I Acceleration From Rest Under Constant Power

[russ_watters]

That is what I find interesting though, any supplied power $P>0$ causes this seemingly paradoxical result. What are the interventions in the physical world that would prevent it from actually happening?

Depends on the device. A car has a transmission with a limited range of gear ratios, a clutch or torque converter that slips and tires with limited grip. If you rev the engine and pop the clutch, you'll spin the tires, stall the engine or break the transmission or drive shaft. The first/most common limiter is tire friction.

 

[Vanadium 50]

I can't believe this is still going on.

1. Building a device where this approximation is valid is impossible. Actual devices do not have constant power at v = 0. Here is the very first graph Google founmd:

As engine speed goes to zero, so does power. This is unavoidable.

2. Constant power implies infinite force. And infinite force does not give you infinite acceleration. It breaks your device. Smacking your car's engine with a sledgehammer does not make it go any faster.

 

[hutchphd]

That is what I find interesting though, any supplied power P>0 causes this seemingly paradoxical result.

I do not understand your particular vexation here. Suppose I collide two billiard balls head-on at t=0. I require energy be conserved. At t=0 the velocity is zero, yet an instant later all is copacetic. Paradox? Of course not because the balls cannot instantaneously reverse. Why? because other physics intervenes in that limit. The (physical) limit is a useful approximate fiction.

 

[Lnewqban]

Thats not the problem I'm having. The problem is $\lim_{t \to 0} \ddot x \rightarrow \infty$ for all $P > 0$.

Perhaps we should move away from transduction capable machines in this discussion.
What if instead we analyze the first instants of the simple free fall of a body, or the highest point reached by a vertical projectile trajectory?

If we consider a simple lever, or gear’s pair, yes, the input force should tend to infinite as the arm distace to the fulcrum or axis of rotation tends to zero.
But at that point, do we still have a lever, or rather a support point unable to transfer our “constant power”?

 

[erobz]

2. Constant power implies infinite force. And infinite force does not give you infinite acceleration. It breaks your device. Smacking your car's engine with a sledgehammer does not make it go any faster.

This is more of hitting the engine with a pea breaks your car... $P$ can be arbitrarily small. Which is why I found it troubling...

I can't believe this is still going on.

1. Building a device where this approximation is valid is impossible.

I'm fine with this...I implied as much in the OP. It was the implied paradox created by the constant power assumption that I was interested in.

Simply letting $P$ be the next simplest model $P = kt$ resolves the issue. $\ddot x (0) = \rm{const.}$

Sorry for bothering everyone, I was genuinely surprised by this...not trolling.

 

[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.