Why doesn't v=a(t) consider a motor's RPM?

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SUMMARY

The discussion centers on the application of the equation v=(f/m)(t) in scenarios involving motors, particularly in electric bicycles. It establishes that this equation does not account for the motor's RPM, which significantly influences velocity. The relationship between force, acceleration, and RPM is highlighted, indicating that as RPM changes, so does the force produced by the motor. The conversation concludes that the gearing between the motor and the bicycle's wheels is crucial for optimizing speed, as it determines the effective relationship between motor RPM and bicycle velocity.

PREREQUISITES
  • Understanding of Newton's second law (f=ma)
  • Familiarity with the concepts of velocity and acceleration
  • Knowledge of motor RPM and its impact on force output
  • Basic principles of mechanical gearing and its effect on speed
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  • Learn about mechanical gearing systems and their impact on vehicle performance
  • Explore the dynamics of constant acceleration versus variable acceleration
  • Investigate the effects of air resistance on moving vehicles
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Engineers, physics students, electric bicycle designers, and anyone interested in understanding the dynamics of motor-driven vehicles and optimizing their performance.

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Given
f=ma, f/m=a
v=a(t), v=(f/m)(t)

Is it wrong to apply "v=(f/m)(t)" to situation where the force is generated by a motor? If no, then how come "v=(f/m)(t)" doesn't consider the motor's RPM? From my understanding, velocity should also be determined by the motor's rpm. For example, an electric bicycle whose motor produces 1000N @ 100 rpm (maximum) will be faster than an electric bicycle whose motor produces 1000N @ 1 rpm (maximum). However per v=(f/m)(t), the velocity of both cases are the same assuming m and t are the same.
 
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Hmm, maybe you meant something else. If v is velocity and a is acceleration then v=a(t) is wrong.

Edit maybe you meant v=a*t
 
yes
 
david90 said:
Given
f=ma, f/m=a
v=a(t), v=(f/m)(t)

Is it wrong to apply "v=(f/m)(t)" to situation where the force is generated by a motor? If no, then how come "v=(f/m)(t)" doesn't consider the motor's RPM? From my understanding, velocity should also be determined by the motor's rpm. For example, an electric bicycle whose motor produces 1000N @ 100 rpm (maximum) will be faster than an electric bicycle whose motor produces 1000N @ 1 rpm (maximum). However per v=(f/m)(t), the velocity of both cases are the same.

Assuming that (as Dale suggests) you meant ##v=at## not ##v=a(t)##... That relation only holds when ##a## is constant for the entire time you're considering (and when the initial speed is zero as well). However, ##a## is only constant if ##F## is constant, and that's not true for the motors you're describing here. Because the force produced by the motor varies with RPM, and RPM is determined by the speed, the force and the acceleration will both change with the speed so you can't use ##v=at##.
 
What if we assume the motor in both cases are operating at their maximum rpm and the force at the bicycle's wheel is constant?
 
If they are at their maximum rpm then a=0
 
david90 said:
What if we assume the motor in both cases are operating at their maximum rpm and the force at the bicycle's wheel is constant?
If the force at the bicyle's wheel remains constant, then we reach a stable equilibrium where that force is exactly equal to the air resistance slowing the bicycle down so the net force on the bicycle is zero, meaning that the acceleration is zero and the speed no longer changes.

The gearing between the motor and bicycle's wheels matters - it controls the relationship between the speed of the bicycle and the RPM of the motor. If you get the gearing exactly right, you can arrange things so that when the engine is turning at the speed at which it generates maximum force the bicycle is moving at the speed at which produces air resistance exactly equal to the output of the motor - that will be the fastest speed achievable until you install a more powerful motor.
 
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