Equations of Motion of a Mass Attached to Rotating Spring

AI Thread Summary
The discussion centers on deriving the equations of motion for a mass attached to a rotating spring in polar coordinates. The radial force is defined as Fr = -k(r-a), accounting for the spring's equilibrium length. Participants clarify the need for proper dimensional analysis and the correct application of Newton's second law for both radial and tangential components. The final equations involve expressing the radial acceleration and angular velocity in terms of the spring's extension and mass. The conversation concludes with a focus on finding the correct expression for angular velocity and the next steps to solve for r(t).
SaraM
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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 Fr= -k⋅r
Radial acceleration ar = mv2/r
Tangential Force Fθ = 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 : mar= -k⋅r ⇒ r = a⋅cos(ωt) (which is not the correct answer)
In θ: 1- mr⋅aθ= r⋅F (got me nowhere)
2- vθ= v/r=ω ; ω=√k/m (how do I recover θ(t) from it?)

Thank you in advance
 
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Welcome to PF!
SaraM said:
2. Homework Equations

Radial Force Fr= -k⋅r
This does not account for the equilibrium length of the spring.
Radial acceleration ar = mv2/r
Does the expression on the right have the correct dimensions for an acceleration?
In polar coordinates, the radial acceleration ar 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 Fθ = Torque = r⋅F
Force and torque do not have the same dimensions. So, a force cannot equal a torque.
 
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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?
 
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SaraM said:
##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?
 
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OK, got it. Thank you so much for your help! Cleared things up.
 
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