Acceleration From Rest Under Constant Power

[jbriggs444]

John Baez wrote a series of insights on theoretical problems with the continuum model. He digs a fair bit deeper than just the infinities being contemplated here.

Let me link the first article in his series: https://www.physicsforums.com/insights/struggles-continuum-part-1/

 

[Vanadium 50]

his is more of hitting the engine with a pea breaks your car

No, it's not. It's really not. Infinite force.

Anyway, I am done here. I am convinced that nothing I write will ever change your mind.

 

[erobz]

No, it's not. It's really not. Infinite force.

Anyway, I am done here. I am convinced that nothing I write will ever change your mind.

Thank you for your patience in sharing your expertise.

 

[erobz]

No, it's not. It's really not. Infinite force.

Just so you know, because I think you misunderstand what I meant. I know the force is infinite. The "pea" was in reference to the size of the applied power required to make the infinite force. I agreed with your every part of your previous post.

 

[bob012345]

As others have stated, I believe there is no paradox here.

Physically we must have an applied force to an object giving an initial acceleration $a_0$ and we can deduce its function (letting $\alpha$ = some constant with units $\frac{m^2}{s^3}$) as $$a(t) = \sqrt{\frac{\alpha}{t + c}}$$

at $t=0$ we can set the constant $c = \frac{\alpha}{a_0^2}$ where $a_0 = \frac{F_{applied}(0)}{m} = \frac{F_0}{m}$ so we have$$a(t) = \sqrt{\frac{\alpha}{t + \alpha a_0^{-2}}}$$ then we can integrate to get $v(t)$

$$v(t) = \int a(t) \,dt$$ $$v(t) = 2\sqrt{\alpha} \sqrt{t + \alpha a_0^{-2}} + C$$ where C is a constant of integration which we set with the condition that $v(0) = 0$ so $C = - 2\alpha a_0^{-1}$ thus we have

$$a(t) = \sqrt{\frac{\alpha}{t + \alpha a_0^{-2}}}$$ and $$v(t) = 2\sqrt{\alpha} \sqrt{t + \alpha a_0^{-2}} - 2\alpha a_0^{-1}$$ thus the mechanical power $\vec{F} ⋅\vec{v}$ is

$$P(t) = m\sqrt{\frac{\alpha}{t + \alpha a_0^{-2}}} \left(2\sqrt{\alpha} \sqrt{t + \alpha a_0^{-2}} - 2\alpha a_0^{-1}\right)$$ or

$$P(t) = 2m \alpha \left( 1 - \frac{1}{a_0} \sqrt{\frac{\alpha}{t + \alpha a_0^{-2}}}\right)$$ as $t → ∞$ we can identify $\alpha$ in the limit as a constant $\alpha = \frac{P_c}{2m}$

Suppose we apply a force to a 1kg mass such that $a_0 = 10 m/s_2$ so $F_{0} = 10N$ and $P_c = 10$ then we have the graph where red is velocity, blue is acceleration due to the applied force and green is power;

Now let $a_0 = 100$

now let $a_0=10000$

It becomes closer and closer to a step function. All this is saying is that you would have to move this with infinite force to make the step function perfect. There is nothing unreasonable or paradoxical with that.

 

[erobz]

As others have stated, I believe there is no paradox here.

Physically we must have an applied force to an object giving an initial acceleration $a_0$ and we can deduce its function (letting $\alpha$ = some constant with units $\frac{m^2}{s^3}$) as $$a(t) = \sqrt{\frac{\alpha}{t + c}}$$

at $t=0$ we can set the constant $c = \frac{\alpha}{a_0^2}$ where $a_0 = \frac{F_{applied}(0)}{m} = \frac{F_0}{m}$ so we have$$a(t) = \sqrt{\frac{\alpha}{t + \alpha a_0^{-2}}}$$ then we can integrate to get $v(t)$

$$v(t) = \int a(t) \,dt$$ $$v(t) = 2\sqrt{\alpha} \sqrt{t + \alpha a_0^{-2}} + C$$ where C is a constant of integration which we set with the condition that $v(0) = 0$ so $C = - 2\alpha a_0^{-1}$ thus we have

$$a(t) = \sqrt{\frac{\alpha}{t + \alpha a_0^{-2}}}$$ and $$v(t) = 2\sqrt{\alpha} \sqrt{t + \alpha a_0^{-2}} - 2\alpha a_0^{-1}$$ thus the mechanical power $\vec{F} ⋅\vec{v}$ is

$$P(t) = m\sqrt{\frac{\alpha}{t + \alpha a_0^{-2}}} \left(2\sqrt{\alpha} \sqrt{t + \alpha a_0^{-2}} - 2\alpha a_0^{-1}\right)$$ or

$$P(t) = 2m \alpha \left( 1 - \frac{1}{a_0} \sqrt{\frac{\alpha}{t + \alpha a_0^{-2}}}\right)$$ as $t → ∞$ we can identify $\alpha$ in the limit as a constant $\alpha = \frac{P_c}{2m}$

Suppose we apply a force to a 1kg mass such that $a_0 = 10 m/s_2$ so $F_{0} = 10N$ and $P_c = 10$ then we have the graph where red is velocity, blue is acceleration due to the applied force and green is power;

View attachment 314354

Now let $a_0 = 100$
View attachment 314355

now let $a_0=10000$View attachment 314356

It becomes closer and closer to a step function. All this is saying is that you would have to move this with infinite force to make the step function perfect. There is nothing unreasonable or paradoxical with that.

Thats a pretty slick analysis (thanks for showing it), but the approximate step function that is contrived is not a part of the initial setup. In any finite initial acceleration, the function you are describing is not constant. That function is never (philosophically) equivalent to $P(0) = P_o$ because it always begins at $P = 0$?

 

[Dale]

At this point we will leave this thread closed.

The sticky part is how does anything begin to move if we can't apply any power without visiting ∞?

Power isn’t force. Force provides acceleration. Power does not.

At $v=0$ the power is $0$ regardless of the force and therefore regardless of the acceleration. Therefore the acceleration is completely independent of the power at $v=0$. There is no paradox, no matter how you slice it, simply a mistake thinking that power does something that it does not do.

If P=0 there is no motion.

That is simply not true.