Physics: You probably have noticed that, as you increase ω, there will be a value, ωcrit, for which r(ω) goes to infinity. find ωcrit. answer in terms of k and m.

You probably have noticed that, as you increase ω,

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lele1522

15.03.2022

For the mass m, attached to a spring, to move in a circle centripetal force and restoring force of the spring must be equal. Centripetal force is given with this formula:
$F_c=\frac{mv^2}{r}=\frac{mw^2r^2}{r}=mw^2r$
Restoring force only occurs if the spring is stretched. If L is the length of unstretched spring we have the following formula for restoring force:
F=k(r-L)

r is the length of a circle that mass m is traveling along.
As said above, these two forces have to be equal:
F=k(r-L)=mw^2r

We solve for r:
$F=k(r-L)=mw^2r\\
kr-kL=mw^2r\\
mw^2r-kr=-kL\\
r(mw^2-k)=-kL\\
r=\frac{-kL}{mw^2-k}\\
r(w)=\frac{kL}{k-mw^2}

$
r(w) will go to infinity when denominator is equal to zero:
$k-mw^2=0\\
k=mw^2\\
w^2=\frac{k}{m}\\
w_{crit}=\sqrt{\frac{k}{m}}$
Please keep in mind that Hooke's law has it's limitations, and before we reach our critial value of angular velocity spring will be strecthed to a point where Hooke's law does not aply anymore.

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You probably have noticed that, as you increase ω, there will be a value, ωcrit, for which r(ω) goes to infinity. find ωcrit. answer in terms of k and m.

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