Physics olympiad problem -- struggling with polar coordinates

[Chestermiller]

Ok that helps a lot @Chestermiller, thanks! One last, quick query- the condition for $v_{0}$ is also that $\theta=0$, right? Why don't we take that into consideration when we write down the solution , i.e.
$v_{0}=r_{0}e^{-k(0)}\dot \theta (1+k^2)^{1/2}=r_{0}\dot \theta(1+k^2)^{1/2}$ where the initial condition is that $v=v_{0}$ at $\theta=0$.


So that the exponential term vanishes?i.e. why do we not take into consideration theta =zero initially?( In any case, you have proved that the speed itself is the constant $C$, so it doesn't really matter; but the speed should also be independent of $\theta$, shouldn't it?)

This is basic to how one solves a differential equation, subject to an initial condition. We don't first substitute the initial condition into the differential equation and then integrate. We first integrate, and then substitute the initial condition into the solution to obtain the constant of integration.

 

[cr7einstein]

No @Chestermiller that is not what I am saying; I know how to solve a differential equation. What I am saying is that the problem states that at $v_{0}$, $\theta =0$; which would mean that the exponential term becomes unity; but in the solution, we still have the exp. term. Why is that?

 

[Chestermiller]

No @Chestermiller that is not what I am saying; I know how to solve a differential equation. What I am saying is that the problem states that at $v_{0}$, $\theta =0$; which would mean that the exponential term becomes unity; but in the solution, we still have the exp. term. Why is that?

I'm really having trouble understanding what you're asking. So, I'm going to try applying the initial condition in a slightly different form:

$$(r_0\sqrt{1+k^2}e^{-k\theta(t)}\dot \theta(t))^2=(r_0\sqrt{1+k^2}\dot \theta(0))^2$$or
$$r_0\sqrt{1+k^2}e^{-k\theta(t)}\dot \theta(t)=r_0\sqrt{1+k^2}\dot \theta(0)$$

I hope this works better for you. I'm out of ideas on what to try next.

 

[cr7einstein]

@Chestermiller Oh oh Ohk! You are writing theta as a function of time and not an independent variable! I am so, so sorry I did not see that! Thanks a lot! I can finally breath freely :)!
Again, thanks a lot for being so patient!

 

[ehild]

What I am saying is that the problem states that at $v_{0}$, $\theta =0$; which would mean that the exponential term becomes unity; but in the solution, we still have the exp. term. Why is that?

The problem states that at t=0, θ=0 and v=Vo. It is not the same what you wrote: "at $v_{0}$, $\theta =0$"
The independent variable is the time t, Vo is the value of the speed at t=0, and all along the track ,at all values of theta. The sentence "at Vo, theta=0" has no sense. The relation between $\theta$,$\dot \theta$, and the time is what you can read in Chest's post.
$r_0\sqrt{1+k^2}e^{-k\theta(t)}\dot \theta(t)=r_0\sqrt{1+k^2}\dot \theta(0)=V_0$

 

[cr7einstein]

@ehild thanks a lot; I have understood it now...thank you for your patience and understanding.

 

[ehild]

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The particle acceleration is the time derivative of the velocity vector, and, using what you have been learning for polar coordinates, is given by:
$$\vec{a}=\frac{d\vec v}{dt}$$
From a force balance on the particle, we know that the tangential component of acceleration is equal to zero (frictionless track), right? The tangential component of acceleration is equal to the acceleration vector dotted with a unit tangent vector to the trajectory. But the velocity vector is tangent to the trajectory. So, the unit vector tangent to the trajectory is equal to the velocity vector divided by its own magnitude. Since the dot product of the acceleration and a unit tangent vector to the trajectory must be equal to zero, the dot product of the acceleration vector with the velocity vector must also be equal to zero.

So, if we take the dot product of the acceleration vector with the velocity vector we obtain zero, for any horizontal frictionless track.
That means the time derivative of the velocity is perpendicular to the velocity. But $\frac{d(\vec v)^2}{dt}=2\vec v \dot{ \vec v} = 0$ as they are perpendicular. That means the time derivative of v² is constant, the speed is constant.

 

[Chestermiller]

So, if we take the dot product of the acceleration vector with the velocity vector we obtain zero, for any horizontal frictionless track.
That means the time derivative of the velocity is perpendicular to the velocity. But $\frac{d(\vec v)^2}{dt}=2\vec v \dot{ \vec v} = 0$ as they are perpendicular. That means the time derivative of v² is constant, the speed is constant.

Yes. That's what Haruspex was hinting at in post #24, but apparently, it did not resonate with the OP. It's not surprising, considering that the OP was considering $\theta$ to be an independent variable rather than the dependent variable (see post #34).