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I have studied that a finite angular displacement $\triangle \theta$ is a scalar. But, $\delta \vec \theta$ is a vector. Now, when it is a uniformly accelerated motion we are dealing with, we use equation:

$$\theta~=~\vec{\omega}_i t + \frac{1}{2} \vec{\alpha} t^2, $$

where $\vec{\theta}_0$ is initial angle and $\vec{\omega}_0$ is intial Now my question is, What is $\vec{\theta}$ here? Vector? If yes, then how can we get a finite value for it? (Since as it is a vector only when it is infinitesimally small.) If no, then how is this equation valid?

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  • $\begingroup$ The equation is not valid. $\endgroup$
    – John Alexiou
    Commented Aug 20, 2013 at 23:58
  • $\begingroup$ It is valid. The only question was about $\theta$ $\endgroup$
    – Saurabh Raje
    Commented Aug 21, 2013 at 2:24
  • $\begingroup$ It is not valid because with rigid body dynamics you deal with a sequence of rotations via. something called generalized coordinates, and not with the componets of vectors directly. $\endgroup$
    – John Alexiou
    Commented Aug 21, 2013 at 13:26

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$\Delta \theta$ doesn't necessarily need to be a scalar. Check this Wikipedia article about angular displacement in 3D. It can be denoted as a vector, having magnitude equal to the radians covered, and direction according to the Right Hand Thumb rule.

Saying that, there is no difference between $\Delta\theta$ and $\delta\vec\theta$.

Maybe you regard $\Delta\theta$ as the magnitude of $\delta\vec\theta$. In that case, the $\theta$ in your equation is a shorthand for the angular displacement vector $\delta\vec\theta$.

Now to answer your second question, $\delta\vec\theta$ can be finite, because its direction is given by the right hand thumb rule and not the direction of the displacement vector of the particle, during an infinitesimal change of $d\theta$.

So if the particle rotates by, say, $\pi$ radians clockwise about an axis denoted by unit vector $\hat a$, the angular displacement is given by $$\delta\vec\theta = -\pi\hat a$$

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  • $\begingroup$ the wiki article specifically says that it is a quantity having direction in 3D. What about 2d? @udiboy $\endgroup$
    – Saurabh Raje
    Commented Jul 21, 2013 at 16:14
  • $\begingroup$ If the circular motion is happening in a 2D plane, the vector is directed out from the plane or into the plane depending upon the sense. Either way it is parallel to the axis, because $\delta\vec\theta$ is defined as $\vec r X \vec v$ $\endgroup$
    – udiboy1209
    Commented Jul 21, 2013 at 16:18
  • $\begingroup$ that is fine, but do you mean to say that even the finite $\triangle \theta$ has a direction? Is it a vector then? If not, do you mean to say that its a tensor? $\endgroup$
    – Saurabh Raje
    Commented Jul 21, 2013 at 16:23
  • $\begingroup$ That's exactly what I say. Finite $\Delta \theta$ has a direction. $\endgroup$
    – udiboy1209
    Commented Jul 21, 2013 at 16:32

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