Without prefix: What is the maximum power and force output of a trolley motor?

[vu10758]

The power output from a trolley motor depends upon velocity P(v) = av(b^2-v^2). The power is 0 for v^2>b and "a" and "b" are constants. Determine the speed of the maximum power output and the speed of the maximum force exerted by the motor.


How do I begin this problem? I don't know how to start.

 

[Mindscrape]

In terms of the calculus, what do you know about power?

 

[vu10758]

It's the derivative of work with respect to time.

 

[Mindscrape]

Okay, good, but let's take it a step further. We know that P = \frac{dW}{dt} = \frac{\vec{F} \cdot \vec{ds}}{dt} = \vec{F} \cdot \frac{ \vec{ds}}{dt} = \vec{F} \cdot \vec{v}. Does this give you a start or any ideas?

 

[OlderDan]

The power output from a trolley motor depends upon velocity P(v) = av(b^2-v^2). The power is 0 for v^2>b and "a" and "b" are constants. Determine the speed of the maximum power output and the speed of the maximum force exerted by the motor.


How do I begin this problem? I don't know how to start.

The first part is a calculus maximization problem (or maybe a graphing calculator problem??). You are given P as a function of v and asked to maximize with respect to v.

I'm still thinking about the second part. The force is not constant here, so you cannot say P = Fv

 

[Mindscrape]

The first part is a calculus maximization problem (or maybe a graphing calculator problem??). You are given P as a function of v and asked to maximize with respect to v.

I'm still thinking about the second part. The force is not constant here, so you cannot say P = Fv

Oh, oops, I overlooked that, but you are right. The force is not constant and can't be taken out of the derivative. Hum, it would be easy to graph into mathematica or another graphical program to find the quantitative solution, but the analytical solution is tough. Yes, I will have to think about this some more too.

 

[OlderDan]

Oh, oops, I overlooked that, but you are right. The force is not constant and can't be taken out of the derivative. Hum, it would be easy to graph into mathematica or another graphical program to find the quantitative solution, but the analytical solution is tough. Yes, I will have to think about this some more too.

The more I think about it, the more I think you were right to begin with, at least for any physically reasonable force. The force related to an incremental displacement ds is not going to change while moving the distance ds. So it's not a product rule derivative when you write
dW = Fds
It's just dividing the incremental work by the incremental time

dW/dt = Fds/dt
P = Fv
F = P/v = a(b^2-v^2) is maximum when v = 0

That seems reasonable actually. I think electric motors typically have maximum torque at zero speed, so you would expect maximum force at zero speed.