Object moving radially outward on Rotating Flywheel

In summary, the conversation discusses a scenario where a point mass is attached to a rotating flywheel, causing a decrease in the rotational kinetic energy of the system. This decrease in energy is due to internal work being done by the cord attached to the object, which pulls towards the axis and increases the tension in the cord. This tension can be compared to a coiled spring, with potential energy increasing as the radius of the object increases. This demonstrates conservation of total energy, where external work can be done by controlling the tension in the cord.
  • #1
julienl07
3
0
Hello. This has been bothering me.

A point mass is on a rotating flywheel that has a constant initial angular velocity, ω0. The object (point mass), initially at some distance r0 from the axis of rotation, now moves out to a further distance rf, and then stops. Say the wheel has a moment of inertia Iw, and the mass is m.

If the system includes the wheel and object, its total moment of inertia is

I = Iw+mr2

By conservation of angular momentum

ωf0[itex]\frac{I_{0}}{I_{f}}[/itex]

If the initial rotational kinetic energy of the system was [itex]\frac{1}{2}I_{0}ω_{0}^{2}[/itex], the final is [itex]\frac{1}{2}I_{f}ω_{f}^{2}[/itex] = [itex]\frac{1}{2}I_{0}ω_{0}^{2}\frac{I_0}{I_f}[/itex] = [itex]E_{rk_0}\frac{I_0}{I_f}[/itex]. Ignoring the gory details, the rotational kinetic energy of the system has decreased, but no external work has been done.

In my ideal situation, I imagine that the object is attached to the axis of rotation by a cord, and that there are 'guide walls' forming an 'aisle' that goes radially outward from the axis. The tension in the cord is momentarily lowered, and the object moves radially outward through this 'aisle', before the cord is held still. The only interaction that I can think of (other than from the cord) is a normal force between the guide wall and the object. This force both increases the speed of the object, and slows the angular speed of the wheel by a torque, so that the angular velocity of the object and wheel remain equal as it moves outward.

So, sure internal work has been done from the torque, but no external work has been done.
How is it that the total rotational/kinetic energy of the system has changed, but no external work has been done? No heat was generated or anything.

Is there something that I have not seen? Can anyone explain this?

Thanks.
 
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  • #2
You forgot the work done by the cord when the particle moves radially. That's where the change of KE comes from.
 
  • #3
Imagine that a coiled like spring (like a clock spring) is being wound up as the cord lenthens. The decrease in kinetic energy of the system would equal the increase in potential energy of the spring. If there was no friction or any losses, such a system could oscillate between a minimum and maximum radius for the object, exchanging kinetic for potential energy and vice versa.
 
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  • #4
Thanks for the replies.

Although I didn't mention it, I know that work is done by the cord. The cord pulls towards the axis while the object moves away from the axis, thus the work is negative.

But again, this is internal work. By definition, if no external work has been done, then the total energy of the system (kinetic, potential, heat) should remain constant, right?

@rcgldr:

Are you suggesting that the cord is elastic? (You say that it would oscillate between maximum and minimum radius.) I meant that the cord does not wind around the axis, and that the object moves "infinitely slowly" outward before stopping (and remaining) at some new, greater radius (theoretical ideal system). Unless I misunderstood you, Is there some sort of potential gain by being at a greater radius?

Thanks.
 
  • #5
julienl07 said:
But again, this is internal work. By definition, if no external work has been done, then the total energy of the system (kinetic, potential, heat) should remain constant, right?

If you define "the system" as including whatever changes the tension in the string, then yes this is internal work.

But conservation of total energy is not the same as conservation of kinetic energy. Some other form of energy is being converted into KE when the string changes length. For example if the length of the string was being controlled by an electric motor/generator. you would need to supply electrical energy to the motor to wind the string in and reduce the radius of the mass (and increase the KE of the rotating system), or the generator would supply electrical energy as the string moved out and the KE of the system decreased.
 
  • #6
julienl07 said:
Are you suggesting that the cord is elastic? (You say that it would oscillate between maximum and minimum radius.)
Not the cord, but a wind up coiled spring attached to post that the cord is wrapped around on, so that the cord can move inwards and unwind the spring, or outwards and wind the spring. Since there's no dampening (friction) the system would tend to oscillate. If there was some type of latch mechanism to release and then stop the spring without any losses, then that would allow the radius to be changed without oscillation.

julienl07 said:
Is there some sort of potential gain by being at a greater radius?
The potential gain only exists because the spring winds up as the radius increases, which increases the potential energy in the spring. The point of using the spring was to create an example where an internal potential energy increases as the radius increases, so that total energy of the system would remain constant.

If you allowed external work to be done, then the cord could pass through a hole at the center of the plane, with the object sliding around the hole at various distances (radius) on a frictionless surface. Then external work could be done by pulling the cord inwards through the hole, or negative work done by allowing the tension to pull the cord outwards. The work done would be equal to the integral of tension as a function of distance over the distance the cord moved. Tension will equal some constant / r3 (if the radius is doubled, the tension is 1/8th of what is was before).
 

Related to Object moving radially outward on Rotating Flywheel

1. How does the motion of an object on a rotating flywheel differ from motion on a stationary surface?

On a rotating flywheel, an object moving radially outward experiences a combination of linear and circular motion, whereas on a stationary surface, the object only experiences linear motion. This is due to the rotational velocity of the flywheel, which causes the object to also move in a circular path.

2. What factors affect the speed of an object moving radially outward on a rotating flywheel?

The speed of an object on a rotating flywheel is affected by the rotational speed of the flywheel, the distance from the center of the flywheel, and the mass of the object. As the rotational speed or distance from the center increases, the speed of the object will also increase. However, as the mass of the object increases, the speed will decrease.

3. How does the direction of rotation of the flywheel affect the motion of an object moving radially outward?

The direction of rotation of the flywheel determines the direction of the object's circular motion. If the flywheel is rotating clockwise, the object will move in a counterclockwise circular path, and vice versa. This direction of rotation also affects the magnitude of the object's speed, as it will either add to or subtract from the rotational velocity of the flywheel.

4. Can an object moving radially outward on a rotating flywheel experience acceleration?

Yes, an object moving radially outward on a rotating flywheel can experience acceleration. This is due to the change in the object's velocity as it moves in a circular path. The object may also experience centripetal acceleration, which is directed towards the center of the flywheel and keeps the object in its circular motion.

5. How does friction between the object and the flywheel affect the motion of the object?

Friction between the object and the flywheel can cause the object to slow down or even stop if the friction is strong enough. This is because the frictional force acts in the opposite direction of the object's motion, causing it to lose speed. However, if the friction is minimal, the object may continue to move at a constant speed due to the rotational velocity of the flywheel.

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