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If $\omega = \frac{v}{r}$, then why do we need torque and angular acceleration? The velocity v can be found just by Newton's second law of motion $F = ma => a = F/m$ and $v = v_0 + at$ . Then we could find the angular velocity by dividing the linear velocity with a radius.

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  • $\begingroup$ I don't understand the question. The equation $F = ma$ doesn't involve the velocity...at best you can say $v = \omega r \rightarrow a = \dot{\omega}r$ (assuming $r$ is constant). Seems like you need the angular acceleration for sure. $\endgroup$ – Jared May 7 '15 at 5:58
  • $\begingroup$ @Jared I've updated to include how the velocity is obtained. $\endgroup$ – lawls May 7 '15 at 6:01
  • $\begingroup$ I don't think what you are describing involves rotation--in which case of course angular quantities do not make sense and are certainly not necessary (although still correct). You cannot, for instance, describe a sphere rotating with simple linear mechanics (the reason is that there are internal centripetal forces which are what cause rotation/conservation of angular momentum). $\endgroup$ – Jared May 7 '15 at 6:05
  • $\begingroup$ Conservation of angular momentum is a convenience and is unnecessary if we accept conservation of linear momentum. The problem is that if you want to discard angular momentum, then, e.g., to describe a spinning ball you have to know the internal forces which provide a centripetal force. From that knowledge, you can show that conservation of linear momentum gives rise to macroscopic spin. But we can still describe the same motion without such knowledge if we accept conservation of angular momentum (i.e. if we accept that the internal forces will be such that the ball doesn't deform). $\endgroup$ – Jared May 7 '15 at 6:15
  • $\begingroup$ And as a caveat to the above, this is purely in the classical sense. Quantum spin is a very different thing which makes angular momentum much less trivial. $\endgroup$ – Jared May 7 '15 at 6:16
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If you have a single particle then you can indeed easily describe its motion using Newton's laws. However in rotating systems we are often dealing with continuous bodies not point particles. These are characterised by a moment of inertia, rather than just by a mass, and we have an equation analogous to Newton's second law:

$$ T = I\dot{\omega} = I\ddot{\theta}$$

that corresponds to:

$$ F = ma $$

This makes torque and angular acceleration very useful concepts for dealing with real world systems. You could describe your object as a integral of infinitesimal volume elements, but that would turn all the equations of motion into integral equations and make life unnecessarily complicated.

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  • $\begingroup$ The moment of inertia comes from assuming that a body does not deform under rotation. The problem isn't that this turns all equations of motions into integral equations (that's already the case), the problem is that you have to come up with internal forces which stop a body from deforming which, much like the normal force, are whatever they need to be (which makes finding them and characterizing them very difficult). In reality, if you spin a disk fast enough, the rotational speed will be too great for the disk to remain intact and at that point, the moment of inertia no longer has meaning. $\endgroup$ – Jared May 7 '15 at 7:08
  • $\begingroup$ Nice answer +1 😀 $\endgroup$ – mick May 7 '15 at 7:17
  • $\begingroup$ @Jared: yes, that's true, but in many engineering problems the moment of inertia can be treated as constant. My point is that in real life (I'm assuming engineers can reasonably be described as life :-) the equations based on the moment of inertia are extremely useful. $\endgroup$ – John Rennie May 7 '15 at 7:24
  • $\begingroup$ @JohnRennie Actually that's exactly my point, it is extremely useful--because it's a simplification (but obviously a reasonable one). $\endgroup$ – Jared May 7 '15 at 7:27
  • $\begingroup$ @Jared $F = ma$ equally assumes that $m$ doesn't change during the application of force. $\endgroup$ – Dai May 7 '15 at 8:46

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