- #1

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- 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:

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