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If we have a rotating object and it's angular velocity is constant, can the angular acceleration ever be different than 0?

EDIT : I know it should be $0$, but my book has this formula $$\frac{d\vec{\omega}}{dt}=\frac{d\omega\vec{a}}{dt}= \frac{d\omega}{dt}\vec{a} + \omega\frac{d\vec{a}}{dt}$$ where $\vec{a}$ is the unit vector of the rotation axis. Then in an example later with constant angular velocity, it goes on and finds the angular acceleration without saying it should be $0$. You have any idea why?

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closed as unclear what you're asking by SRS, ja72, Jon Custer, JMac, Cosmas Zachos Jan 16 '18 at 16:17

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    $\begingroup$ I'm voting to close this question as off-topic because the answer is clearly 'No' which is trivial to see from the definition of angular acceleration. $\endgroup$ – SRS Jan 13 '18 at 14:47
  • $\begingroup$ Please, see my comment below. I have a reason to ask. $\endgroup$ – Thomas Jan 13 '18 at 14:54
  • $\begingroup$ I don't understand. I guess, you know that $\frac{d\boldsymbol{\omega}}{dt}$ represents the angular acceleration. When $\boldsymbol{\omega}$ is constant, the angular acceleration must vanish. @Thomas $\endgroup$ – SRS Jan 13 '18 at 14:57
  • $\begingroup$ Yes, @SRS I thought of that too... $\endgroup$ – QuIcKmAtHs Jan 13 '18 at 14:58
  • $\begingroup$ If the rotation is confined in the xy-plane (for example), then $\vec{a}=\hat{z}$ where $\hat{z}$ is the unit vector along the z-axis. That doesn't change with time. Moreover, you said angular velocity is constant (not angular speed) which is a vector $\boldsymbol{\omega}$. For it to be constant, both its magnitude and direction must be constant. $\endgroup$ – SRS Jan 13 '18 at 15:02
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I think your question broadly is how to derive the angular acceleration vector from the angular velocity vector $\vec{\omega}$?

First decompose the angular velocity vector to a magnitude and a direction $\vec{\omega} = \omega\, \vec{z}$.

The take the derivative of the above

$$ \vec{\alpha}= \frac{{\rm d}}{{\rm d}t} \vec{\omega} = \dot{\omega}\, \vec{z} + \omega \frac{{\rm d} \vec{z}}{{\rm d}t} $$

At this point, there are two options. a) $\vec{z}$ is fixed in space and $\frac{{\rm d} \vec{z}}{{\rm d}t}=0$, or b) $\vec{z}$ is fixed to the rotating body and $\frac{{\rm d} \vec{z}}{{\rm d}t} = \vec{\omega} \times \vec{z}$ based on the law of derivatives on a rotating frame.

It turns out for a single rotation axis, both scenarios yield the same result. You can check that

$$ \require{cancel} \vec{\alpha} = \dot{\omega}\, \vec{z} +\cancel{ \vec{\omega} \times (\omega \vec{z})} $$

In this case if $\omega = \mbox{(constant)}$ and $\dot{\omega}=0$ then $\vec{\alpha}=0$.

But consider the case a body is rotating about an axis that is riding on a rotating frame. $$ \vec{\omega} = \vec{\Omega} + \omega \vec{z}$$ Where $\vec{\Omega}$ is the speed of the base body (the frame), $\vec{\omega}$ the speed of the child (connected) body and $\vec{z}$ the relative rotation axis between the two.

In this case $\frac{{\rm d} \vec{z}}{{\rm d}t} = \vec{\Omega} \times \vec{z}$ since the axis of rotation is riding on the rotating frame.

$$ \vec{\alpha} = \dot{\omega}\, \vec{z} + \vec{\Omega} \times (\omega \vec{z}) $$

and the last part is not necessarily zero.

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  • $\begingroup$ I fully understand it now. Thanks for the detailed answer! $\endgroup$ – Thomas Jan 14 '18 at 16:44
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Angular acceleration is defined as $$a=\frac{d\omega}{dt}$$where $\omega$ is the angular velocity. For constant angular velocity, this will always be $0$. The linear acceleration is not $0$, however.

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  • $\begingroup$ My book has this formula: $\frac{d\vec{omega}}{dt}=\frac{d\omega\vec{a}}{dt}= \frac{d\omega}{dt}\vec{a} + omega\frac{d\vec{a}}{dt}$, where $\vec{a}$ is the unit vector of the rotation axis. Then in an example later with constant angular velocity, it goes on and finds the angular accelaration without saying it should be $0$. You have any idea why? $\endgroup$ – Thomas Jan 13 '18 at 14:52
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    $\begingroup$ This should be edited into your post, so people can answer it more wholly. $\endgroup$ – QuIcKmAtHs Jan 13 '18 at 14:55
  • $\begingroup$ @Thomas In this example surely $\vec a$ is not changing therefore $\frac{d\vec a}{dt}=0$ and so it still holds that $\frac{d\vec\omega}{dt}=0?$ $\endgroup$ – CooperCape Jan 13 '18 at 14:59
  • $\begingroup$ @CooperCape in the relative formula of the book it says that even thought $\vec{a}$ has constant magnitude, it's change is caused by the change in it's orientation. I think that means that the rotation axis is moving, because in the problem there are 2 rotations happening simultaneously and this creates a translation and a rotation. Would this explain it? $\endgroup$ – Thomas Jan 13 '18 at 15:04
  • $\begingroup$ Well I suppose that in this case $\frac{d\vec\omega}{dt}$ will have a $\frac{d\vec a}{dt}\omega$ term which is non-zero for a non-constant $\vec a$. $\endgroup$ – CooperCape Jan 13 '18 at 15:09
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Angular velocity is a vector (strictly speaking a pseudovector) so it has both a magnitude and a direction. That means a change in angular velocity can be a change in the magnitude, or change in the direction or a change in both.

And this is what your equation is telling you. The first term is the change in the rotation rate and the second term is the change in direction. Either can be zero, but the angular acceleration will only be zero if both are zero.

You ask:

If we have a rotating object and its angular velocity is constant, can the angular acceleration ever be different than 0?

and the answer is clearly no. However suppose the rate of rotation about the axis is constant but the direction of the axis is changing. In that case the angular acceleration would indeed be non-zero.

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  • $\begingroup$ You mean in the last line, the angular acceleration will only be zero if both are zero? $\endgroup$ – Thomas Jan 13 '18 at 15:09
  • $\begingroup$ Yes! The direction of the axis is changing. So that means that in my formula, i will only have the second term, since the first one will be $0$, correct? $\endgroup$ – Thomas Jan 13 '18 at 15:12
  • $\begingroup$ @Thomas yes ${}{}$ $\endgroup$ – John Rennie Jan 13 '18 at 15:15
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Yes

The angular acceleration $\vec \alpha$ can be non-zero even when the angular velocity $\vec \omega$ remains constant. Consider the following example for two equal masses connected by a rod and spinning along an axis that makes an angle $\theta$ with the z-axis.

Simple Rotor

Clearly $d \vec \omega / dt = 0$ but $\vec L$ precesses about $\vec w$. Thus there is a net torque: $$ \frac{d \vec L}{dt} = \Sigma \vec N = I \vec \alpha$$

and $\vec \alpha$ is non-zero.

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  • $\begingroup$ NO! The angular momentum vector is parallel to $\omega$. The angular momentum of each particle is perpendicular to the circle of motion you have drawn. $\endgroup$ – Bill N Jan 14 '18 at 4:22
  • $\begingroup$ @BillN the r in L = r x p is the vector from an origin not an axis. You aren’t correctly calculating the total angular momentum of the system if you calculate L from the center of one circle then from a different circle. $\endgroup$ – cms Jan 14 '18 at 22:55
  • $\begingroup$ My apologies! You're right about $\vec{L}$ about a point. You might want to add something about the moment of inertia tensor being time dependent and the forces related to the torque. ( If $\vec{L}$ is precessing, but $\vec{\omega}$ is constant, then what is the torque changing and where are the forces?) (I can't undo the downvote until you edit the answer again.) $\endgroup$ – Bill N Jan 15 '18 at 14:54

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