlier. As I understand it, we have a wheel rotating in the same direction as the Earth. By conservation of angular momentum, the rotation of the Earth itself slows significantly in response. That much of the new theory is entirely correct. The rotation of the Earth does slow as the Ferris wheel spins up. Not by much.

Let us take a moment to figure out how big that "not by much" is. We want to compare the angular momentum of the Ferris wheel with that of the Earth.

We have a Ferris wheel which weighs, let us say, 100 metric tons (100,000 kg) and has a radius of 10 meters and which is spun up to something atrocious like 1000 radians per second (10,000 rpm or so). Assume for the sake of argument that the mass is all concentrated at the rim. The angular momentum, $L$, of the Ferris wheel is given by $L=I \omega$. The moment of inertia, $I$, of the Ferris wheel is given by $I=mr^2$. Put it all together and you have: $$L_{\text{wheel}}=I\omega=mr^2\omega=100,000 \times 100^2 \times 1000 = 1 \times 10^{12} \text{ kg m}^2\text{/s}$$
We also have an Earth which weighs about 6 x 10²⁴ kilograms and has a radius of 4000 kilometers. It is spun up to a rotation rate of only about 7 x 10^-5 radians per second. Assume for the sake of argument that its density is uniform. The angular momentum, $L$, of the Earth is given by $L=I\omega$. The moment of inertia of the Earth is given by $I=\frac{2}{5}mr^2$. Put it all together and you have: $$L_{\text{Earth}} = I\omega=\frac{2}{5}mr^2\omega = \frac{2}{5} \times 6 \times 10^{24} \times 4,000,000^2 \times 7 \times 10^{-5}$$ $$= 4.4 \times 10^{33} \text{ kg m}^2\text{/s}$$

The ratio is a factor of 4 x 10²¹ or so. Even if I messed up by a few powers of ten, the Earth still won't slow down by very much. [Edit: Googled it. I was off by only a factor of two]

If the Earth did slow down by much, the effect would be to reduce the centrifugal force associated with the rotating Earth and, therefore, increase the apparent force of local gravity on the Ferris wheel.

This effect makes the spinning wheel heavier than the stationary wheel.

[But not by enough to measure]There is no law of conservation of centripetal force. The centripetal force associated with the rotation of the Earth does not have to be redistributed when the Earth slows down. It simply disappears.

You went through all these calculations and eventually were able to contradict me by a tiny tiny amount [but not enough to matter]. You'll go through efforts to find a tiny tiny effect as long as you're satisfied with it being there.

But you didn't go back to see if the effect you assumed a priori would never matter, would actually end up mattering more than the tiny tiny effect which you celebrated finding.

Maybe as the slowing of the Earth was tiny tiny (tiny-squared), the weight loss of the Ferris wheel was just single-tiny. But you didn't bother calculating it because you started off assuming it wasn't there. You didn't actually calculate that it wasn't there. You just assumed

By the way, centripetal force is proportional to angular momentum, and angular momentum is conserved, so how could centripetal forces just disappear?