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This is inspired by the following question.

Consider some object which I want to lift from rest with a constant power throughout the whole process; the power I apply when lifting the object from rest is the same power I apply to keep lifting it. The force $F$ and the speed $v$ may change, but power may not.

The power is $P=Fv$. If we want constant power, then $dP/dt=0$.

First, differentiate wrt time, $$\dfrac{dP}{dt} = F \dfrac{dv}{dt}+v \dfrac{dF}{dt}.$$

Set equal to zero, this guarantees constant power, which implies:

$$F \dfrac{dv}{dt} = -v\dfrac{dF}{dt}.$$

From there, I use $F=\dfrac{d(mv)}{dt}\implies F = m\dfrac{dv}{dt}$ and get $$\left( \dfrac{dv}{dt} \right)^2 + v\dfrac{dv}{dt} = 0.$$

How may I solve this non-linear ODE?


EDIT: Initially, I got to the wrong conclusion that $F(v-v_0) = -v(F-F_0)$ is a solution. That's what the answer by Vilvanesh addresses.

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  • $\begingroup$ If power is constant then this must mean that the quantity Fv is constant. What does this tell you about the derivatives above? $\endgroup$
    – joseph h
    Mar 19 at 5:41
  • $\begingroup$ if F increased by a factor of x, then v is multiplied by a factor of 1/x. However, this becomes tricky when the initial speed is 0, which is why I ask the question $\endgroup$
    – user256872
    Mar 19 at 5:43
  • $\begingroup$ So the F and v are changing, so how can dP/dt=0? $\endgroup$
    – joseph h
    Mar 19 at 5:47
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    $\begingroup$ Why did you assume the force is not a function of the velocity? $\endgroup$
    – Yejus
    Mar 19 at 5:59
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    $\begingroup$ Your question seems contradictory. $\endgroup$
    – joseph h
    Mar 19 at 6:03
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$$\left( \dfrac{dv}{dt} \right)^2 + v\dfrac{dv}{dt} = 0$$ $$\frac{dv}{dt}\times\Big[\frac{dv}{dt}+v\Big]=0$$

When $\displaystyle{\frac{dv}{dt}=0}$, $v$ has to be a constant. But that isn't possible because initially it was at rest but while being lifted, it is not. $$\frac{dv}{dt}+v=0$$ $$\frac{dv}{v}=-dt$$ $$\int\limits_{v_o}^v\frac{dv}{v}=-\int\limits_0^tdt$$ $$\ln{v}-\ln{v_o}=-t$$ But since $v_o=0$, $\ln{v_o}$ won't be defined.

Hence a solution doesn't exist for the above differential equation.

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    $\begingroup$ so it is therefore impossible to lift an object with constant power. thank you. $\endgroup$
    – user256872
    Mar 19 at 6:48
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Your discussion is valid till $\displaystyle{F\frac{dv}{dt}=-v\frac{dF}{dt}}$.

I think the only way to reach $F(v-v_o)=v(F-F_o)$, is by assuming $F$ is constant while integrating on the left side and by assuming $v$ as constant while integrating on the right side. This is inconsistent because on the left side $F$ is being treated as a constant and on the right side it is treated as a variable. Same is applicable for $v$.

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  • $\begingroup$ Updated my question. I accepted your post since it did answer my initial question, but any advice for my updated post is much appreciated $\endgroup$
    – user256872
    Mar 19 at 6:12

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