Connection between spring extension and angular acceleration?

Question

A spring has been attached to a pole. At the other end of it is a mass. If the pole were to start rotating, how should I go about trying to find a mathematical relation between the extension of the spring and the angular acceleration of the system?

I know that angular acceleration increases centrepetal acceleration, causing the mass to exert a force on the spring, resulting in an extension in the spring, proportional to the spring constant. Um...I don't exactly know how to make the mathematical connections. I suppose it might involve a differential equation. Can you please help me?

I figured that much, Mr Edward, but angular acceleration causes increase in centripetal force, which causes an extension in the spring...

...which reduces the centripetal force.

It's all in a big loop. I don't quite know how to get around it.

Answer 2 is ok when there is no angular acceleration. What happens when there is?

Answer 1

Not sure why you think you need differential equations. Basically this problem is about finding out what is the length of the spring when the centripedal force balances the force exerted by the spring constant:

(1) Centripedal force: F = ma, where "m" is the mass at the end of the spring and "a" is the centripedal acceleration

(2) Spring constant: F = kx, where "k" is the spring constant, and "x" is the length of the spring

Here we want to set the two forces to equal each other because that's when the centripedal force exactly counterbalances the force due to the spring constant, i.e. the spring will extend no longer and "x" has reached a maximum value.

ma = kx, but we know that centripedal acceleration is v^2/r, and x is also r, the radius of the circle or the length of the spring

mv^2/r = kr => r = sqrt(m/k)v, where v is the instantaneous velocity of mass, m, at some distance r from the center of the pole.

Answer 2

Here is more info

Wikipedia in link number one (http://en.wikipedia.org/wiki/Centripetal_force#References_.26_External_Links) has the explanation both graphical and analytical.
Accentually we have two components perpendicular to each other. One is tangential (perpendicular to the radius also called ‘alpha’ ) and the other is parallel to the radius and extended outward ( (v^2)/r ). For better understanding visualize that it is the resultant that gets decomposed into tangential and centripetal components.

Also check this link out http://hyperphysics.phy-astr.gsu.edu/hbase/rotq.html#rq
And
http://hyperphysics.phy-astr.gsu.edu/hbase/cf.html#cfc
If you continue to have problems try to reach me again.

I had a a drawing but I could not email it back to you .

Original posting
Okay here are some equations for you to consider

F=kx=fc+ft
fc- centripetal force component
ft – tangential force component
ft=Iat=I(dw/dt) I- moment of inertia of a point mass (spring is assumed to be mass less)
I=mr^2
w=v/r (angular velocity)
r=R+x where
R – the initial radius (constant)
x- spring extension (function of time)
r radis of rotation (function of time)

We have a differential equation

-kx= fc + ft
-kx = mw^2 r + mr^2 (dw/dt)
-k(r-R) = mw^2 r + mr^2 (dw/dt)
-kR = (k +mw^2)r+ mr^2 (dw/dt)

Ah and v/r=w


More info at
1.http://en.wikipedia.org/wiki/Centripetal_force#Derivation_using_calculus
2.http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html
3.http://hyperphysics.phy-astr.gsu.edu/hbase/rotq.html#drot


I hope it helps
Let me know.