Torque/speed of motor relation

[aadharmg]

Just encountered a question which has led me to absolute craziness. I simply don't understand the relation between the speed of a motor and the torque generated. The question is "Comment on the magnitude of the torque of a motor when its velocity is constant". Someone pleaaaase explain.

[jamonwindeyer]

Just encountered a question which has led me to absolute craziness. I simply don't understand the relation between the speed of a motor and the torque generated. The question is "Comment on the magnitude of the torque of a motor when its velocity is constant". Someone pleaaaase explain.

Hey! Cool little question this, really highlights a critical aspect of motors! Pretty easy if you break it down:

- Constant angular velocity means that there is no angular acceleration
- By F=ma (or in an angular sense, \(\tau=m\alpha\)), no angular acceleration means no angular force.
- Angular force is torque! So if a motor is turning at a constant speed (with no load attached), then that motor isn't generating any significant torque. There might be a little, to overcome friction

[aadharmg]

Hey! Cool little question this, really highlights a critical aspect of motors! Pretty easy if you break it down:

- Constant angular velocity means that there is no angular acceleration
- By F=ma (or in an angular sense, \(\tau=m\omega\)), no angular acceleration means no angular force.
- Angular force is torque! So if a motor is turning at a constant speed (with no load attached), then that motor isn't generating any significant torque. There might be a little, to overcome friction

That makes a lot more sense! Just to make sure, can F = ma be applied to all phenomena involving force? I'm trying to break this down. I understand that if a = 0 then F = 0, but I want to make sure that based on T = Fd, I can write it as T = (ma)d? and hence show that Torque is basically 0? Or is the 'F' from T = Fd basically coming from F = BILSin(x)?

[jamonwindeyer]

That makes a lot more sense! Just to make sure, can F = ma be applied to all phenomena involving force? I'm trying to break this down. I understand that if a = 0 then F = 0, but I want to make sure that based on T = Fd, I can write it as T = (ma)d? and hence show that Torque is basically 0? Or is the 'F' from T = Fd basically coming from F = BILSin(x)?

You can't really write that. The issue is the \(m\). The \(m\) isn't the same for the angular version, the real thing is actually:



Where \(\tau\) is torque, \(\alpha\) is angular acceleration, and \(I\) is the moment of inertia. This is just a quantity analogous to mass in linear problems - There are formulas that let you calculate \(I\) for certain objects, but they (and all this message) are not assessable

The angular acceleration and \(a\) are also different! Writing \(\tau=Mad\) is basically saying, if I have a mass accelerating at a certain rate, and tie it to something else at a set distance that then causes it to spin, then you get torque. But \(a\neq\alpha\), and \(m\neq I\), because one is linear and one is rotational

(Again, not assessable! This is uni stuff )

You can apply the idea in principal even without the maths though. No angular acceleration means no angular force, which means no torque (same thing)

Oh, and the force in the torque formula does come from \(F=BIL\sin{\theta}\) in a motor, but since you've got multiple sides and varying angles as the motor spins, you can't say, \(\tau=BIL\sin{\theta}\times d\)

[aadharmg]

You can't really write that. The issue is the \(m\). The \(m\) isn't the same for the angular version, the real thing is actually:



Where \(\tau\) is torque, \(\alpha\) is angular acceleration, and \(I\) is the moment of inertia. This is just a quantity analogous to mass in linear problems - There are formulas that let you calculate \(I\) for certain objects, but they (and all this message) are not assessable

The angular acceleration and \(a\) are also different! Writing \(\tau=Mad\) is basically saying, if I have a mass accelerating at a certain rate, and tie it to something else at a set distance that then causes it to spin, then you get torque. But \(a\neq\alpha\), and \(m\neq I\), because one is linear and one is rotational

(Again, not assessable! This is uni stuff )

You can apply the idea in principal even without the maths though. No angular acceleration means no angular force, which means no torque (same thing)

Oh, and the force in the torque formula does come from \(F=BIL\sin{\theta}\) in a motor, but since you've got multiple sides and varying angles as the motor spins, you can't say, \(\tau=BIL\sin{\theta}\times d\)

Okay, that makes an incredibly greater amount of sense! I fully understand your point, thank you so much for helping me through this!