# Thread: Conservation of Energy vs. Conservation of Angular Momentum

1. Hello friends:

Recently I've been studying the physics of the rotational motion of rigid bodies. I understand that rotational kinetic energy is expressed as:

EK = ½ mω2R2

Where m is the mass of the rotating point-mass body, ω is the angular velocity, and R is the radius of the circular path.

I also understand that angular momentum is expressed as:

L = mR2ω

According to the Law of the Conservation of Angular Momentum, if no resultant torque acts on this system, then angular velocity remains constant. If angular velocity increases, then the radius of the circular path must decrease enough to disallow any change in angular momentum. For example, if a 2-kilogram point mass rotates at 1 radian per second on a circular path of .707 meters, the angular momentum L = 1. (I will omit units for the sake of brevity.) If the same point-mass rotates at 10 radians per second, then the radius R must decrease to 0.224 meters to allow L to remain equal to 1.

Now here's the problem I'm running into. I calculated this point-mass's initial kinetic energy as 0.5 Joules. Using the final values of ω = 10 rad/s and R = 0.224 meters, the kinetic energy increases to 5 Joules. How is this increase in energy possible? Where is the extra 4.5 Joules of energy coming from?

I'm making a mistake somewhere. Can anybody here explain this paradox to me?

Thanks!

Jagella  2.

3. Hello,

You said the "angular-velocity increases" and then you said you need to trim down the radius to compensate for increase in angular-momentum. I think that's where the extra-energy coming from, the increase in velocity.  4. The extra energy comes from the work that has to be applied to the point mass in order to reduce the radius of its circular motion. Imagine your mass swinging around at the end of a massless string. The string provides the centripetal force needed to hold the mass in a circular path. So let's say that you want to decrease the radius of the path. You have to draw the string in, To do so, you have to exert an inward force on the string and therefore the mass. This force is exerted on the mass over the distance that it moves in towards the center. Force times distance is work, so work is performed on the mass. This work will be equal to the kinetic energy difference between the the different radii. The extra energy comes from whatever moved the mass into the smaller radius. IOW, outside energy has to be applied to the mass to get it to reduce the radius of its path and this energy accounts for the increase of rotational kinetic energy.  5. Originally Posted by msafwan You said the "angular-velocity increases" and then you said you need to trim down the radius to compensate for increase in angular-momentum. I think that's where the extra-energy coming from, the increase in velocity.
OK, great. I understand now. Thanks a lot for the help!

Jagella  6. Originally Posted by Janus The extra energy comes from the work that has to be applied to the point mass in order to reduce the radius of its circular motion. Imagine your mass swinging around at the end of a massless string...
In the classic example of an ice skater spinning on the ice, the skater pulls her arms in to increase her angular velocity. Although her angular momentum is conserved, her rotational kinetic energy increases. The increase in rotational kinetic energy equals the work she exerts to pull her arms in to her sides. Is that correct?

Anyway, I've found that in physics, some seemingly intractable problems end up having simple solutions. Like T.H. Huxley said when Darwin discovered natural selection: "How extremely stupid not to have thought of that!"

Thanks a lot for the help!

Jagella  7. Originally Posted by Jagella  Originally Posted by Janus The extra energy comes from the work that has to be applied to the point mass in order to reduce the radius of its circular motion. Imagine your mass swinging around at the end of a massless string...
In the classic example of an ice skater spinning on the ice, the skater pulls her arms in to increase her angular velocity. Although her angular momentum is conserved, her rotational kinetic energy increases. The increase in rotational kinetic energy equals the work she exerts to pull her arms in to her sides. Is that correct?
Yep. You've got it! Good work!   8. Originally Posted by pmb Yep. You've got it! Good work! Am I a big boy or what? It's the biggest discovery in physics since the Higgs Boson.

Jagella  9. Originally Posted by Jagella  Originally Posted by pmb Yep. You've got it! Good work! Am I a big boy or what? It's the biggest discovery in physics since the Higgs Boson.

Jagella
Yep. You sure are a big boy and today you can start to where big boy pants too!   10. Originally Posted by Jagella In the classic example of an ice skater spinning on the ice, the skater pulls her arms in to increase her angular velocity. Although her angular momentum is conserved, her rotational kinetic energy increases. The increase in rotational kinetic energy equals the work she exerts to pull her arms in to her sides. Is that correct?
You wrote:
EK = ½ mω2R2

which mean: when R decrease -> ω increase. But then why do you said reducing R also increase -> EK and ω?  11. Originally Posted by msafwan  Originally Posted by Jagella In the classic example of an ice skater spinning on the ice, the skater pulls her arms in to increase her angular velocity. Although her angular momentum is conserved, her rotational kinetic energy increases. The increase in rotational kinetic energy equals the work she exerts to pull her arms in to her sides. Is that correct?
You wrote:
EK = ½ mω2R2

which mean: when R decrease -> ω increase. But then why do you said reducing R also increase -> EK and ω?
He also wrote

L = mR2ω.

Since L is angular momentum, and is a conserved property, this is the formula used to determine the increase in ω.

If you then plug the new values of ω and R into You find that the kinetic energy increases.  12. Originally Posted by msafwan But then why do you said reducing R also increase -> EK and ω?
That's a good question. As long as no resultant torque is applied to the system, the angular velocity ω will increase with a decreased radius R to maintain the angular momentum. The work required to decrease R increases the system's kinetic energy. Jagella  13. Originally Posted by pmb Yep. You sure are a big boy and today you can start to where big boy pants too! I've noticed that physicists don't like hubris in others. Ironic, is it not?Jagella  14. Originally Posted by Jagella  Originally Posted by pmb Yep. You sure are a big boy and today you can start to where big boy pants too! I've noticed that physicists don't like hubris in others. Ironic, is it not?Jagella
Yes. Quite ironic indeed~  15. Janus, I applied some integral calculus to this problem. I'm getting work = 4.48 Joules. I found that at any value for the radius r (.707 => r => .224), work = the centripetal force times r = 2 / (4r3). If I integrate 2 / (4r3) as r goes from .224 to .707, then the result is about 4.48 Joules. This value is close to the 4.5 Joule value I calculated for work using the difference in rotational kinetic energy between r = .707 meters and r = .224 meters.

Isn't math fun and sometimes helpful?

Jagella  16. Originally Posted by Janus The extra energy comes from the work that has to be applied to the point mass in order to reduce the radius of its circular motion. Imagine your mass swinging around at the end of a massless string. The string provides the centripetal force needed to hold the mass in a circular path. So let's say that you want to decrease the radius of the path. You have to draw the string in, To do so, you have to exert an inward force on the string and therefore the mass. This force is exerted on the mass over the distance that it moves in towards the center. Force times distance is work, so work is performed on the mass. This work will be equal to the kinetic energy difference between the the different radii. The extra energy comes from whatever moved the mass into the smaller radius. IOW, outside energy has to be applied to the mass to get it to reduce the radius of its path and this energy accounts for the increase of rotational kinetic energy.
Contrary to the above highlighted popular belief, Force times distance is NOT Work.

Force times Time is Work (the quantity of energy exerted). Then the distance will be determined by the amount of Resistance (the opposing Force times Time) or Friction (the rate of energy dissipation). The opposing Forces are constantly at odds, so Work is done on both sides, even if there is no distance traversed. Applied Force on a mass without movement will manifest itself as stress within the body of the mass, just as you yourself will exhaust yourself and overheat if you push on a large tree all day.  17. Originally Posted by MagnaMoRo Force times Time is Work
No, force time time is impulse, not work.  18. Originally Posted by MagnaMoRo Contrary to the above highlighted popular belief, Force times distance is NOT Work.

Force times Time is Work (the quantity of energy exerted).
So, if I lean against the wall all day, I'm working? I am putting force on the wall.  19. Originally Posted by Harold14370 So, if I lean against the wall all day, I'm working?
Hey, it's what I get paid for. Don't knock it.   20. Originally Posted by Strange  Originally Posted by MagnaMoRo Force times Time is Work
No, force time time is impulse, not work.
Really!

In the SI system of measurement, Work is measured in joules.

The joule is a derived unit of energy, work equal to (1 kg·m^2/s^2).

This literally means that if a force of (1 kg) is applied to a mass of (1 kg) for 1 second, witout any outside friction or gravitaional influence, then it will be influenced to accelerate and move 1 meter within that 1 second.

Remember, the Kilogram is itself the representation of the force required to hold a specific quantity of mass stationary, against the acceleration of a Earth Gravity.

So, if an upward thrust or force equal to (1 kg·m^2/s^2) is applied to a mass weighing 1 kg against 1 Earth Gravity for 1 second then it will not accelerate for that 1 second. Yet 1 Joule of Work will have been expended. If over 2 seconds then 2 Joules of Work will have been expended.

(1 kg·m^2/s^2) · (1 s) = (1 Joule)
(Force) · (Time) = (Total Joules) or (Work)  21. Originally Posted by MagnaMoRo This literally means that if a force of (1 kg) is applied to a mass of (1 kg) for 1 second, witout any outside friction or gravitaional influence, then it will be influenced to accelerate and move 1 meter within that 1 second.
The (kg) is not a force, it is a unit of mass; the above statement is thus meaningless.
But regardless, force over time is impulse - you will find the definition and further explanations here.

Really !
Yes, really, as any textbook on classical mechanics will tell you. Would you like some references ?

Remember, the Kilogram is itself the representation of the force required to hold a specific quantity of mass stationary, against the acceleration of a Earth Gravity.
The kg is a unit of mass, and just as valid in gravity-free space ( or anywhere else for that matter ) as it is on Earth. To reflect this, the official definition of the kg will soon ( 2014 ) be changed into multiples of the Planck constant, quite independent of any forces.

So, if an upward thrust or force equal to (1 kg·m^2/s^2) is applied to a mass weighing 1 kg against 1 Earth Gravity for 1 second then it will not accelerate for that 1 second.
Of course it will - it accelerates at a=F/m for as long as the net force is non-zero.

(1 kg·m^2/s^2) · (1 s) = (1 Joule)
(Force) · (Time) = (Total Joules) or (Work)
Force x time = Impulse
Force x distance = Work

You should realise that this is all classical mechanics, well understood and well defined for the past 400 years. There are no "common misconceptions" so far as impulse and work is concerned.  22. @MagnaMoRo, you're confusing mass and weight. Confusingly, the pound can be used as either, but the kilogram is only a measure of mass.  23. There is a unit of force called the kilogram-force, which is equal to the weight of a 1 kilogram mass on the earth's surface, or about 9.8 newtons. However MagnaMoRo has misunderstood the concept entirely.  Bookmarks
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