0
$\begingroup$

I was attempting a rotational mechanics question where it was mentioned that:

Due to friction about the axis, the wheel experiences an angular retardation proportional to its angular velocity. If its angular velocity falls to half while it makes n rotations, how many more rotations will it make before coming to rest?

I was writing differential equations for this (they turned out to be very similar to first-order chemical kinetics):

$$\omega = \omega_0e^{-kt}, \alpha = -k\omega$$ (where $k$ is a positive constant)

This doesn't make sense with the idea of angular velocity ever being zero, for if it does become zero then $e^{-kt}$ would have to become zero, which is not possible.

Am I making any conceptual/physical integration mistake here?

$\endgroup$

2 Answers 2

2
$\begingroup$

As soon as you have an expression of the form $\omega(t)=\omega_0e^{-k\,t}$ then Mathematically, when $t\to \infty$ the $\omega\to 0$, and so $\omega$ would never reach zero in a finite time.

However, in the real world it is often the case that the final value of $\omega$ has to be (significantly) smaller than the initial value $\omega_0$.
Then the decision has to be made as to what is meant by significantly smaller.

For example you might consider the final value being one percent of the initial value as sufficiently small, ie $0.01 = e^{-kt_{1\%}} \Rightarrow t_{1\%} \approx 4.6/k. $

In the case of radioactive decay, the time to reach $1\%$ of the original activity would be $0.01 = \left(\frac 12 \right)^n \Rightarrow n\approx 6.6$ where $n$ is the number of half lives.

$\endgroup$
1
  • $\begingroup$ Thanks, I didn't consider the negative sign in the exponent of e. So I can't solve this for t, but'll have to use velocity and displacement instead. $\endgroup$
    – zxayn
    Commented Oct 18, 2023 at 15:34
2
$\begingroup$

Note that if it has the angular velocity $\omega(t)=\omega_0 e^{-kt}$, the total amount of rotation will be $\theta=\int_0^\infty \omega(t) dt= \omega_0 / k$.

$\endgroup$
1
  • $\begingroup$ This is a nice point. $\endgroup$
    – Farcher
    Commented Oct 18, 2023 at 17:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.