6
$\begingroup$

What is the difference between angular frequency and angular velocity? I think one is used for SHM and the other for circular motion? Also can both be used for centreptal accelartion? I think angular freq$=2\pi f$ and angular velocity$=d\theta /dt$? Please confirm or expalin. Are they the same thing for circular motion??

Improve this question
$\endgroup$

2 Answers 2

Reset to default
6
+50
$\begingroup$

Well, the key difference here is that one is a vector quantity while the other is a scalar.

If your angle is measured in radians then angular frequency $\omega$ is given by

$$ \omega = 2 \pi f \space \mbox{(rad)} s^{-1} $$

while angular velocity is

$$ \vec{\Omega} = \frac{d \vec{v}}{dt} \mbox{m} \space s^{-1} $$

What you have above is the magnitude of the angular velocity (which I am assuming is expressed in radians).

$$ \vert \omega \vert = \frac{d \theta}{dt} \mbox{rad} \space s^{-1} $$

Often people leave out the radian, since it's just a number. The radian is engineering-dimensionless.

Indeed usually you would use $\omega$ to talk about oscillators, and $\vec{\Omega}$ for circular motion. You need to be careful if your equations are vector equation, in which the direction is important, or scalar equations, where you're only looking for a magnitude.

I assume you are aware of the difference between distance and displacement, or speed and velocity, yes?

Improve this answer
$\endgroup$
8
  • $\begingroup$ so if the motion is in one direction and in a circle does dθ/dt=2pi*f? $\endgroup$
    – user43487
    Commented Apr 12, 2014 at 19:22
  • $\begingroup$ @Joseph I guess by 'one direction' here you mean that there is one degree of freedom, namely the body moves along a circle. In that case, then yes, $ \frac{d \theta}{dt} = 2 \pi f $ But as I say, this is a scalar equation, so it would not work with, forces say. You may need to give us some more details of the full question that you are working on, because I do not want to accidently lead you towards a future mistake. $\endgroup$
    – Flint72
    Commented Apr 12, 2014 at 19:29
  • $\begingroup$ I am an a-level student teaching this to myself, all need to know is that both dΘ/dt and 2.pi.f can be used for centripetal acceleration when there are no vectors involved and they can be interchanged in cirular motion when it is just purely circular motion? $\endgroup$
    – user43487
    Commented Apr 12, 2014 at 19:31
  • 1
    $\begingroup$ @Joseph Yes, indeed they both can be used. Err, I am already aware that I seem to be making it overly confusing, I just don't want to put you wrong, but I'm going to add another piece anyway; So one more thing to be aware of, this type of circular motion, such as a beed on a hoop, or a ball at the end of a string, is one dimensional. Think about walking along a tightrope. This would be one-dimensional, yes? Now, if the tightrope goes in a circle and joins up withself, itsn't it stil going to be one-dimensional, yes? It's a relatively subtle, but important distinction. $\endgroup$
    – Flint72
    Commented Apr 12, 2014 at 19:43
  • 1
    $\begingroup$ Your definition of angular velocity is incorrect. For one thing, dv/dt has dimensions of acceleration, so the units you've indicated don't go with your expression. But the dimensions of angular velocity should be the same as those of angular speed anyway; it should have SI units of rad/s, not m/s. $\endgroup$
    – pwf
    Commented Jun 10, 2015 at 18:13
2
$\begingroup$

Consider a torsional pendulum with a torsional constant $\kappa$ and moment of inertia, $\mathcal{I}$ about the rotational axis. The function for the angular position, $\Theta$, can be written $$\Theta = A\cos(\omega t + \phi_o)$$ where

  • $A$ is the angular amplitude,
  • $\phi_o$ is the initial phase of the oscillation so that $\Theta(0)$ has the correct value, and
  • $\omega = \sqrt{\frac{\kappa}{\mathcal{I}}}$ is the angular frequency of oscillation, and is generally a constant of motion unless something actively modifies the system (changes the moment of inertia or the torsional constant). The period of oscillation would be $2\pi/\omega$.

If we take the time derivative of this angular function we get $$ \frac{\mathrm{d}\Theta}{\mathrm{d}t}= - \omega A \sin(\omega t + \phi_o).$$

This tells us the instantaneous angular speed of the pendulum and it is constantly changing. This is different from the angular frequency of oscillation, and that difference can be confusing unless one takes great care to keep the concepts separate. This is also the quantity that one would use for centripetal acceleration calculations: $$a_c=\left( \frac{\mathrm{d}\Theta}{\mathrm{d}t}\right)^2 r$$

This difference usually happens when a system has an oscillation of angular position.

For uniform circular motion, we have $$\Theta(t) = \omega t + \Theta_o$$ and $$\frac{\mathrm{d}\Theta}{\mathrm{d}t} = \omega.$$ In this case, the two concepts are equal to each other.

Improve this answer
$\endgroup$