Force/torque in a differential / gear ring.

In summary: Other times it is pretty easy to see that energy has been conserved (e.g. a weight dropped from a height). But in either case, it is still an article of faith. Where else could the energy go? - is the crucial question which justifies that step in reasoning.
  • #1
hihiip201
170
0
Hi guys:



I have 3 questions:


Imagine an open differential that looks like a gear ring :

and now let's call the rings (orbital gear) the wheel gear, the gear in the center the center gear.


1. is the reason why torque from input shaft is evenly split to both wheels regardless of different speed because : As long as the speed difference is constant, the differential has a constant angular velocity and hence net torque must equal to zero?



2. In the gear ring video, if i am to apply a torque not from the differential(center gear in track/sun gear) but now either : hold one of the wheel ring(orbital gear) fixed and turn the other one, or if I turn both of them at the opposite direction.

will it be correct to say that under no friction, there will only be forces between the wheel ring and track gear be non-zero during acceleration? in other word if i continue to exert a force from one wheel ring I will be accelerating the angular speed and speed of the center gear? and just the angular acceleration if i turn both wheel ring together?



3. Finally, a more fundamental question:

why is horse power equal to force times velocity (or T omega)? is it because we first defined kinetic energy to be 1/2mv^2? What prompted the people in the old times to define energy in such a way?

thanks!
 
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  • #2
why is horse power equal to force times velocity

Power is work per unit time (Definition of power)
Work is force times distance
so
Power is force times distance per unit time
Distance per unit time is velocity
Re-grouping gives:
Power is force times velocity
QED
 
  • #3
sophiecentaur said:
Power is work per unit time (Definition of power)
Work is force times distance
so
Power is force times distance per unit time
Distance per unit time is velocity
Re-grouping gives:
Power is force times velocity
QED


Oops sorry I think my question should have been why force is force times distance then.

I recognize that it was introduced by a French mathematician as weight lifted through a height, but how do we know that work, which was introduced in terms of potential energy works for kinetic energy as well?


thank you
 
  • #4
hihiip201 said:
Oops sorry I think my question should have been why force is force times distance then.

I recognize that it was introduced by a French mathematician as weight lifted through a height, but how do we know that work, which was introduced in terms of potential energy works for kinetic energy as well?


thank you

You don't mean that, do you? lol
'work = force times distance' ties directly with the Equations of motion, Newton's laws and the conservation of Energy
You do work by accelerating an object and that work turns up as KE - the sums all work.
Raise an object to a height of h and the work done is mgh . That is the GPE you have given the object. Drop it from that height and the KE at the bottom will be the same as the GPE at the top. "Where else could the energy go? - is the crucial question which justifies that step in reasoning.
 
  • #5
sophiecentaur said:
You don't mean that, do you? lol
'work = force times distance' ties directly with the Equations of motion, Newton's laws and the conservation of Energy
You do work by accelerating an object and that work turns up as KE - the sums all work.
Raise an object to a height of h and the work done is mgh . That is the GPE you have given the object. Drop it from that height and the KE at the bottom will be the same as the GPE at the top. "Where else could the energy go? - is the crucial question which justifies that step in reasoning.

ya i didn't mean "that" lol, fail.
when you say gpe is converted into KE when dropped, aren't we still using the work energy theorem in that case? I'm trying to understand how that theorem is coming from in the first place.

I can see how kinetic energy and work can tie together via GPE, but there's got to be some physics reasoning to justify using work energy purely on kinetic energy.

thank you for your reply!
 
  • #6
The "physics reasoning" is that Energy is conserved. If you can't come up with somewhere else the energy can go, all the GPE must 'go into' the resulting KE. Once it gets back down to the ground there is no GPE left. Are you suggesting there should be (even in our ideal case) some other form of energy in the resulting situation? Would that make sense?

It is an article of faith that energy in = energy out but better than that, as it is confirmed by experiment many times. Sometimes the energy is hard to identify (e.g. experimental error or the dreaded E = mc2. You never get more energy out than you would have expected, if you did the budget correctly.

I don't know what more you could want. Sorry. You may just have to live with this and it will hit you as reasonable, eventually.
 
  • #7
sophiecentaur said:
The "physics reasoning" is that Energy is conserved. If you can't come up with somewhere else the energy can go, all the GPE must 'go into' the resulting KE. Once it gets back down to the ground there is no GPE left. Are you suggesting there should be (even in our ideal case) some other form of energy in the resulting situation? Would that make sense?

It is an article of faith that energy in = energy out but better than that, as it is confirmed by experiment many times. Sometimes the energy is hard to identify (e.g. experimental error or the dreaded E = mc2. You never get more energy out than you would have expected, if you did the budget correctly.

I don't know what more you could want. Sorry. You may just have to live with this and it will hit you as reasonable, eventually.



believe or not I know how stubborn I am, and I completely understand the concept of COE and the fact that it is proven by experiment, but I always just thought there are some other ways that you can look at it. instead of just saying "Oh,conservation of energy", It is reasonable to me, but I just don't feel like it is enough.
 

Related to Force/torque in a differential / gear ring.

1. What is force/torque in a differential/gear ring?

Force/torque refers to the amount of rotational power or twisting force that is applied to the differential/gear ring in a mechanical system. It is typically measured in units of newtons (force) or newton-meters (torque).

2. How is force/torque calculated in a differential/gear ring?

The force/torque applied to a differential/gear ring is calculated by multiplying the distance from the center of the differential/gear ring to the point of force application by the magnitude of the force or torque applied. This is known as the moment of force/torque.

3. What is the purpose of a differential/gear ring in a mechanical system?

The differential/gear ring is an essential component in a mechanical system as it allows for the transfer of power and torque from the engine to the wheels, while also allowing for different rotational speeds between the left and right wheels during turns.

4. How does the size of the differential/gear ring affect force/torque?

The size of the differential/gear ring can affect the amount of force/torque that can be transmitted through it. A larger differential/gear ring can withstand higher levels of force/torque without breaking, while a smaller one may not be able to handle as much.

5. What are some common applications of force/torque in a differential/gear ring?

Force/torque in a differential/gear ring is commonly used in vehicles, such as cars and trucks, to transfer power from the engine to the wheels. It is also used in industrial machinery, such as conveyor systems, to transfer power and torque between different mechanical components.

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