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View Full Version : What is the actual weight of a rotating object?

KhashayarShatti
2012-Jan-03, 11:47 AM
I honestly don't know if this is ATM. I think of it as a simple engineering problem.
My question is: Is the weight of a rotating object equal to its stationary weight?
For simplicity please consider a rotating ring of radius r, Gravitational constant G, speed of w1 vertically downwards(horizontal ring plane), ring cross section to be very small.

I've done some calculations as follows and i've come to the conclusion that the weight is less. Experimentally i don't have an accurate weighing set, so my argument is mathematical as follows:

1- Consider any particle on the ring with a mass of m,

5- So the tangential speed of the particle=r*w1

6- Now this particle is moving in the gravity field of earth and this gravity force causes it to experience a curved path if it was allowed to by the ring . For this reason i've taken w2 as the speed of this particle around earth at that moment. w2 axis will be horizontal.

7- Now if an object spins and is given a torque to rotate(caused by gravity) then a third speed w3 will be induced on the particle by the formula T=I*w1*w2 . This torque induces w3. Direction of w3 is also horizontal and at righe angles to w2 and w1.

8- This torque induces an upward force on the particle=T/r.

9- So the actual weight of the object=mg-T/r.

All my question is on item no. 6. Is this correct? If not, why not?

HenrikOlsen
2012-Jan-03, 12:36 PM
Ignoring relativistic effects, the weight of a rotating object is the same as that of a non-rotating object.

You've completely misunderstood how things in orbit stays up.

Strange
2012-Jan-03, 03:59 PM

swampyankee
2012-Jan-03, 05:07 PM
If you're not convinced, get a well-balanced disk, an electric motor, with power supply, a tachometer, and a high accuracy scale.

Weigh the disk/motor assembly when it's stationary.

Turn the motor on to start the disk/motor assembly spinning. Record the angular velocity and the weight.

Repeat until the disk flies apart.

snowcelt
2012-Jan-04, 03:45 PM
I was reading Graham Green's latest book and I am positive that he states that moving and/ or spinning objects gain mass.

Strange
2012-Jan-04, 03:50 PM
All my question is on item no. 6. Is this correct? If not, why not?

I don't understand if your ring is vertically or horizontally aligned (but that doesn't make any difference).

In steps 6, 7 and 8 you consider a single particle at a moment in time. You need to integratye over all the particles in the ring (which will all cancel out) and over a complete rotation (which will cause the effects for single particle to cancel out).

So the spinning ring has the same mass as a stationary one. Apart from relativistic effects: adding (kinetic) energy will increase its effective mass by a tiny amount.

swampyankee
2012-Jan-04, 10:30 PM
I was reading Graham Green's latest book and I am positive that he states that moving and/ or spinning objects gain mass.

E=mc2. Since E is now increased -- by E=(1/2)Iω2 -- m will increase, so he's right. Of course, chances are the increase in mass will be very difficult to measure.

Ivan Viehoff
2012-Jan-05, 09:30 AM
I was reading Graham Green's latest book and I am positive that he states that moving and/ or spinning objects gain mass.
I think you mean Brian Greene. Graham Greene's books include "The Power and the Glory", "Travels with My Aunt", and "The Heart of the Matter", but none of his books dealt with physics.

Clearly any object in motion has a relativistic mass greater than its rest mass due to the effects of special relativity. However OP was referring to weight in a classical framework, like we might say that you are "weightless" when falling under gravity, or "weigh less" when immersed in a dense fluid.

Strange
2012-Jan-05, 09:48 AM
I think you mean Brian Greene. Graham Greene's books include "The Power and the Glory", "Travels with My Aunt", and "The Heart of the Matter", but none of his books dealt with physics.

Well, two of those sound like they could do...

Ivan Viehoff
2012-Jan-05, 02:55 PM
Well, two of those sound like they could do...
I should have left the second one out: the word "aunt" makes clear that "travel" has nothing to do with the physics of motion.

KhashayarShatti
2012-Jan-05, 07:37 PM
E=mc2. Since E is now increased -- by E=(1/2)Iω2 -- m will increase, so he's right. Of course, chances are the increase in mass will be very difficult to measure.
We are concerned with weight. In fact mass is quantity of matter but as you know weight is force. So whatever the mass may be, weight can be from zero to infinity depending on the acceleration of mass.

HenrikOlsen
2012-Jan-05, 07:46 PM
snowcelt et al.: KhashayarShatti is asking the question because he doesn't know enough elementary physics to know what he's missing.
Introducing relativistic effects is not going to make him less confused.

KhashayarShatti: Don't get sidetracked by the talk about things gaining weight when they move, it's mostly irrelevant for your question because it mainly applies to very fast speeds.

The relevant thing to know is that rotating an object does NOT make it lighter. In neither weight nor mass.

SkepticJ
2012-Jan-06, 03:55 AM
E=mc2. Since E is now increased -- by E=(1/2)Iω2 -- m will increase, so he's right. Of course, chances are the increase in mass will be very difficult to measure.

Actually, mass doesn't increase with speed. As an object approaches the speed of light, it gets harder and harder to push to a higher speed. It's as if its mass increases, but it doesn't actually: http://en.wikipedia.org/wiki/Mass_in_special_relativity#Modern_view

KhashayarShatti
2012-Jan-06, 06:30 AM
I don't understand if your ring is vertically or horizontally aligned (but that doesn't make any difference).

In steps 6, 7 and 8 you consider a single particle at a moment in time. You need to integratye over all the particles in the ring (which will all cancel out) and over a complete rotation (which will cause the effects for single particle to cancel out).

So the spinning ring has the same mass as a stationary one. Apart from relativistic effects: adding (kinetic) energy will increase its effective mass by a tiny amount.

I did it. The centre of mass lies in gravity field but i can't prove that centre of gravity exists for the whole system at orbital speed as a case study. Centre of mass exists, but CG doesn't seem to exist as every particle seems to be at orbital speed or higher(perhaps a kind of -ve logic and illusion).

swampyankee
2012-Jan-07, 04:22 AM
I did it. The centre of mass lies in gravity field but i can't prove that centre of gravity exists for the whole system at orbital speed as a case study. Centre of mass exists, but CG doesn't seem to exist as every particle seems to be at orbital speed or higher(perhaps a kind of -ve logic and illusion).

Center of mass is not effected by the velocities of connected components. If this wasn't true, it would be impossible to balance any rotating machinery. Perhaps no rotating machinery has peripheral velocities as high as orbital velocity -- about 7900 m/s -- but there is a lot of rotating machinery which is made and used every day which has peripheral velocities of about 500 m/s, and nobody has noticed that the center of mass changes location due to spinning.

Jerry
2012-Jan-10, 05:01 AM
spinning is a way to locate the center of mass of a non-spherical body...notice that the shape of the body is not relevant, and it is not possible to use some force acting within the mass to counter gravity (unless there is a secondary external force in the loop).

eburacum45
2012-Jan-10, 08:10 PM
Actually, mass doesn't increase with speed. As an object approaches the speed of light, it gets harder and harder to push to a higher speed. It's as if its mass increases, but it doesn't actually: http://en.wikipedia.org/wiki/Mass_in_special_relativity#Modern_view
I suspect swampyankee was referring to the fact that you add energy to an object at rest in order to make it spin, therefore the overall mass increases slightly as mass and energy are equivalent.

snowcelt
2012-Jan-11, 11:22 AM
Khash states in his first point that a "particle" does whatever. Therefor that "particles'" spin is relevant. Brian (thank you Ivan,) Greene points out that the more agitated some thing is, the more massive it is. There are quantum mechanical aspects to this because any question of position, location, spin etc are part of understanding what the mass will be.

swampyankee
2012-Jan-11, 11:44 AM
I suspect swampyankee was referring to the fact that you add energy to an object at rest in order to make it spin, therefore the overall mass increases slightly as mass and energy are equivalent.

I was. Leonard Susskind had said (in one of his class lectures on iTunes) that increasing the temperature of an object, which increases its internal energy, will increase its mass. Since temperature is just another form of kinetic energy, the same thing would happen if the object is spun up.

KhashayarShatti
2012-Jan-17, 04:55 PM
snowcelt et al.: KhashayarShatti is asking the question because he doesn't know enough elementary physics to know what he's missing.
Introducing relativistic effects is not going to make him less confused.

KhashayarShatti: Don't get sidetracked by the talk about things gaining weight when they move, it's mostly irrelevant for your question because it mainly applies to very fast speeds.

The relevant thing to know is that rotating an object does NOT make it lighter. In neither weight nor mass.

HenrikOlsen:regarding elementary physics, you said to Jeff Root: (regarding force of a chain or belt around a pulley)
"The problem seems to be that you don't have the automatic appreciation that the smaller the radius at the end the larger the forces applied there becomes".

I didn't want to make that thread active, so i referred to it here for the knowledge of elementary physics.
HenrikOlsen, I had some time to compute the force around a pulley. I think there are other parameters to consider.
In fact force =(mass per unit time) * (velocity). [change of momentum per unit time]

Now if radius is reduced, I think mass per unit time also reduces in the same proportion. You can work it out for yourself. So i think there is no change of force but the size of pulley can be made very small.
To simplify, assume the belt goes round the pulley for 180 degrees. Whatever the radius of the pulley, the net change of momentum per unit time is the same when speed is the same. I just want to know which one is correct.

Now my question: If a belt rotates around a pulley very fast, is there any centrifugal force? I honestly would like to know the answer when a ring is supposed to rotate.

HenrikOlsen
2012-Jan-17, 10:02 PM
In fact force =(mass per unit time) * (velocity). [change of momentum per unit time]

Now if radius is reduced, I think mass per unit time also reduces in the same proportion. You can work it out for yourself. So i think there is no change of force but the size of pulley can be made very small.
Time is reduced and since you divide by the time, mass*velocity per time increases (force is increased).
Very small pulley means very large force.

Incidentally it's silly to talk about mass per time since the mass is constant, it's the velocity change per time AKA acceleration that's relevant, and that it's PER TIME, i.e. smaller time -> larger acceleration.

This is really elementary physics which gives the impression that you've started physics by reading the wrong level books.
You need the fundamentals first, starting at the middle means you're making very elementary mistakes like this one.

KhashayarShatti
2012-Jan-19, 03:07 PM
Time is reduced and since you divide by the time, mass*velocity per time increases (force is increased).
Very small pulley means very large force.

Incidentally it's silly to talk about mass per time since the mass is constant, it's the velocity change per time AKA acceleration that's relevant, and that it's PER TIME, i.e. smaller time -> larger acceleration.

This is really elementary physics which gives the impression that you've started physics by reading the wrong level books.
You need the fundamentals first, starting at the middle means you're making very elementary mistakes like this one.

If you are familiar with engineering, mass per unit time=m dot

If you are familiar with engineering, force on pulley depends on the tension in the belt or chain.

Get the fundamentals on engineering first:p, you will see how simple it is. Most probably you are trying to send the thread to off topic threads; perhaps that is your physics profession.

Swift
2012-Jan-19, 04:27 PM
Get the fundamentals on engineering first:p, you will see how simple it is. Most probably you are trying to send the thread to off topic threads; perhaps that is your physics profession.
KhashayarShatti

Be very careful. Watch what you accuse other members of doing. If you think someone is trying to drive the thread off topic, you Report the post, you don't respond to them in thread. But, Henrik's post was fine and people have been trying to answer your questions.

Hornblower
2012-Jan-19, 06:16 PM
If you are familiar with engineering, mass per unit time=m dotIf you are familiar with engineering, force on pulley depends on the tension in the belt or chain.

Get the fundamentals on engineering first:p, you will see how simple it is. Most probably you are trying to send the thread to off topic threads; perhaps that is your physics profession.

My bold. Can you explain, in appropriate mathematical detail, what you mean by "m dot"? I am not familiar with that expression.

caveman1917
2012-Jan-19, 07:59 PM
My bold. Can you explain, in appropriate mathematical detail, what you mean by "m dot"? I am not familiar with that expression.

A dot above a variable is usually shorthand for the time derivative of that variable, ie \dot{x} = v = dx/dt. Not sure why anyone would want to have the time derivative of mass though, since that's obviously constant in classical mechanics.

swampyankee
2012-Jan-20, 01:07 AM
A dot above a variable is usually shorthand for the time derivative of that variable, ie \dot{x} = v = dx/dt. Not sure why anyone would want to have the time derivative of mass though, since that's obviously constant in classical mechanics.

Mass flow -- denoted by \dot{m}=dm/dt -- occurs in engineering and physics quite often. Indeed, there exist many journals devoted to this part of classical physics: fluid dynamics.

HenrikOlsen
2012-Jan-20, 01:14 AM
For this particular application it's silly to use though, as the mass is constant and the velocity is changing.

KhashayarShatti
2012-Jan-20, 07:51 AM
A dot above a variable is usually shorthand for the time derivative of that variable, ie \dot{x} = v = dx/dt. Not sure why anyone would want to have the time derivative of mass though, since that's obviously constant in classical mechanics.

Good question. I would like to know if i'm right.
For clarity i refer to F=m(dv/dt)+v(dm/dt)
Force is generated from two dissimilar conditions: Sorry if the subject is repeating

1- You accelerate a fixed mass of m, so you get f=m(dv/dt)

2- Sometimes you accelerate some mass and expel it, so you get f=v(dm/dt) = v(m dot)

Now imagine a straight belt touches a pulley at speed v. The mass of the belt that touches the pulley is small. As the belt is pulled in tension around the pulley to 90 degrees, the mass of the belt that touches the pulley changes and this mass follows a curved path.

I think there are two ways to calculate the force.
1- static force; the component of tension force that gradually changes from 0 to 90 degrees while maintaining the tension fixed in the belt.
2- Rate of change of momentum of that mass of belt that gradually enters a circular path around the pulley while maintaining the tension fixed in the belt.
Can this be assumed that a kind of moving mass goes round the pulley?

Now my question is what happens to the force on pulley if the size of pulley gradually reduces while maintaining the tension fixed in the 90 degrees belt? Could this be considered that, while the size of the pulley reduces, the mass that touches the pulley also reduces in the same proportion?

swampyankee
2012-Jan-20, 11:14 AM
The appropriate uses \dot{m} are restricted to fluid flow problems where the system is a control volume with fluid flowing through it. Using a control volume analysis for a problem with a rope and pulley is possible, but it would have to be very careful formulated, and probably be done wrong. In any case, having masses moving around in circles or something topological equivalent to circles is a common dynamic problem; the only reason to use \dot{m} in this case would be if there was significant mass transfer from the belt to the pulley.

caveman1917
2012-Jan-20, 03:49 PM
Mass flow -- denoted by \dot{m}=dm/dt -- occurs in engineering and physics quite often. Indeed, there exist many journals devoted to this part of classical physics: fluid dynamics.

You're right, i should have qualified that as "not sure why anyone would want to have \dot{m} in the definition of force in classical mechanics".

BabyOxide
2014-Oct-09, 01:49 AM
i was searching on google, for an answer to spinning objects having less weight. and i have came across this, i dun fully understand all the equations. and it does not seem to be conclusive in this thread either.

the reason, this is intriguing bcos i was thinking about rotors and helicopters... no dun get me wrong i know there is lift and pressure involved, but supposed its just the rotor ?

weight != mass .... i supposed weight depends on gravity... u can have mass but u might be weightless in space.

i have a fan , supposed its symmetrical on the blades... now if i broke 1 blade off ,,,when the fan spins ,, it would jerk in a certain direction bcos the weight distribution is no longer correct..
this is interesting to note, supposed u have a beaker of water and u spin it at its absolute center, the water will spread outwards and upwards to the side, while the middle sinks down a little...

this tells me that when spinning , the weight ?? / or is it mass ? distribution tries to span outwards ....

supposed u put this on a weighing scale , while still doing the experiment , the mass probably remains unchanged, bcos this would calculate the mass of the object and technically nothing is loss, hence mass should remain the same...

however, with gravity in effect, and the idea of mass being pushes outwards , like in the beaker, would not the object be lighter ? somehow ?? but perhaps only in terms of the direction of the force ?

supposed a object is still ... the mass is same, the weight unsure, but the force due to gravity is downwards

when u spin it , the force is now transitioning from downwards to side wards ,, like in the case of the fan with broken blade,
i'm not sure but does this mean the force has lessen on downward and more on side wards
and since weight is a measurement of downward force in respect to gravity.
this object becomes lighter as the force is no longer downward the faster it spins ?

Grey
2014-Oct-09, 06:10 PM
Hi, BabyOxide, welcome to the board. It's true that there may be other forces on a spinning object, some of which may offset the weight to some extent (for example, the lift exerted by a helicopter rotor), but the weight of the object itself is unchanged by the rotation.

NEOWatcher
2014-Oct-09, 07:43 PM
Put your spinning beaker on a scale. It will weigh the same spinning or not.

KhashayarShatti
2014-Oct-13, 04:24 PM
The answer to this question has already been given in 3 different threads in this forum especially for a ring rotating at orbital speed.