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My textbook says:

Angular displacement is the angle through which position vector of a moving particle rotates.

What if the particle goes around a circle two and a half times?

In this case, I thought angular displacement should be 5π. Because if there is an observer at the center of the circular trajectory, he will have to rotate through 5π if he is observing the moving particle (from $t$ = $0$ to $t$ = the time it takes for the particle to complete 2.5 revolutions)

Is it correct that angular displacement in this case would be 5π, and not just π?

The reason I am asking this question is because I stumbled on this question

Angular distance Vs Angular displacement

(Check out the accepted answer. It says that $2n\pi$ angular distance corresponds to zero angular displacement, where $n$ is an integer. Is it correct?)

That brings me to ask another question. Is angular distance just the magnitude of angular displacement?

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2 Answers 2

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Angular distance : total angle in radians covered during the period of observation. it could be greater than $2\pi$

Angular displacement : angle in rad by which the particle (or object) is rotated from its initial position.

Since after covering $2n\pi$ rads ie: after n complete rotations the particle returns to its initial position, the angle by which it rotated from initial position = 0. and hence angular displacement is zero.

For 2.5 revolutions, angular distance actually is $5\pi$ but after first two revolutions, the particle is back on same place. So angular displacement is only the radian corresponding to last half revolution which is $\pi$

hope that clears it

Additionally, if we wish, we may call usual clockwise measured angles as positive and downward (anticlockwise) measured angles negative. Then, if we rotate by say $7\frac{\pi}{4}$ rad (which is 45 or $\frac{\pi}{4}$ degrees behind the initial position), we may call the angularbm displacement as $-\frac{\pi}{4}$

last part of the question has an inaccuracy... hope that you meant angular displacement instead of angular velocity. Angular distance is not the magnitude of angular displacement. Angular displacement $=$ angular distance $-$ maximum possible $2n\pi$ ie : remainder when angular distance is divided by $2\pi$.

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  • $\begingroup$ I'm still not fully convinced that if a particle completes 2.5 revolutions, its angular displacement is $\pi$, and not $5\pi$. Should it not be $5\pi$ because if it is $\pi$, then its average angular velocity would be $\frac{π}{t}$, which does not seem useful at all because it does not tell us about its average angular velocity as it completed its first revolution around the circle, it can not tell us the time taken to complete the first revolution, for example. On the other hand, if its average angular velocity is $5\frac{π}{2}$, it gives us all those information $\endgroup$
    – 4d_
    Commented Apr 28, 2020 at 6:21
  • $\begingroup$ "If we rotate by say, $\frac{7π}{4}$ rad, we may call the angular displacement as $\frac{-π}{4}$" Can we? How? Both are opposite in direction. One is clockwise and the other is counterclockwise. How are both the same? $\endgroup$
    – 4d_
    Commented Apr 28, 2020 at 6:24
  • $\begingroup$ Thank you, You're Right that angular displacement doesn't seem to be a useful quantity. an average angular velocity of $\frac{\pi}{t}$ doesn't seem to be useful at all... But you have to realise that so is actual linear displacements. '' you went to school and came back, your average velocity is zero.... how did you get there then..! '' It is seeming to be more odd because it is a circular motion. And justnas you have said, it is more useful to use angular distance in place of angular displacement for almost all purposes. It is 'cause of the way displacements are defined. $\endgroup$
    – Rishab Navaneet
    Commented Apr 28, 2020 at 12:56
  • $\begingroup$ 2nd part : see, Displacement is just by how much angle you shifted finally...unlike distances its depends only on where you reach at the end. a rotation of $7\frac{\pi}{4}$ anti-clockwise would take you to the same position as a rotation of $-\frac{\pi}{4}$ clockwise. So both are equivalent. It seems odd maybe because it doesn't happen in linear motion. There a displacement forward can never be a displacement backward... but on a circular track you will find that both would take you to same final point $\endgroup$
    – Rishab Navaneet
    Commented Apr 28, 2020 at 13:02
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Angular displacement is the angle through which position vector of a moving particle rotates.

the rotational Matrix to rotate the position vector is:

$$R=\left[ \begin {array}{ccc} \cos \left( \varphi \right) &-\sin \left( \varphi \right) &0\\ \sin \left( \varphi \right) &\cos \left( \varphi \right) &0\\ 0&0&1 \end {array} \right] $$

thus

$$\frac{d\varphi}{dt}=\omega(t)\quad \varphi(t)=\int_{t_i}^{t_f}\ \omega(t)\,dt$$

so to get the angular displacement ($\varphi)$ you need to know the angular velocity $(\omega(t))$ and the time that your disc rotate

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