# Spiral motion and the centripetal force

• mathman44

#### mathman44

Major brainfart here.

Consider the spiral motion of a particle such that the distance from the origin to the particle is

r(t) = e^t

with a constant angular velocity.

Now since the particle rotating with a constant angular velocity, I would think that the net force on the particle would have to be an inward centripetal force in the radial direction.

However, there must also be a net outward force, in the radial direction, to account for the accelerating distance between the origin and the particle.

Obviously this can't be right, but I can't pin down where I'm going wrong.

Cheers!

hi mathman44!

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the radial acceleration is r'' - θ'2r

(and can you prove that? )

Isn't r'' the radial acceleration? Did you mean that r'' = -w^2 * r ?

And isn't this just equivalent to saying that the net force is equal to the centripetal force?

ps for a polar system I'm getting

mrw^2 - dV/dr = mr''

as the force equation using a simple lagrangian.

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Consider the spiral motion of a particle such that the distance from the origin to the particle is

r(t) = e^t

with a constant angular velocity.

Now since the particle rotating with a constant angular velocity, I would think that the net force on the particle would have to be an inward centripetal force in the radial direction.
No. Consider a particle on a straight course missing the origin. It has an angular velocity around the origin, and if its speed changes appropriately it may even have a constant angular velocity for while (but will have to shoot off to infinity in a finite time).
However, there must also be a net outward force, in the radial direction, to account for the accelerating distance between the origin and the particle.
It's not at all clear that it would be a net outward or net inward. That will depend on the details of r(t).
Radial acceleration means the component of the acceleration which is directed towards/away from the origin (not, acceleration in the value of the radius). As Tiny Tim says, this is given by r'' - θ'2r. E.g. in circular motion about the origin, r'' = 0, but the particle will have acceleration towards the origin.
$$\boldsymbol{a}\ =\ (\ddot{r}-r\dot{\theta}^2)\hat{\boldsymbol{r}}\ +\ (r\ddot{\theta}+2\dot{r}\dot{\theta}) \hat{\boldsymbol{\theta}}$$