Simulatenous Conservation of Momentum and Conservation of Energy

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Say I have three wheels. Their characteristics are the following:

WHEEL ONE
20 kg mass
10 cm radius
Accelerated from 0 to 10 m/s in one second

WHEEL TWO
10 kg mass
20 cm radius
Accelerated from 0 to 10 m/s in one second

WHEEL THREE
10 kg mass
10 cm radius
Accelerated from 0 to 20 m/s in one second

We assume all wheels are close enough to uniform mass density and that their moment of inertia is calculated according to (mass*radius^2), the case for a cylinder. The values above were specifically chosen so that the final angular momentum for all three wheels are the same. We now determine the rotational kinetic energy for all three wheels.

WHEEL ONE
20 kg mass
10 cm radius
Accelerated from 0 to 10 m/s in one second
Angular Momentum = 20 kg*m/s^2
Final Rotational Kinetic Energy = (1/2) * (20 kg * (10 cm)^2) * ((10 m/s)/(10 cm))^2 = 1000 Joules

WHEEL TWO
10 kg mass
20 cm radius
Accelerated from 0 to 10 m/s in one second
Angular Momentum = 20 kg*m/s^2
Final Rotational Kinetic Energy = (1/2) * (10 kg * (20 cm)^2) * ((10 m/s)/(20 cm))^2 = 500 Joules

WHEEL THREE
10 kg mass
10 cm radius
Accelerated from 0 to 20 m/s in one second
Angular Momentum = 20 kg*m/s^2
Final Rotational Kinetic Energy = (1/2) * (10 kg * (10 cm)^2) * ((20 m/s)/(10 cm))^2 = 2000 Joules

Consider that angular momentum was transferred from one wheel to the other. Apparently what I said suggests that the rotational kinetic energy does not follow a one-to-one relationship with angular momentum. What other forms of energy should be present in this scenario?
 

Answers and Replies

  • #2
Redbelly98
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> "What other forms of energy should be present in this scenario?"

Heat.

This is similar to linear momentum problems involving collisions between different masses. Linear momentum is conserved. But kinetic energy is either conserved (elastic collision) or reduced (inelastic), with the lost kinetic energy converted to heat.

p.s. FYI, units of angular momentum are kg*m^2 / s, as seen from the equation
(moment of inertia) x (angular speed)
= (kg*m^2) x (1/s)
 
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How about regular kinetic energy? You say you set it up so they would have the same angular momentum at the end, that means that they're all traveling at different speeds. So (1/2) M V^2 where v is the center of mass.
 
  • #4
Redbelly98
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Wow, I didn't even think of translational kinetic energy. I interpreted the description as the wheels were stationary but rotating, and the given speeds were just the speed at the wheels' rims. But yes, if he meant that the wheels were moving translationally as well as rotating, regular kinetic energy has to be accounted for too.
 
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> "What other forms of energy should be present in this scenario?"

Heat.

This is similar to linear momentum problems involving collisions between different masses. Linear momentum is conserved. But kinetic energy is either conserved (elastic collision) or reduced (inelastic), with the lost kinetic energy converted to heat.

p.s. FYI, units of angular momentum are kg*m^2 / s, as seen from the equation
(moment of inertia) x (angular speed)
= (kg*m^2) x (1/s)
If I am a figure skater and I reduce my moment of inertia by a factor of two while spinning, I would have to spin twice as fast right? If so, my rotational kinetic energy would have to double. I presume that because of the so-called centripetal forces, that it would require work in order to reduce my moment of inertia. Is that basically why?
 
  • #6
Redbelly98
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If I am a figure skater and I reduce my moment of inertia by a factor of two while spinning, I would have to spin twice as fast right?
Yes.
If so, my rotational kinetic energy would have to double. I presume that because of the so-called centripetal forces, that it would require work in order to reduce my moment of inertia. Is that basically why?
Yes, the muscles in your shoulders and elbows do work on your arms, increasing the kinetic energy.

To simplify the picture, think of a ball swinging around on a rope. Imagine the rope is being drawn in while the ball circles around; the ball's path is then an inward spiral. There is work done on the ball because the angle between rope tension and the ball's direction is not 90 degrees (as it is for purely circular motion), so the work formula:
W = F x D x cos(angle)
is nonzero.
 
  • #7
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Yes.


Yes, the muscles in your shoulders and elbows do work on your arms, increasing the kinetic energy.

To simplify the picture, think of a ball swinging around on a rope. Imagine the rope is being drawn in while the ball circles around; the ball's path is then an inward spiral. There is work done on the ball because the angle between rope tension and the ball's direction is not 90 degrees (as it is for purely circular motion), so the work formula:
W = F x D x cos(angle)
is nonzero.
Ok. Then what is:
1) The significance, if any, of F x D?
2) The significance, if any, of F x D minus the work done on the ball?
3) The significance, if any, of F x D x sin(angle)?
4) The significance, if any, of F x D x sin(angle) plus the work done on the ball?
 
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  • #8
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Wow, I didn't even think of translational kinetic energy. I interpreted the description as the wheels were stationary but rotating, and the given speeds were just the speed at the wheels' rims. But yes, if he meant that the wheels were moving translationally as well as rotating, regular kinetic energy has to be accounted for too.
I only meant free-spinning wheels.
 
  • #9
Redbelly98
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Ok. Then what is:
1) The significance, if any, of F x D?
2) The significance, if any, of F x D minus the work done on the ball?
3) The significance, if any, of F x D x sin(angle)?
4) The significance, if any, of F x D x sin(angle) plus the work done on the ball?
The only expression of significance is
F x D x cos(angle)
and this equals the work done on the ball when it travels a short distance D in its trajectory.

See the included figure. Pulling in on the rope results in a non-zero value for the cos(angle) factor, so that work is done on the ball. The work increases the kinetic energy, and thus also the velocity of the ball.

How much does the velocity increase by? By exactly enough so that angular momentum is conserved! :smile:
 

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application engineering in hydrodynamic fluid film bearings

Request to explain as to how to start on these subjects.

The bearing calculations, FEA, bearing design etc.
 
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