Require help with angular velocity and people flying off the planet

In summary: You could use a test mass of ##100kg## if you...If you want to use a test mass, you could use one of those. They are quite accurate.
  • #1
RemotePhysics
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Homework Statement
The diameter of earth is 12,700 km. Calculate :
a> The angular velocity at which earth is spinning
b> The speed at which someone standing on the equator is rotating relative to the pole

Weighting scales do not measure your mass, just the reaction force R that the scales push back on you.
c> Write an equation for the centripital force F in terms of R and weight W for someone rotating
on earth
d> Hence show that the scales would give a reading of approximately 99.7% of the person's
actual mass (ignoring any inacuracy due to the scales)
e> How fast in m/s, would the earth have to be spinning, for someone at the equator to "fly off"?
How long would a day be in this situation?
Relevant Equations
w=v/r
F = W - R ???
My solutions (attempts) :
a> w=v/r | r=6.35x10^6m | therefore V=7.04x10^-5 m/s
b> speed of rotation is faster at the equator than the pole as w=v/r. As w remains constant, as r increases towards the pole V has to decrease.
c> F = W - R

d> Stuck here. I presume that I have to use the equation from part C and gravity must be a factor but I am unsure of how to work this one out!
e> No idea how to work this out? Am I meant to google an equation for speed of rotation and the length of a day!? What would you do?
 
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  • #2
For a) you need the angular velocity ##\omega##. For b) you need the speed ##V##. For c) I think they require more than that.

I think you are expected to know the length of a day on Earth. Although if you want a really accurate answer you could look it up.
 
  • #3
This problem is not much different than the one for the car on the curve of your previous thread.
From that exercise you now know how to calculate the "centrifugal force", which will try to move the body away from the center of the spinning planet (no spinning, no centrifugal effect) and will eventually overcome the gravity force that tries to move the body towards the center of the planet.

Before problems become more complex, it is time to learn about free body diagram:
https://en.wikipedia.org/wiki/Free_body_diagram

Please, take a look at how angular and tangential velocities, as well as centripetal acceleration are related to distance from the axis of rotation:
http://hyperphysics.phy-astr.gsu.edu/hbase/rotq.html

This is bonus information, that is relevant to this problem, just in case you have time to read it:
https://en.wikipedia.org/wiki/Weightlessness

:cool:
 
  • #4
Lnewqban said:
This problem is not much different than the one for the car on the curve of your previous thread.
From that exercise you now know how to calculate the "centrifugal force", which will try to move the body away from the center of the spinning planet (no spinning, no centrifugal effect) and will eventually overcome the gravity force that tries to move the body towards the center of the planet.

Before problems become more complex, it is time to learn about free body diagram:
https://en.wikipedia.org/wiki/Free_body_diagram

Please, take a look at how angular and tangential velocities, as well as centripetal acceleration are related to distance from the axis of rotation:
http://hyperphysics.phy-astr.gsu.edu/hbase/rotq.html

This is bonus information, that is relevant to this problem, just in case you have time to read it:
https://en.wikipedia.org/wiki/Weightlessness

:cool:
Thank you very much for this in-depth answer! Other than the standard equations for circular motion, centripital force, circular velocity, and linear velocity, are the any other that I should be using?
 
  • #5
PeroK said:
For a) you need the angular velocity ##\omega##. For b) you need the speed ##V##. For c) I think they require more than that.

I think you are expected to know the length of a day on Earth. Although if you want a really accurate answer you could look it up.
What do you think is required for C? I am struggling to think of what information may be relavent. This question is odd as according to my spec, my exam questions will contain all relavent figure and information outside what I am expected to learn. Its pretty odd that the question expect me to start googling figures that the exam board will provide!
 
  • #6
RemotePhysics said:
Thank you very much for this in-depth answer! Other than the standard equations for circular motion, centripital force, circular velocity, and linear velocity, are the any other that I should be using?
You are welcome :smile:
Those will suffice.
 
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  • #7
RemotePhysics said:
What do you think is required for C? I am struggling to think of what information may be relavent.
Perhaps you are right and ##F = W - R## is all they wanted.
 
  • #8
PeroK said:
Perhaps you are right and ##F = W - R## is all they wanted.
So i am working through part D, using the formula (mv^2/r)=W-R.
Just wondering if I am suppose to use a made up mass and then show that the reading would be around 99.7% of it. Don't see any other way to get a value for W. Any chance you could work this out so we can compare?
 
  • #9
RemotePhysics said:
So i am working through part c, using the formula (mv^2/r)=W-R.
Just wondering if I am suppose to use a made up mass and then show that the reading would be around 99.7% of it. Don't see any other way to get a value for W. Any chance you could work this out so we can compare?
You could use a test mass of ##100kg## if you want.
 
  • #10
PeroK said:
You could use a test mass of ##100kg## if you want.
Yep, will do. How does gravity factor into all of this?
 
  • #11
RemotePhysics said:
Yep, will do. How does gravity factor into all of this?
I suspect you are supposed to know the surface gravity of the Earth - at least approximately.
 
  • #12
PeroK said:
I suspect you are supposed to know the surface gravity of the Earth - at least approximately.
The value should be 9.81 according to my specification. Just wondering how to include it in this equation?
 
  • #13
RemotePhysics said:
The value should be 9.81 according to my specification. Just wondering how to include it in this equation?
In what way is there a problem?
 
  • #14
PeroK said:
In what way is there a problem?
What do you mean? I was asking where should gravity be placed in the equation that I was using?
Also with a value of 100kg, (mv^2/r) is giving me 7.78x10^-14=W-R which is wrong i think? How would you work this question out?
 
  • #15
RemotePhysics said:
What do you mean? I was asking where should gravity be placed in the equation that I was using?
Also with a value of 100kg, (mv^2/r) is giving me 7.78x10^-14=W-R which is wrong i think? How would you work this question out?

RemotePhysics said:
My solutions (attempts) :
a> w=v/r | r=6.35x10^6m | therefore V=7.04x10^-5 m/s

Maybe this is the problem. Do you really mean ##V = 7 \times 10^{-5}m/s##?
 
  • #16
PeroK said:
Maybe this is the problem. Do you really mean ##V = 7 \times 10^{-5}m/s##?

Aha, 7 × 10-5 is in rad/s not m/s! Does this mean that V should be the linear velocity of earth?

EDIT: I think that I have it! Thank you very much!
 
  • #17
RemotePhysics said:
... part D, using the formula (mv^2/r)=W-R.
Just wondering if I am suppose to use a made up mass and then show that the reading would be around 99.7% of it. Don't see any other way to get a value for W. Any chance you could work this out so we can compare?
Just call the mass M or something similar.

I don't know about other people, but I find your lack of working makes it difficult to follow.
If I get different answers from you, I don't know if you've used different values or done a different method, which may be not right.

(a) is not far from my answer, but I can't see why. But you may have used a different value for the period of rotation. If you showed your working, I could use the same data as you.
(b) puzzled me. It gave me a numerical value in m/sec , but you gave a verbal description, no numbers.
(c) I agree.
(d) I just used M for mass. Then gravity comes in automatically when convert it to weight.

The question setter deliberately tries to confuse you here by saying,
" Weighting scales do not measure your mass, just the reaction force R " which is true.
Then he says, "... the scales would give a reading of ... actual mass ... "

So, although they measure weight, they give a reading of mass. So you have to be very careful about whether R refers to mass or wieght.

When you say "(mv^2/r)=W-R." you are clearly using R as a force in Newtons.
So to answer the question you will either need to convert this to ##R_m## , the mass reading that the scales give, to compare with the real mass M, or compare R with the real weight W

With the level of working you show, I can't tell whether you are understanding this issue or not.
If you show working, we can tell the precise point where you get stuck or go wrong (and probably, you will be able to tell as well!)
 
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  • #18
Merlin3189 said:
The question setter deliberately tries to confuse you here by saying,
" Weighting scales do not measure your mass, just the reaction force R " which is true.
Then he says, "... the scales would give a reading of ... actual mass ... "

While we're on the subject, the question setter may be conceptually confused themselves and getting into a tangle trying to conceptually distinguish one force from another. In my view, the whole concept of weight is superfluous. There is mass and there is force. Why do we need a third concept?
 
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  • #19
@Merlin3189 @PeroK
Here are the images of the questions and my working out thusfar. Whats your thoughts on the wording of the questions? Methinks that they are not clear as they could be...
20200707_140106.jpg
20200707_140046.jpg


Hope that this makes more sense! :)
 
  • #20
For part a) you can get the angular velocity directly from the period (one day). The answer to part a) should be in radians per second.

The answer to part b) should be the speed you found in a).

Part c) is correct with ##W = mg##, of course, where ##g = \frac {GM}{R^2}##. The point I would make is that we measure ##g## on a spinning Earth! So, what is ##g##? Is ##g## what we would measure on a non-spinning Earth or is it what we actually measure on a spinning Earth (and varying from place to place)? That's a question for the question setter. And, likewise, how is weight actually defined on the surface of the Earth?

Part d) looks right. But, again, the ##9.81m/s^2## is what we measure on a spinning Earth. The actual gravity ought to be something like ##9.84 m/s^2##, 99.7% of which is what we measure.
 
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  • #21
PeroK said:
For part a) you can get the angular velocity directly from the period (one day). The answer to part a) should be in radians per second.

The answer to part b) should be the speed you found in a).

Part c) is correct with ##W = mg##, of course, where ##g = \frac {GM}{R^2}##. The point I would make is that we measure ##g## on a spinning Earth! So, what is ##g##? Is ##g## what we would measure on a non-spinning Earth or is it what we actually measure on a spinning Earth (and varying from place to place)? That's a question for the question setter. And, likewise, how is weight actually defined on the surface of the Earth?

Part d) looks right. But, again, the ##9.81m/s^2## is what we measure on a spinning Earth. The actual gravity ought to be something like ##9.84 m/s^2##, 99.7% of which is what we measure.

Thanks, this clears up a lot of my questions!

What about part e? I would assume that a person would fly off Earth when gravity would be equal to 0. Therefore I must find the speed at which gravity is equal to zero? This is purely my own thoughts as there is nothing on this in my syllabus! The issue with this is that I cannot find any formula in my text-book that would work for this.

Another idea is that a person may fly off Earth when the radius (r) in the equation V=wr (angular velocity) is larger than the radius of earth. My reasoning is that it a person cannot fly off the Earth when the radius is zero as V=w*0=0.

I cannot seem to find a speed for Earth's rotation at which a person would fly off online else I would have given my forumla's a go and see if they matched up.

Whats your thoughts on this?
 
  • #22
RemotePhysics said:
Thanks, this clears up a lot of my questions!

What about part e? I would assume that a person would fly off Earth when gravity would be equal to 0. Therefore I must find the speed at which gravity is equal to zero? This is purely my own thoughts as there is nothing on this in my syllabus! The issue with this is that I cannot find any formula in my text-book that would work for this.

Another idea is that a person may fly off Earth when the radius (r) in the equation V=wr (angular velocity) is larger than the radius of earth. My thinking here is that it a person cannot fly off the Earth when the radius is zero as V=w*0=0.

Whats your thoughts on this?
There's no conceptual problem here. The faster the Earth spins, the less effective gravity we feel. Eventually, if it spins fast enough gravity isn't enough to keep up on the surface.
 
  • #23
PeroK said:
There's no conceptual problem here. The faster the Earth spins, the less effective gravity we feel. Eventually, if it spins fast enough gravity isn't enough to keep up on the surface.
Trying to think of which formula I would use for this... I suppose it must angular velocity (w) and gravity (g)? Maybe the centripital force equation F=(mv^2)/r where F=0?
 
  • #24
RemotePhysics said:
Trying to think of which formula I would use for this... I suppose it must angular velocity (w) and gravity (g)? Maybe the centripital force equation F=(mv^2)/r where F=0?
The equation is simply ##g = a_c##, where ##g## is the purely gravitational acceleration at the surface of the Earth and ##a_c## is the centripetal acceleration of the rotation.

That should be clear.
 
  • #25
PeroK said:
The equation is simply ##g = a_c##, where ##g## is the purely gravitational acceleration at the surface of the Earth and ##a_c## is the centripetal acceleration of the rotation.

That should be clear.

This is a tad odd as there is no mention of this in my course. I suppose that what you are saying is that when when the centripital acceleration > gravitational acceleration then the person would fly off?
 
  • #26
RemotePhysics said:
This is a tad odd as there is no mention of this in my course. I suppose that what you are saying is that when when the centripital acceleration > gravitational acceleration then the person would fly off?
You are supposed to be able to apply your knowledge to a range of problems. In this case you are expected to work out for yourself that condition, using your knowledge of forces.
 
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  • #27
RemotePhysics said:
This is a tad odd as there is no mention of this in my course. I suppose that what you are saying is that when when the centripital acceleration > gravitational acceleration then the person would fly off?
That's right, though I might say that, when
the force needed to give the required centripetal acceleration > the force provided by gravity
then the person will not be held in a circular path on the surface of the earth.

Changing to refer to accelerations, answers your earlier problem about the person's mass: both forces are proportional to mass, so you cancel the mass from the forces and are left with the accelerations.Edit: Anyhow, what result do you get when you solve your equation,
centripital acceleration > gravitational acceleration
 
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What is angular velocity?

Angular velocity is a measure of the rate at which an object rotates or revolves around a fixed point or axis. It is typically measured in radians per second or degrees per second.

How is angular velocity related to people flying off the planet?

Angular velocity is related to people flying off the planet because it determines the centrifugal force acting on an object as it rotates. If the angular velocity is too high, the centrifugal force can overcome the gravitational force and cause objects, including people, to fly off the planet.

What factors affect angular velocity?

The factors that affect angular velocity include the radius of rotation, the mass of the rotating object, and the speed at which it is rotating. The greater the radius, mass, or speed, the higher the angular velocity will be.

How can angular velocity be calculated?

Angular velocity can be calculated by dividing the change in angular displacement by the change in time. It can also be calculated by dividing the linear velocity by the radius of rotation.

Can angular velocity be controlled?

Yes, angular velocity can be controlled by changing the radius of rotation, the mass of the rotating object, or the speed at which it is rotating. By adjusting these factors, the angular velocity can be increased or decreased as desired.

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