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Suppose an object is subjected to a force of constant magnitude, which is always directed to the origin. And suppose we know the initial position of the object relative to the origin, and the initial velocity of the object, can we determine if the object will perform uniform circular motion? If so, what conditions are necessary? Can we determine its position as a function of time from these givens?

I know that if we know that an object performs uniform circular motion, and we have the equations which describe its motion, for example $$ \mathbf r= \begin{bmatrix} \cos(t)\\ \sin(t)\\ \end{bmatrix} $$ we can find the velocity, and acceleration simply by taking derivitives. But can we go the other way around and deduce the equation of motion as I described above? Perhaps by solving the differential equation $$ m \ddot{\mathbf r} = - \lVert \mathbf F \rVert \frac{\mathbf{r}}{\lVert \mathbf r \rVert}$$ where $\lVert \mathbf F \rVert$ is constant?

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  • $\begingroup$ When you write: "Suppose an object is subjected to a force of constant magnitude, which is always directed to the origin..." how do you determine that it is "directed to the origin" ? Do you use a light beam to aim at the origin or an abstract geometric line that aims at the origin with an infinite speed ? $\endgroup$ Oct 21 '20 at 0:40
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But can we go the other way around and deduce the equation of motion as I described above? Perhaps by solving the differential equation

$$m \ddot{\mathbf r} = - \lVert \mathbf F \rVert \frac{\mathbf{r}}{\lVert \mathbf r \rVert}$$ where $||\mathbf F||$ is constant

This would not be the correct equation of motion.

You can just apply Newton's second law in Polar Coordinates:

$$\mathbf F=m\mathbf a=m(\ddot r-r\dot\theta^2)\,\hat r+m(r\ddot\theta+2\dot r\dot\theta)\,\hat\theta$$

For a force of constant magnitude always pointing towards the origin we have $\mathbf F=-F\,\hat r$, and so the equations of motion become

$$m\ddot r-mr\dot\theta^2+F=0$$ $$r\ddot\theta+2\dot r\dot\theta=0$$

which hold for any initial conditions.

In order to have uniform circular motion, we need

  1. $\dot r$, $\ddot r$, and $\ddot\theta$ to all be $0$ and,
  2. $r$ and $\dot \theta$ to be non-zero (they also need to be constant, but that follows from point 1).

This occurs when

$$F=mr\dot\theta^2$$ $$\dot r=0$$

So this shows that in order to have uniform circular motion we need for our initial conditions

  1. The force magnitude is equal to $mr(0)\cdot(\dot\theta(0))^2$
  2. $\dot r(0)=0$

If these two properties are not true of the initial conditions then you will not get uniform circular motion. You can determine what the motion will be from the general equations of motion we obtained above.

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  • $\begingroup$ Thank you for your answer!! There are two points I want to clarify: The first is, if I am given the initial position and velocity in cartesian coordinates, how can I calculate $\dot \theta(0)$ and $\dot r(0)$? The second is, Are those conditions only necessary or also sufficient, meaning if I know that these initial conditions hold, can I be sure that the motion will be uniform circular motion? $\endgroup$
    – Roee
    Oct 21 '20 at 7:16
  • $\begingroup$ @user800827 I actually showed they are sufficient here, but they are also necessary. To determine the initial polar coordinates just transform your initial velocity vector into polar coordinates based on the initial position. $\endgroup$ Oct 21 '20 at 12:51
  • $\begingroup$ Could you explain exactly why they are sufficient? I can't imagine why $\dot r(0)=0$ would imply $\dot r = 0$ or why $F=mr(0)\cdot(\dot\theta(0))^2$ would imply $F=mr\dot\theta^2$, unless we also know that $\ddot r=0$ and $\ddot \theta=0$, which is not necessarily given. $\endgroup$
    – Roee
    Oct 21 '20 at 12:56
  • $\begingroup$ @user800827 Look at the equations of motion and think about what happens when those two conditions are true. $\endgroup$ Oct 21 '20 at 13:24
  • $\begingroup$ I would appreciate it if you could write down a proof of these implications, as I haven't been able to prove them. $\endgroup$
    – Roee
    Oct 21 '20 at 17:59
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These are the equations of motion

$${\ddot{r}}\,m-m{\dot\varphi }^{2}r+F=0\tag 1$$ $$ \ddot\varphi \,r+2\,{\dot r}\,\dot\varphi=0\tag 2 $$

equation (2) is also; $$\frac{d}{dt}\left(\,r^2\dot\varphi\right)=0$$

thus :

$$\dot{\varphi}=\frac{L}{r^2}$$ where L is a constant. substitute $\dot\varphi$ in equation (1)

$$m\,\ddot{r}-m\,\frac{L^2}{r^3}+F=0\tag 3$$

the conditions for a uniform circular motion are: $~\ddot r=0$ and $r=\text{constant}=r_0~\Rightarrow $ the initial conditions are

$$r(0)=r_0,~D(r)(0)=0$$

and the force F is:

$$F=m\,\frac{L^2}{\,r_0^3}$$

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