Calculate what would happen to the rpm of the large gear

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The discussion revolves around calculating the impact of a 50lb load on the rpm of a large gear when paired with a smaller gear in a 1:3 gear ratio. Initially, the large gear spins at 1,000 rpm, causing the small gear to spin at 3,000 rpm without load. When a load is applied to the small gear, the relationship between torque and speed becomes complex, as increasing the load will reduce the speed of the large gear due to the conservation of power. Participants emphasize that while torque is transmitted equally, the system's power must remain constant, leading to a decrease in speed when load is added. Understanding these dynamics requires a clear grasp of the forces and moments involved, often illustrated through free body diagrams.
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I have a picture of two gears, but I need some help with mechanical advantage. Let's say the larger gear is providing the work, and if there was no load supplied to the smaller gear, the large gear be spinning at 1,000 rpm, which would make the small gear spin at 3,000 rpm since the gear ratio is 1:3. Now let's say we add a load of 50lbs resistance to the small gear. Since the small gear is spinning 3 times as fast do we multiply that by 3? To 150lbs. Is that enough data to calculate what would happen to the rpm of the large gear?

Even though the small gear is spinning faster, there is 3 times the torque on it because of the gear ratio. So would that cancel out the excess torque and equal increase in speed without the large gear slowing down?..confusing. I know the system would obviously slow down as soon as you add the 50lb load, but the added speed and torque relationship is what's confusing.
 

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In any system forces and moments are conserved. This means that the torque applied by the gear to the pinion is the same. Force however, will be multiplied because it's being applied at a smaller distance.
 
minger said:
In any system forces and moments are conserved. This means that the torque applied by the gear to the pinion is the same. Force however, will be multiplied because it's being applied at a smaller distance.

Ok so the force of the pinion would be multiplied, I am guessing by a factor of 3 in this scenario. But it would take more force to move the pinon because it has the move the load 3 times as fast...right? So what happens?
 
minger said:
In any system forces and moments are conserved.

Huh? Mass and energy are conserved, not forces or moments.
 
njguy said:
Now let's say we add a load of 50lbs resistance to the small gear. Since the small gear is spinning 3 times as fast do we multiply that by 3?
No. You are directly applying load to the small gear, so whatever the load you apply is the load.

njguy said:
Is that enough data to calculate what would happen to the rpm of the large gear?
What do you mean? What is driving the large gear? The additional load to the pinion means that the driver of the larger gear (usually called the bull gear) has to also pick up the reflected load or else the speed will decrease. It all depends on what is driving the large gear.

njguy said:
Even though the small gear is spinning faster, there is 3 times the torque on it because of the gear ratio. So would that cancel out the excess torque and equal increase in speed without the large gear slowing down?..confusing. I know the system would obviously slow down as soon as you add the 50lb load, but the added speed and torque relationship is what's confusing.
Think of it this way...in terms of power. You have a torque and a speed. If one goes up, the other has to go down to maintain the same power (forgetting about some losses). So if the speed of the pinion is 3X the bull gear, that means it has to have less torque at the pinion to maintain the constant input power to the bull gear.

So you have:
\tau * \omega = P

F * d * \omega = P
 
Cyrus said:
Huh? Mass and energy are conserved, not forces or moments.

Conserved may not have been the word I was going for. Since there is no acceleration, forces and moments sum to zero.

The important thing in introductory gearing is to draw a nice free body diagram. Once you understand that gears transmit torque equally from one to another, then determining forces on the gears and the shafts become apparent.
 
minger said:
Conserved may not have been the word I was going for. Since there is no acceleration, forces and moments sum to zero.

The important thing in introductory gearing is to draw a nice free body diagram. Once you understand that gears transmit torque equally from one to another, then determining forces on the gears and the shafts become apparent.

I just felt like giving you a hard time. :approve:
Thats what she said
 
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