# Central force on a particle following a logarithmic spiral

JD_PM
Homework Statement:
Given a central force field that allows to keep particles in the following logarithmic spiral ##r( \theta ) = k e^{\alpha \theta}## (where ##k## and ##\alpha## are constant).

a) Determine ##r(t)## and ##\theta (t)##.

b) Give the equation of the force.

c) Discuss the nature of the force fields associated to the obtained force.
Relevant Equations:
$$m( \ddot r - r \dot \theta^2 ) = F(r)$$
$$m( r \ddot \theta + 2 \dot r \dot \theta) = 0$$
I want to focus this question on understanding the force ##F(r)## I get (thus, I want to focus on c) ). However, below the dashed line, I included steps on how I derived ##F(r)##.

We are going to work in polar coordinates.

Knowing that the acceleration is:

$$a = \Big( \ddot r - r \dot \theta^2 \Big) \hat r + \Big( r \ddot \theta + 2 \dot r \dot \theta \Big) \hat \theta$$

Newton's second law ends up with the form:

$$F(r) = m \Big( \ddot r - r \dot \theta^2 \Big) \ \ \ \ (1)$$

$$m \Big( r \ddot \theta + 2 \dot r \dot \theta \Big) = 0 \ \ \ \ \ \ \ (2)$$

a)

$$\theta (t) = \frac{(At + B)}{mr^2} \ \ \ \ \ (3)$$

$$r (t) = k \ exp \Big(\alpha \frac{(At + B)}{mr^2} \Big) \ \ \ \ \ (4)$$

b)

$$F(r) = m \Big[ k\Big( \frac{A \alpha}{m r^2} \Big)^2 e^{\frac{B \alpha}{m r^2}} e^{\frac{A \alpha t}{m r^2}} - \frac{1}{r^3} \Big(\frac{A}{m}\Big)^2 \Big] \ \ \ \ \ (5)$$

c)

Let me define the following quantities to simplify the equation for ##F(r)##

$$C = \frac{A}{m}$$

$$D = \frac{B}{m}$$

So we get a central force of the form:

$$F(r) = m \Big[ k\Big( \frac{C \alpha}{ r^2} \Big)^2 e^{\frac{D \alpha}{r^2}} e^{\frac{C \alpha t}{ r^2}} - \frac{C^2}{r^3}\Big] \ \ \ \ \ (6)$$

Alright so one notices that the repulsive inverse square field and the attractive inverse cube field are involved.

However, what can we say about the exponents?

Thanks.

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a)

- Getting ##\theta (t)##

OK so ##(2)## can be rewritten as follows:

$$\frac{d}{r dt} \Big( m r^2 \dot \theta \Big) = 0$$

Integrating twice with respect to ##t## we get:

$$\theta (t) = \frac{(At + B)}{mr^2} \ \ \ \ \ (3)$$

Where ##A## and ##B## are constants.

- Getting ##r (t)##

By plugging in equation ##(3)## into the given equation ##r( \theta ) = k e^{\alpha \theta}## we get ##r (t)##

$$r (t) = k \ exp \Big(\alpha \frac{(At + B)}{mr^2} \Big) \ \ \ \ \ (4)$$

b)

So we just have to use equation ##(1)## to get:

$$F(r) = m \Big[ k\Big( \frac{A \alpha}{m r^2} \Big)^2 e^{\frac{B \alpha}{m r^2}} e^{\frac{A \alpha t}{m r^2}} - \frac{1}{r^3} \Big(\frac{A}{m}\Big)^2 \Big] \ \ \ \ \ (5)$$

Homework Helper
$$m( r \ddot \theta + 2 \dot r \dot \theta) = 0$$ ...
(2) can be rewritten as follows ...
Integrating ##{\displaystyle {d\over dt}}\Bigl ( mr^2\dot \theta \Bigr) = 0\ ## once with respect to ##t## we get
##\Bigl ( mr^2\dot \theta \Bigr) = A\ ##, i.e. angular momentum is constant.
How do you get (3) from that ?

• JD_PM
Abhishek11235
Can you elaborate more on your question? Do you want a graph of function?

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BvU is pointing out that you made a mistake in going from

$$\frac{d}{ dt} \Big( m r^2 \dot \theta \Big) = 0$$
to
$$\theta (t) = \frac{(At + B)}{mr^2} \ \ \ \ \ (3)$$

You said that you integrated the first equation twice with respect to time. But it appears that you treated ##r## as a constant in the second integration.

• JD_PM
JD_PM
BvU is pointing out that you made a mistake ... you treated ##r## as a constant in the second integration.

You both are right, let's fix it.

Let me use:

$$C = \frac{A}{m}$$

The idea is to integrate the following equation with respect to time (where both ##r## and ##\theta## are functions of time):

$$\int d \theta= \int \frac{C}{r^2} dt \ \ \ \ (7)$$

We do not know what is the explicit formula for ##r(t)## but we know what's it for ##r(\theta )##

$$r( \theta ) = k e^{\alpha \theta}$$

My idea is to use the chain rule to get ##r(t)##:

$$\frac{d}{dt} r(t) = \frac{dr}{d \theta}\frac{d \theta}{dt} = k \alpha e^{\alpha \theta} \dot \theta \ \ \ \ (8)$$

The next step is plugging ##(8)## into ##(7)## and integrate.

Multiplying top and bottom of the RHS of ##(7)## by ##(\frac{d}{dt})^2##, I get:

$$\frac{(\frac{d}{dt})^2 C dt}{(\frac{d}{dt} r)^2} = \frac{(\frac{d}{dt})^2 C dt}{ (k \alpha e^{\alpha \theta} \dot \theta)^2} = \frac{ C dt}{ (k \alpha e^{\alpha \theta} d\theta)^2}$$

$$\int k \alpha e^{\alpha \theta} (d \theta)^3 = \int C dt$$

I've done something wrong because ##d \theta## shouldn't be to the power of three.

Do you agree on the idea of using the chain rule though?

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Do you agree on the idea of using the chain rule though?

Another idea is to use or derive the general central force equation:
$$\frac{d^2}{d\theta^2}(\frac 1 r ) + \frac 1 r = -\frac{mr^2}{l^2}F(r)$$

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• JD_PM and TSny
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My idea is to use the chain rule to get ##r(t)##:

$$\frac{d}{dt} r(t) = \frac{dr}{d \theta}\frac{d \theta}{dt} = k \alpha e^{\alpha \theta} \dot \theta \ \ \ \ (8)$$
OK. Note that this may be written as $$\dot r = \alpha r \dot \theta \ \ \ \ (8')$$

The next step is plugging ##(8)## into ##(7)## and integrate.

Multiplying top and bottom of the RHS of ##(7)## by ##(\frac{d}{dt})^2##, I get:

$$\frac{(\frac{d}{dt})^2 C dt}{(\frac{d}{dt} r)^2} = \frac{(\frac{d}{dt})^2 C dt}{ (k \alpha e^{\alpha \theta} \dot \theta)^2} = \frac{ C dt}{ (k \alpha e^{\alpha \theta} d\theta)^2}$$
What you did here is a wild ride! But it's not correct. You cannot treat ##(\frac{d}{dt})^2## this way.

In your first post, you arrived at ##\frac{d}{rdt} (mr^2\dot\theta) = 0## which may be simplified to ##\frac{d}{dt}(r^2 \dot \theta)=0##.
So ##r^2 \dot \theta = C##, where ##C## is a constant. (##C## is the angular momentum per unit mass.) Solve this for ##\dot \theta## and substitute into equation (8') above.

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• JD_PM
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OK. Note that this may be written as $$\dot r = \alpha r \dot \theta \ \ \ \ (8')$$

So ##r^2 \dot \theta = C##, where ##C## is a constant.

If you've got this far, perhaps solving for ##r## first is simplest!

JD_PM
What you did here is a wild ride! But it's not correct. You cannot treat ##(\frac{d}{dt})^2## this way.

I have the impression that mathematicians treat ##\frac{d}{dt}## differently than physicists, don't they?

Try solving this for ##\dot \theta## and then substitute into equation (8') above.

We have:

$$\dot \theta = \frac{C}{r^2} \ \ \ \ (9)$$

Plugging ##(9)## into ##(8')## we get:

$$\dot r - \frac{\alpha C}{r} = 0$$

This is a linear first order homogeneous ODE whose solution is (if you do not agree I'll include the steps):

$$r(t) = \pm \sqrt{2C(\alpha t + E)} \ \ \ \ (10)$$

Where ##E## is the constant of integration.

Plugging ##(10)## into ##(9)## we get:

$$d \theta = \frac{d t}{2C(\alpha t + E)}$$

This is a first order homogeneous ODE whose solution is (if you do not agree I'll include the steps):

$$\theta (t) = \frac{ln(\alpha t + E)}{2 \alpha} + F$$

Where ##F## is the constant of integration.

The double derivative of ##(10)## with respect to time is:

$$\ddot r = \frac{-2 \alpha^2 C^2}{4(2C(\alpha t + E))^2} = \frac{-\alpha^2 C^2}{2r^2}$$

The force ##F(r)## I get is:

$$F(r) = m \Big( \frac{-\alpha^2 C^2}{2r^2} - \frac{C}{r}\Big)$$

There may be something wrong. Checking dimensional analysis ...

EDIT: As PeroK pointed out, this result is not OK. Dimensional analysis of ##\frac{-\alpha^2 C^2}{2r^2}## yields ##L^2##, while it should yield the dimension for acceleration: ##\frac{L}{T^2}##

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$$r(t) = \pm \sqrt{2C(\alpha t + E)} \ \ \ \ (10)$$

Where ##E## is the constant of integration.

First, you should be able to simplify this immediately to:

$$r(t) = \sqrt{r_0^2 + 2C\alpha t}$$

Second, you already have an equation relating ##r## and ##\theta##, so use that instead of integrating ##\theta##.

To get ##f(r)## you can again shortcut things by using the constant angular momentum to avoid differentiating ##\theta##.

PS I'd replace ##C = l/m## where ##l## is the angular momentum.

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The force ##F(r)## I get is:

$$F(r) = m \Big( \frac{-\alpha^2 C^2}{2r^2} - \frac{C}{r}\Big)$$

There may be something wrong. Checking dimensional analysis ...

This can't be correct.

• JD_PM
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To get ##f(r)## you can again shortcut things by using the constant angular momentum to avoid differentiating ##\theta##.

By this I mean that, using ##l = mr^2 \dot \theta##, we have:
$$f(r) = m(\ddot r - r \dot \theta^2) = m(\ddot r - \frac{l^2}{m^2 r^3})$$

JD_PM
Second, you already have an equation relating ##r## and ##\theta##, so use that instead of integrating ##\theta##.

Which one?

The only equation I see is ##\dot r = \alpha r \dot \theta##, and we need to integrate it.

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Which one?

The only equation I see is ##\dot r = \alpha r \dot \theta##, and we need to integrate it.

You forgot this one!

Homework Statement:: Given a central force field that allows to keep particles in the following logarithmic spiral ##r( \theta ) = k e^{\alpha \theta}## (where ##k## and ##\alpha## are constant).

• JD_PM
JD_PM
OK I made a mistake here:

$$\ddot r = \frac{-2 \alpha^2 C^2}{4(2C(\alpha t + E))^2} = \frac{-\alpha^2 C^2}{2r^2}$$

It is

$$\ddot r = \frac{-\alpha^2 C^2}{r^{3/2}}$$

Thus the force is:

$$F(r) = m \Big( \frac{-\alpha^2 C^2}{r^{3/2}} - \frac{C}{r}\Big)$$

But this is again wrong (I checked dimensions and do not match).

I do not see why, as I checked several times my computations and they are OK.

EDIT: I am going to try the procedure PeroK suggested

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OK I made a mistake here:

It is

$$\ddot r = \frac{-\alpha^2 C^2}{r^{3/2}}$$

Thus the force is:

$$F(r) = m \Big( \frac{-\alpha^2 C^2}{r^{3/2}} - \frac{C}{r}\Big)$$

But this is again wrong (I checked dimensions and do not match).

I do not see why, as I checked several times my computations and they are OK.

Sadly, I think both terms are wrong. If ##r = (a + bt)^{1/2}##, then time derivatives of ##r## involve (odd) integer powers of ##r##.

Alternatively, try implicitly differentiating ##r^2##.

I don't see how you got the second term. See post #12.

PS If you get ##F(r)## using post #6, you can see what you're aiming for!

• JD_PM
JD_PM
Alright, so from post #12 we see that ##F(r)## is:

$$F(r) = m \Big( \ddot r - \frac{C^2}{r^3}\Big)$$

By differentiating equation (10) we get:

$$\ddot r = \frac{-\alpha^2 C^2}{r^{3/2}}$$

Thus:

$$F(r) = m \Big( \frac{-\alpha^2 C^2}{r^{3/2}} - \frac{C^2}{r^3}\Big)$$

Which can be rewritten as:

$$F(r) = \frac{A^2}{mr^2} \Big( \frac{-\alpha^2 }{r^{-1/2}} - \frac{1}{r}\Big)$$

I am closer to the equation you gave at #6;

I am sure there has to be a way to go from ##\frac{-\alpha^2 }{r^{-1/2}}## to ##\frac{d^2}{d\theta^2}(-\frac 1 r )##... but how? I have been trying to use the chain rule. We also know ##r(\theta)## from the exercise statement. I suspect we could also use it to get rid of ##\alpha##

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Alright, so from post #12 we see that ##F(r)## is:

$$F(r) = m \Big( \ddot r - \frac{C^2}{r^3}\Big)$$
OK

By differentiating equation (10) we get:

$$\ddot r = \frac{-\alpha^2 C^2}{r^{3/2}}$$
The power of ##r## in the denominator is not correct. Once you fix that, you will see that ##F(r)## has a nice dependence on ##r##.

• JD_PM and BvU
JD_PM
The power of ##r## in the denominator is not correct. Once you fix that, you will see that ##F(r)## has a nice dependence on ##r##.

$$\ddot r = \frac{-\alpha^2 C^2}{r^3}$$

Thus ##F(r)## is:

$$F(r) = \frac{A^2}{mr^2} \Big( \frac{-\alpha^2 }{r} - \frac{1}{r}\Big)$$

This expression is OK. I've noticed that the expression ##\frac{-\alpha^2 }{r}## is OK, as ##\alpha## is dimensionless.

To put it in the same form PeroK suggested in #6, we see that:

$$\frac{d^2}{d \theta^2} \Big(\frac{1}{r}\Big) = \frac{\alpha^2 }{ke^{\alpha \theta}}$$

Thus:

$$F(r) = \frac{A^2}{mr^2} \Big( -\frac{d^2}{d \theta^2} \Big(\frac{1}{r}\Big) - \frac{1}{r}\Big)$$

Thank you all, it was fun!

Nature of the obtained force

Only the attractive inverse cube field is involved.

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$$F(r) = \frac{A^2}{mr^2} \Big( \frac{-\alpha^2 }{r} - \frac{1}{r}\Big)$$

$$\frac{d^2}{d \theta^2} \Big(\frac{1}{r}\Big) = \frac{-\alpha^2 }{ke^{\alpha \theta}}$$

$$F(r) = \frac{A^2}{mr^2} \Big( -\frac{d^2}{d \theta^2} \Big(\frac{1}{r}\Big) - \frac{1}{r}\Big)$$

Just an observation. It looks odd to me that you leave expressions in the form that you do. Maybe it's just style, but I would have written:
$$F(r) = \frac{A^2}{mr^2} \Big( \frac{-\alpha^2 }{r} - \frac{1}{r}\Big) = -\frac{A^2}{mr^3}(\alpha^2 +1)$$
$$\frac{d^2}{d \theta^2} \Big(\frac{1}{r}\Big) = \frac{-\alpha^2 }{ke^{\alpha \theta}} = -\frac{\alpha^2 }{r}$$
$$F(r) = \frac{A^2}{mr^2} \Big( -\frac{d^2}{d \theta^2} \Big(\frac{1}{r}\Big) - \frac{1}{r}\Big) = -\frac{A^2}{mr^3}(\alpha^2 +1)$$

It's almost like you switch off one step before the end. It's a minor point, perhaps.

• JD_PM
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I think the issue is a bit overcomplicated in this thread. The point is that you have given the trajectory in terms of ##r=r(\theta)##. You don't care about how the trajectory is run through as a function of time. The two "homework equations" are all you really need.

The 2nd equation is nothing else than angular-momentum conservation, i.e., you can integrate it to
$$r^2 \dot{\theta}=\ell=\text{const}. \quad (*)$$
Then you introduce the central potential
$$V(r)=-\int \mathrm{d} r F(r)$$
and use the energy-conservation law,
$$\frac{m}{2} \dot{\vec{x}}^2 + V(r)=E=\text{const},$$
$$\frac{m}{2} (\dot{r}^2 + r^2 \dot{\theta}^2)+V(r)=E. \qquad (**)$$
Now you can eliminate ##\dot{\theta}##, using (*) and get a differential equation for ##\mathrm{d}r/\mathrm{d} \theta=\dot{r}/\dot{\theta}## from (**), which let's you derive ##V(r)## from the given solution ##r=r(\theta)##.

• • BvU and JD_PM
JD_PM
Thank you for your post vanhees71! :)

You don't care about how the trajectory is run through as a function of time.

But note that ##a)## is asking for ##r(t)## and ##\theta (t)##.

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Sure, but this you can address easier as soon has you have found ##F(r)## or ##V(r)## ;-).

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I think the issue is a bit overcomplicated in this thread. The point is that you have given the trajectory in terms of ##r=r(\theta)##. You don't care about how the trajectory is run through as a function of time.
Part a) in the OP asked for the trajectory as a function of time!

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Yes, and this you get after you've found ##V(r)## by solving the equation of motion for ##r##.

$$r = ke^{\alpha \theta}$$
$$\dot r = \alpha r \dot \theta = \frac{\alpha l}{mr}$$
$$\frac{d(r^2)}{dt} = 2r \dot r = \frac{2\alpha l}{m}$$
$$r = (r_0^2 + \frac{2\alpha l}{m}t)^{1/2}$$
• 