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Question about gears and torque

  1. Jan 11, 2010 #1
    So this is something I've been thinking about...

    Let's say you have an engine and a motor.

    The engine has high torque (let's just say 3,000 ft/lb) and low RPM (2 RPM) and you attach a huge gear to the end of it (10 ft circumference). A very small gear (1 inch circumference) is attached to the motor. So the engine turns this huge gear 2 times a minute and the huge gear turns the small gear 240 times a minute.

    So, the part I'm wondering now is that is there a loss of torque now that the new RPM is 240 and no longer 2 going into the motor? I'm thinking that since there is no slippage of the gears the torque will remain at 3,000 ft/lb going into the motor and it will now be at 240 RPM as well. Is this right or am I missing something?
     
  2. jcsd
  3. Jan 11, 2010 #2
    If the torque were a static 3000lb/ft then yes, it would be divided by 120 on the static smaller gear.

    However, since you are talking about torque over time with motion you're talking about energy. If you could design magical gears that keep the torque loss at 0 but were equally efficient then you would have to have an RPM loss in the smaller gear to adjust for the small amount of energy lost in friction of the gear teeth meshing and what not.
     
  4. Jan 11, 2010 #3
    In the first part what do you mean by static? I think you might be getting at what I'm looking for in the second part. How is there torque loss?
     
  5. Jan 11, 2010 #4

    rcgldr

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    If there are no losses, then the torque on the small gear will be 2/240 of the large gear.

    In reality the torque is transfered between the gears via a linear force, which is opposed by friction between the gears. The force on the smaller gear's teeth will be reduced by the opposing friction force, and the torque on the smaller gear will end up being reduced.

    What he meant is if the motor were producing a torque without movement. If there is no movement, then there would be no losses due to friction.
     
  6. Jan 11, 2010 #5

    russ_watters

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    I'm not sure what ChmDude is getting at, but gearing provides mechanical advantage, just like a lever: when you multiply distance, you divide force (torque) and vice versa. So if you start with 3000 ft-lb (not ft/lb) and go from 2 to 240 rpm, the torque is 3000/240=12.5 ft-lb of torque.
     
  7. Jan 11, 2010 #6

    It would be 3000/120 which is 25lb/ft if the gears weren't moving.

    Once the gears start moving you will see losses in torque due to friction between the surfaces of the gears, movement of lubrication on the surface, etc...

    That lever you cited has 100% force transfer until you start to move the load up or down, then some of the force is lost to friction at the fulcrum.
     
  8. Jan 11, 2010 #7

    russ_watters

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    Basically correct, but since we know nothing about what the friction is going to be and it isn't helpful in explaining the theory, we assume a lossless situation for the purpose of the explanation.
     
  9. Jan 11, 2010 #8
    Yes, that's what my professors always did - they got their gears from the Ideal Hardware store. You know the place, they sell frictionless pullies, massless ropes, etc. Absolutely essential stuff to teach physics.
     
  10. Jan 11, 2010 #9

    russ_watters

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    ....um, ok.....sure, why not.
     
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