Physics olympiad problem -- struggling with polar coordinates

[cr7einstein]

Homework Statement

This is a physics olympiad problem; and I am still struggling with it. I will quote it here:

" A particle moves along a horizontal track following the trajectory $$r=r_{0}e^{-k\theta}$$, where $$\theta$$ is the angle made by the position vector with the horizontal. Recall that the velocity in polar coordinates is $$\frac{d\vec r}{dt}=\dot r \hat r+r \dot \theta \hat \theta$$. If at $$t=0, \theta=0$$; the velocity is $$v_{0}$$, Find $$\theta$$ dependence of velocity and the angle the velocity vector makes with the radial vector."

The answer-$$v(\theta)=v_{0}$$ and independent of $$\theta$$; and $$\alpha(\theta)=tan^{-1}(+_{-}\frac{1}{k})$$ and independent of $$\theta$$.

It makes sense, as the equation resembles a logarithmic spiral, and the results hold for it; but how do I prove it?

Homework Equations

The Attempt at a Solution

I tried differentiating it all I could, but I am always getting an extra $e^{-k\theta}$ in the term for velocity.

$$\vec v=\frac{d}{dt}(r_{0}e^{-k\theta}) \hat r + r_{0}e^{-k\theta}\frac{d\theta}{dt}=-kr_{0}e^{-k\theta}\dot \theta\hat r + r_{0}e^{-k\theta}\dot \theta \hat \theta=r_{0}e^{-k\theta}\dot \theta(-k\hat r+\hat \theta)--------(1)$$

At $$\theta=0$$,
$$\vec v= \vec v_{0}=r_{0}\dot \theta(-k\hat r+\hat \theta)------------------(2)$$But then how do I prove that (1) and (2) are the same; i.e. how do I get rid of $$e^{-k\theta}$$ in the first equation? Or do I have to approach it differently( but I want to stick to polar coordinates).

Please help; I am posting this after struggling for 2 straight days; and now I am completely frustrated. What's even more even annoying that it is only a 4 mark (2 marks for each part) problem.

Thanks in advance!

 

[Chestermiller]

To do this problem, you need to know how θ is varying with t, or if the magnitude of v is constant, or if the magnitude of dθ/dt is constant. Do they tell you any of these things?

 

[cr7einstein]

That is exactly my problem! If they mention any of the following, then it is just an equiangular logarithmic spiral and I can make a transformation $$\theta=\omega t$$ and $$k=cot \alpha$$ and get my answer; but they haven't mentioned any such thing. And I don't think this will be very complicated, as it was only a two marker.

 

[Chestermiller]

People who make up problems leave out important information all the time. It's very hard to remember to include everything when you are formulating a problem. I'm guessing that they meant to say that dθ/dt is constant.

Chet

 

[haruspex]

I'm guessing that they meant to say that dθ/dt is constant.

That would not yield a constant speed, but dr/dt constant would.
@cr7einstein , are you quite sure you quoted the problem word for word? There are a few phrases that read oddly. E.g., it says the track is horizontal, then specifies theta as the angle the radius makes to the horizontal.

 

[Chestermiller]

That would not yield a constant speed, but dr/dt constant would.
@cr7einstein , are you quite sure you quoted the problem word for word? There are a few phrases that read oddly. E.g., it says the track is horizontal, then specifies theta as the angle the radius makes to the horizontal.

Yes. Maybe they meant that the magnitude of the velocity vector is constant.

 

[cr7einstein]

I believe by 'horizontal track' they mean fixed in the horizontal plane. Also, I have proved that the direction is independent of $$\theta$$. However, I just need to show that the magnitude is independent of the angle, and at any instant the velocity would be the same as the velocity at t=0 and $$\theta=0$$. But for that, I would need to get rid of the exponential term. And I don't see how can I do that. Just in case, I will post the link to the question paper- http://olympiads.hbcse.tifr.res.in/uploads/ino-2015-qp-and-solutions/inpho-2015-qp-sol/at_download/file — PDF document, 1993Kb.
It also has a diagram (which is not very helpful though!).

 

[Chestermiller]

You left out some key words of the problem statement from your OP: "of mass m slides along a frictionless track." This is not strictly a kinematics problem. It is also a dynamics problem. You need to consider a force balance on the particle; what is the tangential acceleration equal to?

 

[cr7einstein]

Sorry! My bad!. I thought it was irrelevant.
Now usually in such cases the friction provides the centripetal force; but the track is frictionless; how can there be any force (other than the weight and the normal reaction, which cancel out) to cause any unbalanced forces? But then, what would cause the mass to move along the (almost)circular trajectory? I am even more confused now ! Any hints, please @Chestermiller ? I am guessing I should double differentiate the given equation of trajectory, and write down the 2nd law in polar coordinates.( But can't see how it helps me).

 

[TSny]

how can there be any force (other than the weight and the normal reaction, which cancel out)

Careful. You seem to be making the assumption that the normal reaction must be vertical.

 

[Orodruin]

You need to consider a force balance on the particle; what is the tangential acceleration equal to?

This is not necessary. You could simply refer to that no energy is dissipated from the system (force is orthogonal to velocity) and use energy conservation.

 

[Chestermiller]

This is not necessary. You could simply refer to that no energy is dissipated from the system (force is orthogonal to velocity) and use energy conservation.

What I was getting at was that, without friction, the tangential component of the velocity vector (i.e., the component in the direction tangent to the particle trajectory) does not change in magnitude from its initial value v₀.

@ cr7Einstein The tangential component of velocity vector is not the same as the theta component of the velocity vector (because the trajectory is not circular). Do you know how to determine the component of the velocity vector in the direction tangent to the particle trajectory? Of course, the tangential component of the velocity vector in the direction tangent to the particle trajectory is equal to the magnitude of the velocity vector, because the component normal to the particle trajectory is zero (since the particle is on a track).

Chet

 

[cr7einstein]

@Chestermiller Sorry I really don't follow you here. The velocity vector IS tangent to the particle trajectory, isn't it? If it isn't then how will you calculate the tangential component? As far as I have done, the direction of the velocity vector is independent of theta(look at the unit vectors); but the magnitude always seems to have the irritating exponential term. So I can't really prove that the velocity at theta=0 is the same as elsewhere, can I? I too am beginning to think that this is more like a dynamics problem, but it beats me how to move on furthur.

@TSny I don't see any reason for the normal reaction to not be vertical; unless of course I 'create' a component to provide the centripetal force; is that what you are saying?

@Orodruin what you said actually makes sense because as there is no PE term or dissipation, the only energy term will be a constant KE throughout ( and hence the velocity is constant too). But how do I quantitatively prove that $$v=v_{0}$$?

I solved the whole paper; but haven't been able to do this one from 4 days...and this supposed to an easy problem! totally frustrated. Sorry if I am being totally unreasonable. Appreciate the help guys, but I think that I would really need a solution now or I'm going to go crazy! Thanks in advance!

 

[Chestermiller]

@Chestermiller Sorry I really don't follow you here. The velocity vector IS tangent to the particle trajectory, isn't it?

Yes.

As far as I have done, the direction of the velocity vector is independent of theta(look at the unit vectors); but the magnitude always seems to have the irritating exponential term. So I can't really prove that the velocity at theta=0 is the same as elsewhere, can I? I too am beginning to think that this is more like a dynamics problem, but it beats me how to move on furthur.

In your original post, you wrote: $$\vec v=r_{0}e^{-k\theta}\dot \theta(-k\hat r+\hat \theta)\tag{1}$$
The magnitude of this velocity vector is $$v=r_{0}\sqrt{1+k^2}e^{-k\theta}\dot \theta\tag{2}$$
and the vector is, of course, tangent to the particle trajectory. Since the track is frictionless, the component of acceleration along the track is zero, and so the magnitude of the velocity vector is constant at its initial value throughout the motion. So,$$r_{0}\sqrt{1+k^2}e^{-k\theta}\dot \theta=v_0\tag{3}$$

Chet

 

[Orodruin]

But how do I quantitatively prove that

v=v0v=v0​

v=v_{0}?

What is the expression for the kinetic energy? Is there any potential energy?

 

[cr7einstein]

@Chestermiller but $$v=v_{0}$$ at $$\theta=0$$; the exponential term should vanish, shouldn't it?

@Orodruin No there is no PE, so $$1/m v^2=1/2 mv_{0}^2$$ ( for any two points) gives $$v=v_0$$(which is the solution); that is understood. What I am asking is that even though $$v_{0}$$ is independent of theta, why doesn't $$v$$ become independent of $$\theta$$ after all that calculus. ( as the problem suggests to use polar coordinates, I wanted to stick to it and get a 'rigorous' proof).

 

[Orodruin]

which is the solution); that is understood. What I am asking is that even though

v0v0​

v_{0} is independent of theta, why doesn't

vv​

v become independent of

θθ​

\theta after all that calculus. ( as the problem suggests to use polar coordinates, I wanted to stick to it and get a 'rigorous' proof).

$v$ is independent of $\theta$, but you can also express it as a function of $\theta$ and $\dot\theta$. This gives you a differential equation for $\theta(t)$.

Edit: Hint: You can write inline LaTeX by using double hashes (##) instead of double dollars ($$). Your posts will become much more readable.

 

[Chestermiller]

@Chestermiller but $$v=v_{0}$$ at $$\theta=0$$; the exponential term should vanish, shouldn't it?

Only at time t = 0. The initial condition is $\theta = 0$ at t = 0.

You have a variable-separable first order ordinary differential equation for dθ/dt with a specified initial condition. So how do you solve it?

 

[ehild]

No there is no PE, so $$1/m v^2=1/2 mv_{0}^2$$ ( for any two points) gives $$v=v_0$$(which is the solution); that is understood. What I am asking is that even though $$v_{0}$$ is independent of theta, why doesn't $$v$$ become independent of $$\theta$$ after all that calculus. ( as the problem suggests to use polar coordinates, I wanted to stick to it and get a 'rigorous' proof).

vo is the speed at t=0, when theta is also 0. The problem asks the θ dependence of the speed. The speed v is the same vo all along the track, that is independent of theta. Do not mix speed v with velocity $\vec v$.
It has a radial component and the component normal to the radius, along the direction of $\hat \theta$ . You get the cosine of the angle the velocity makes with the radial direction by using the dot product with the radial unit vector and dividing by the magnitude of the velocity.
To get the normal force, you can use the acceleration vector. It has a tangential and a normal component. The tangential component of the acceleration is parallel with the velocity vector. You get it if you dot-product the velocity and the acceleration and divide by the magnitude of the velocity. To get the normal component of the acceleration subtract its tangential component from it.

 

[cr7einstein]

Only at time t = 0. The initial condition is θ=0θ=0\theta = 0 at t = 0.

Yes, that is understood, @Chestermiller. But the problem asks to show that the velocity ANYWHERE is independent of theta; how to prove that?

You have a variable-separable first order ordinary differential equation for dθ/dt with a specified initial condition. So how do you solve it?

Which ODE are you talking about? The only one I can form here is $|v|=r_{0}e^{-k\theta}d\theta/dt \sqrt{1+k^2}$; but we don't know the speed as a function of time, do we? If you have some other equation in mind, post it please!

vvv is independent of θθ\theta, but you can also express it as a function of θθ\theta and ˙θθ˙\dot\theta. This gives you a differential equation for θ(t)

@Orodruin could you please send the equation you just mentioned? I cannot find a relation between velocity and theta unless I can express velocity as a function of time; and I don't see how can I do it. (Thanks for the tip on formatting! Didn't know I could do that).

vo is the speed at t=0, when theta is also 0. The problem asks the θ dependence of the speed. The speed v is the same vo all along the track, that is independent of theta. Do not mix speed v with velocity ⃗v

@ehild But that is what we are supposed to "prove", isn't it? It is quite easy to prove that the velocity's DIRECTION is independent of theta( As i did in the question); but even the magnitude(speed) has an exponential term which has $\theta$; which is not there for $v_{0}$ (i.e. at $\theta=0$). So, how do we show that the speed is independent of theta and the same as $v_{0}$ everywhere?
I have honestly tried my best to solve this, but I am really, really stuck. I don't know if this is appropriate, but I think it would be really nice if someone could give a full solution now. Thanks in advance!

 

[Orodruin]

could you please send the equation you just mentioned

It has already been posted in this thread by Chestermiller.

I suggest you go through the thread again and try to implement the hints you have been given and then come back to show us your attempt and where you get stuck.

 

[ehild]

But that is what we are supposed to "prove", isn't it? It is quite easy to prove that the velocity's DIRECTION is independent of theta( As i did in the question); but even the magnitude(speed) has an exponential term which has $\theta$; which is not there for $v_{0}$ (i.e. at $\theta=0$). So, how do we show that the speed is independent of theta and the same as $v_{0}$ everywhere?

The direction of the velocity does depend on theta. It's angle enclosed by the radius is independent. See figure.
The magnitude of the velocity (the speed) is constant because of conservation of energy. See Chestermiller's Post#14. There is no friction, the only force acting on the particle is the normal force and it does not do work.

The speed contains the product $e^{-k\theta}\dot \theta$ and it must be constant, because the speed is independent of theta. That means the angular velocity increases with theta.

 

[Chestermiller]

Yes, that is understood, @Chestermiller. But the problem asks to show that the velocity ANYWHERE is independent of theta; how to prove that?

Which ODE are you talking about? The only one I can form here is $|v|=r_{0}e^{-k\theta}d\theta/dt \sqrt{1+k^2}$; but we don't know the speed as a function of time, do we? If you have some other equation in mind, post it please!

OK, I thought that saying that the track is frictionless was enough to convince you that the tangential velocity should be constant. And apparently Orodruin's approach to showing that the kinetic energy was constant, also implying that the tangential velocity is constant, did not work for you either. I guess the only other thing that might work would be to determine the tangential component of acceleration and set that equal to zero (since from a force balance, if the track is frictionless, that must be equal to zero). Setting the tangential component of acceleration equal to zero will definitely show that the tangential velocity is constant.

Chet

 

[haruspex]

@cr7einstein , do you agree that if the track is smooth then the only force it can exert on the particle is at right angles to the track? $\vec v.\vec v=|v|^2$, yes? What do you get if you differentiate that?

 

[cr7einstein]

In your original post, you wrote:

⃗v=r0e−kθ˙θ(−k^r+^θ)​

(1)(1)v→=r0e−kθθ˙(−kr^+θ^)\vec v=r_{0}e^{-k\theta}\dot \theta(-k\hat r+\hat \theta)\tag{1}
The magnitude of this velocity vector is

v=r0√1+k2e−kθ˙θ​

(2)(2)v=r01+k2e−kθθ˙v=r_{0}\sqrt{1+k^2}e^{-k\theta}\dot \theta\tag{2}
and the vector is, of course, tangent to the particle trajectory. Since the track is frictionless, the component of acceleration along the track is zero, and so the magnitude of the velocity vector is constant at its initial value throughout the motion. So,

r0√1+k2e−kθ˙θ=v0​

(3)(3)r01+k2e−kθθ˙=v0r_{0}\sqrt{1+k^2}e^{-k\theta}\dot \theta=v_0\tag{3}

@Chestermiller, If you look at the equation(3)in your post #14 for $v_{0}$ is incorrect, as $\theta=0$ at $v=v_{0}$ so the exponential term vanishes. The equation you have given is for $v$ and not $v_{0}$(or am I missing something?). So we have $v_{0}=r_{0}\dot \theta ({1+k^2})^{1/2}$ and $v=-kr_{0}e^{(-k\theta)}\dot \theta ({1+k^2})^{1/2}$ , which obviously don't look the same, but we have to PROVE that they are.

Physically, many arguments have been given(by @Orodruin, @ehild, @haruspex) and I am satisfied by them all; and it is clear that the velocity must be constant and independent of theta( I can also take it as a result for equiangular logarithmic spirals, which is represented by the given equation). What I am asking, however, is that why are we not able to show the same by differentiating the given trajectory( as is hinted in the problem) in polar coordinates; how to 'show' that the velocity is independent of theta by using the approach suggested in the problem ( i.e. differentiating in polar coordinates). I believe the final two options left to do it this way are to somehow show that $\dot \theta$ contains an exponential term which cancels out the exponential arising in velocity term; or that $dv/d\theta=0$; neither of which I have been able to establish. Again, I want to do this problem using differentiating in polar coordinates; the way the problem suggests( for the fun of it). So basically, I could rephrase it as-
show that $v=v_{0}$ using differentiation in polar coordinates.

I hope I am clearer now. Thanks in advance!

 

[ehild]

Physically, many arguments have been given and I am satisfied by them all; and it is clear that the velocity must be constant and independent of theta( I can also take it as a result for equiangular logarithmic spirals, which is represented by the given equation).

The velocity is a vector, and changes direction along the path. It is not constant, but the speed is.

What I am asking, however, is that why are we not able to show the same by differentiating the given trajectory( as is hinted in the problem) in polar coordinates; how to 'show' that the velocity is independent of theta by using the approach suggested in the problem ( i.e. differentiating in polar coordinates). I believe the final two options left to do it this way are to somehow show that $\dot \theta$ contains an exponential term which cancels out the exponential arising in velocity term; or that $dv/d\theta=0$; neither of which I have been able to establish. Again, I want to do this problem using differentiating in polar coordinates; the way the problem suggests( for the fun of it). So basically, I could rephrase it as-
show that $v=v_{0}$ using differentiation in polar coordinates.

Only the form of the trajectory is given, and the particle can move along it "by itself" if it gets an initial push; and then its KE and speed are constant in lack of any loss. But it can move under the effect of a force, and then the angular velocity can be anything, and the speed, which is proportional to $e^{-k\theta} \dot \theta$ won't be constant.
At the same time, the constancy of the linear speed means, that the angular speed is proportional to e^kθ.
That the motion is in horizontal plane and there is no friction - it is crucial. The speed is constant because there is no loss of KE: You can not derive it from the form of the trajectory.

 

[cr7einstein]

That the motion is in horizontal plane and there is no friction - it is crucial. The speed is constant because there is no loss of KE: You can not derive it from the form of the trajectory.

So you mean to say that we cannot derive the $\theta$ dependence of speed from the equation of trajectory? But then what is the point of giving all that information on polar coordinates in the problem?

 

[ehild]

So you mean to say that we cannot derive the $\theta$ dependence of speed from the equation of trajectory? But then what is the point of giving all that information on polar coordinates in the problem?

We cannot derive the theta dependence of speed from the trajectory. Think: You drive a car along a bending road. Can you derive the speed from the radius of the turn? You can drive with 1 km/h or 50 km/h... there is only an upper limit if you want to stay on track.
You could not derive the x or y dependence of the speed either, if the form of the track would be given in Cartesian coordinates.

 

[Chestermiller]

@Chestermiller, If you look at the equation(3)in your post #14 for $v_{0}$ is incorrect, as $\theta=0$ at $v=v_{0}$ so the exponential term vanishes. The equation you have given is for $v$ and not $v_{0}$(or am I missing something?). So we have $v_{0}=r_{0}\dot \theta ({1+k^2})^{1/2}$ and $v=-kr_{0}e^{(-k\theta)}\dot \theta ({1+k^2})^{1/2}$ , which obviously don't look the same, but we have to PROVE that they are.

Physically, many arguments have been given(by @Orodruin, @ehild, @haruspex) and I am satisfied by them all; and it is clear that the velocity must be constant and independent of theta( I can also take it as a result for equiangular logarithmic spirals, which is represented by the given equation). What I am asking, however, is that why are we not able to show the same by differentiating the given trajectory( as is hinted in the problem) in polar coordinates; how to 'show' that the velocity is independent of theta by using the approach suggested in the problem ( i.e. differentiating in polar coordinates). I believe the final two options left to do it this way are to somehow show that $\dot \theta$ contains an exponential term which cancels out the exponential arising in velocity term; or that $dv/d\theta=0$; neither of which I have been able to establish. Again, I want to do this problem using differentiating in polar coordinates; the way the problem suggests( for the fun of it). So basically, I could rephrase it as-
show that $v=v_{0}$ using differentiation in polar coordinates.

I hope I am clearer now. Thanks in advance!

The methodology I recommended in my post #23 should do what you are trying to do. So here goes.

In your original post, the velocity vector at any location along the track is given by Eqn. 1:

$$\vec v=r_{0}e^{-k\theta}\dot \theta(-k\hat r+\hat \theta)\tag{1}$$

The particle acceleration is the time derivative of the velocity vector, and, using what you have been learning for polar coordinates, is given by:
$$\vec{a}=\frac{d\vec v}{dt}=\frac{d(r_{0}e^{-k\theta}\dot \theta)}{dt}(-k\hat r+\hat \theta)-(r_{0}e^{-k\theta}\dot \theta)(k\hat \theta+\hat r)\tag{2}$$
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 from Eqn. 2 with the velocity vector from Eqn. 1, we obtain:

$$\vec{a} \cdot \vec{v}=(r_{0}e^{-k\theta}\dot \theta)\frac{d(r_{0}e^{-k\theta}\dot \theta)}{dt}(1+k^2)=\frac{1}{2}\frac{d(r_0\sqrt{1+k^2}e^{-k\theta}\dot \theta)^2}{dt}=0\tag{3}$$
The solution to this differential equation is:
$$(r_0\sqrt{1+k^2}e^{-k\theta}\dot \theta)^2=C\tag{4}$$where C is the constant of integration. The constant of integration is obtained by applying the initial condition $(r_0\sqrt{1+k^2}e^{-k\theta}\dot \theta)^2=v^2_0$ @ t = 0. So, $C=v^2_0$. Finally,
$$r_0\sqrt{1+k^2}e^{-k\theta}\dot \theta=v_0\tag{5}$$at all times.

 

[cr7einstein]

The solution to this differential equation is:

(r0√1+k2e−kθ˙θ)2=C​

(4)(4)(r01+k2e−kθθ˙)2=C(r_0\sqrt{1+k^2}e^{-k\theta}\dot \theta)^2=C\tag{4}where C is the constant of integration. The constant of integration is obtained by applying the initial condition (r0√1+k2e−kθ˙θ)2=v20(r01+k2e−kθθ˙)2=v02(r_0\sqrt{1+k^2}e^{-k\theta}\dot \theta)^2=v^2_0 @ t = 0. So, C=v20C=v02C=v^2_0. Finally,

r0√1+k2e−kθ˙θ=v0​

(5)(5)r01+k2e−kθθ˙=v0r_0\sqrt{1+k^2}e^{-k\theta}\dot \theta=v_0\tag{5}at all times.

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 $\theta=0$.

(r0√1+k2e−kθ˙θ)2=v20(r01+k2e−kθθ˙)2=v02(r_0\sqrt{1+k^2}e^{-k\theta}\dot \theta)^2=v^2_0 @ t = 0. So, C=v20C=v02C=v^2_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?)

 

[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.
$r_0\sqrt{1+k^2}e^{-k\theta(t)}\dot \theta(t)=r_0\sqrt{1+k^2}\dot \theta(0)=V_0$