Equations of Motion of a Mass Attached to Rotating Spring

[SaraM]

1. Homework Statement

A particle of mass m is attached to the end of a light spring of equilibrium length a, whose other end is fixed, so that the spring is free to rotate in a horizontal plane. The tension in the spring is k times its extension. Initially the system is at rest and the particle is given an impulse that starts its movement at right angles to the spring with velocity v. Write down the equations of motion in polar co-ordinates.

2. Homework Equations

Radial Force F_r= -k⋅r
Radial acceleration a_r = mv²/r
Tangential Force F_θ = Torque = r⋅F

3. The Attempt at a Solution

I tried applying Newton's second law for both the radial and tangential components.
In r : ma_r= -k⋅r ⇒ r = a⋅cos(ωt) (which is not the correct answer)
In θ: 1- mr⋅a_θ= r⋅F (got me nowhere)
2- v_θ= v/r=ω ; ω=√k/m (how do I recover θ(t) from it?)

Thank you in advance

 

[TSny]

Welcome to PF!

2. Homework Equations

Radial Force F_r= -k⋅r

This does not account for the equilibrium length of the spring.

Radial acceleration a_r = mv²/r

Does the expression on the right have the correct dimensions for an acceleration?
In polar coordinates, the radial acceleration a_r has two terms. For a review of velocity and acceleration in polar coordinates, see http://faculty.etsu.edu/gardnerr/2110/notes-12e/c13s6.pdf

Tangential Force F_θ = Torque = r⋅F

Force and torque do not have the same dimensions. So, a force cannot equal a torque.

 

[SaraM]

Thank you for your reply!

Following your hints, here's what I ended up with:
In $r$: $~m( \ddot r -r \dot \theta^2)=- k(r-a)$
In $\theta$: $~\dot\theta=\frac{v}{r}=-\frac{k}{mv}(r-a)$ (which I got from the centripetal force $m\frac{v^2}{r}=-k(r-a)$)

What's left to do is plug in $\dot\theta$ in the equation in $r$ and find $r(t)$
Is there still something incorrect in my reasoning?

 

[TSny]

$m( \ddot r -r \dot \theta^2)=- k(r-a)$

OK, this looks good. It expresses $\sum F_r = ma_r$.

$\dot\theta=\frac{v}{r}$

This is not quite correct. Note that $r \dot{\theta}$ represents the $\theta$ component of $\vec{v}$.

$m\frac{v^2}{r}=-k(r-a)$

No, the correct relation is the relation that you got above: $m( \ddot r -r \dot \theta^2)=- k(r-a)$.

What's left to do is plug in $\dot \theta$ in the equation in $r$ and find $r(t)$.

OK. To get the correct expression for $\dot \theta$ in terms of $r$ you can do either of the following:
(1) Set up $\sum F_\theta = ma_\theta$. Consult the link for the expression for $a_\theta$.

(2) Decide if there is any torque acting on the particle relative to the origin. If not, what quantity is conserved?

 

[SaraM]

OK, got it. Thank you so much for your help! Cleared things up.