Relating radius and angular freq of an rotating object

[serverxeon]

Homework Statement

an ball of mass m is connected by a string with spring constant k, to a rotating shaft.

Find a relation between the radius of the circle, and the angular frequency.

Homework Equations

The Attempt at a Solution

Let:
Natural length of spring = x₀
Extension of spring = x
Radius of circle = r = x + x₀

Centripetal force = Force due to spring
mrω² = kx = k(r-x₀)
Make r the subject,
mrω² = kr-kx₀
kx₀=kr-mrω²
kx₀=r(k-mω²)
r=\frac{kx_{0}}{k-m\omega^{2}}
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The expression indicates that there is an asymptote when \omega=\sqrt{\frac{k}{m}}

Does this means the rotation can never hit this angular velocity?

 

[ehild]

r=\frac{kx_{0}}{k-m\omega^{2}}
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The expression indicates that there is an asymptote when \omega=\sqrt{\frac{k}{m}}

Does this means the rotation can never hit this angular velocity?

Yes. You can also express ω in terms of r:

\omega=\sqrt{\frac{k}{m}(1-\frac{x_0}{r})}
ω would be equal to the "natural frequency" at infinite radius.

ehild

 

[serverxeon]

but expressing ω in terms of radius, i get an asymptote.
what is the physical meaning of this mathematical constraint?

 

[ehild]

If the object performs SHM, the angular frequency is √(k/m) neither higher nor less. Now the object moves along a circle. r is its distance from the origin. It can not be negative. So ω can not exceed √(k/m)


ehild

 

[serverxeon]

sorry, I don't understand your explanation.
You provided a mathematical description, but I am looking for a physical explanation.

I know that r cannot be negative and hence ω can not exceed √(k/m). I can see that mathematically.

So what happens if I attach a motor to the rotating shaft and drive the shaft at an angular freq higher than √(k/m)?

 

[mishima]

So what happens if I attach a motor to the rotating shaft and drive the shaft at an angular freq higher than √(k/m)?

The spring would break or deform. Hooke's law isn't an exact law, its just an experimental approximation for stresses below a certain threshold. You would cross that threshold by pumping omega, since the spring can only stretch out so far. Springs stop being springs (perhaps permanently) when you stretch them beyond their limit.

If you had an infinitely long wire and wound it into an infinitely long spring and could somehow drive it up to that omega with the mass stretching out that far I don't know what would happen since that it isn't really physical.

 

[ehild]

sorry, I don't understand your explanation.
You provided a mathematical description, but I am looking for a physical explanation.

I know that r cannot be negative and hence ω can not exceed √(k/m). I can see that mathematically.

So what happens if I attach a motor to the rotating shaft and drive the shaft at an angular freq higher than √(k/m)?

The relation between r and ω holds for the stationary solution. In real life, everything is in rest initially, before switching the motor on.
The motor exerts some torque τ, and the torque accelerates rotation: dω/dt=τ/I. The moment of inertia I=mr² increases with r as ω increases, and the applied torque becomes less and less effective to increase ω.There is a limit for ω that can not be exceeded by a given torque.

ehild