 #1
JD_PM
 1,128
 158
 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.

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

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