2
$\begingroup$

Basically what I am trying to ask is if a body has an angular velocity $\omega$ or angular acceleration $\alpha$ about an axis then will it have the same angular velocity and acceleration along any other axis? I am really confused . As we know torque $\tau$ of all the forces acting on the body about the earlier axis is $$\tau=I\alpha$$ Where I is the moment of inertia of the body about that axis. ( $\alpha$ depends on $\tau$ and $I$ . If we change the axis then the radius vector of each particle from the force will change about the new axis and hence torque will change (because $\tau=r\times F$) . Also the moment of inertia of the body about that axis will change. Hence $\alpha$ may change or may not ( according to me) because $\tau$ and $I$ may change.

$\endgroup$
  • $\begingroup$ you need to specify if the body is a rigid body or a particle? $\endgroup$ – Saurabh Raje Jul 28 '15 at 7:32
  • $\begingroup$ You can take any. What I am asking is whether the particle or rigid body undergoing pure rotation will have same angular vel or accln about diffetent axes. $\endgroup$ – Tripti Khulbe Jul 28 '15 at 7:41
0
$\begingroup$

If the body is a point mass, and even if frames are stationary, still the angular velocity and accelerations may or may not come out to be the same. The question is a little too generic.

Consider for example, the case of a uniform circular motion, if the angular velocity is measured about an axis which does NOT pass through the centre, it will be a function of time, this is because the torque of the centripetal force will NOT be zero about such an axis.

On the other hand, consider a point mass moving in a straight line. The angular velocity about axes (passing through all points on a parallel line and perpendicular to the plane of motion) will be the same.

It mainly has to do with the expression for torque which comes from what Force the point mass is experiencing. If the torque depends on which axis is chosen, then $\omega$ and $\alpha$ will change.

If it is not a point mass, then firstly you can not measure $\omega$ and $\alpha$ about points which are not on the object/rigid body (unless its acting as the Instantaneous Centre). But, about all points on the rigid body, $\omega$ and $\alpha$ will be the same.

$\endgroup$
  • $\begingroup$ And if the new axis is itself rotating with some ang vel or accln ( not having tranlational vel and accln) then what ( consider a particle only having ang vel and accln a and w about some axis) $\endgroup$ – Tripti Khulbe Jul 28 '15 at 8:40
  • $\begingroup$ If the new axis is rotating with respect to the old one, then definitely angular velocity and acceleration will change (unless of course the axis does not coincide with the object). $\endgroup$ – Saurabh Raje Jul 28 '15 at 8:47
  • $\begingroup$ If we want to FIND ang vel or accln about the new axis then can we simply do w+w' or a+a' ( where w' and a' are ang vel aaccln of the new axis , given that one is having anticlockwise and other clockwise rotation) $\endgroup$ – Tripti Khulbe Jul 28 '15 at 9:03
  • $\begingroup$ @TriptiKhulbe, no, that will not work. To find relative ang vel or accn, find the relative linear vel or accn and then divide by the modulus of radius vector from the axis to the object. $\endgroup$ – Saurabh Raje Jul 28 '15 at 9:12
  • $\begingroup$ But the axis is not moving linearly ( consider an infinitesimally thin rod as the axis) $\endgroup$ – Tripti Khulbe Jul 28 '15 at 9:35
3
$\begingroup$

This issue is a bit confusing because there are two types of angular momentum. There's spin, where a rigid body rotates about an axis through its center of mass, and there's orbital, where the center of mass of a rigid body rotates about an axis. For example, the Earth spins about its axis and rotates around the Sun. The total angular momentum can always be decomposed into a sum of these two terms.

You probably have never heard of this, because the definition of the torque is just $\mathbf{r} \times \mathbf{F}$, which doesn't have any reference to "spin or orbital". But when you write the equation $\tau = I \alpha$, you're implicitly choosing to talk about one or the other, or else $\tau$, $I$, and $\alpha$ are meaningless. (For example, the Earth takes 1 day to spin but 1 year to orbit. So you can't just say "the" angular velocity of the Earth.)

If you're talking about spin, $I$ means the moment of inertia about the center of mass, $\alpha$ means the angular acceleration about an axis through the center of mass, and $\tau$ means the torque about the center of mass. There's no meaningful way to change axes or origin because it's always the center of mass.

Now let's talk about the orbital part. The instantaneous angular velocity can always be defined, even if the object isn't moving in a circle about some point, using the equation $$\mathbf{v} = \boldsymbol{\omega} \times \mathbf{r}$$ However, you can see this quantity is frame dependent: if we just move the origin, $\mathbf{r}$ will change but $\mathbf{v}$ won't, so $\boldsymbol{\omega}$ will change. As an example, if you see an airplane from the ground, its angular velocity appears very low, but if you're hovering next to it, it zips around really fast. So in this case, everything does change.


Okay, maybe neither of these examples were really what you wanted: the first was trivial, and the second didn't say much. We can get more insight by not doing the spin-orbital decomposition at all, which requires tossing out $\boldsymbol{\omega}$ and $\alpha$. The only rotational equation we have left is $$\tau = \frac{dL}{dt}$$ which expanded out is $$ \sum \mathbf{r} \times \mathbf{F} = \frac{d}{dt}\left( \sum \mathbf{r} \times \mathbf{p}\right)$$ Let's transform to another frame. To make it easy, let's just shift the origin by $\mathbf{r}_0$. Now if physics works, the resulting equation should be equivalent.

Let's confirm it. In the other frame, we have $$ \sum (\mathbf{r} + \mathbf{r}_0) \times \mathbf{F} = \frac{d}{dt}\left(\sum (\mathbf{r} + \mathbf{r}_0) \times \mathbf{p}\right)$$ If we subtract out our previous equation, we're left with $$ \sum \mathbf{r}_0 \times \mathbf{F} = \frac{d}{dt} \left( \sum \mathbf{r}_0 \times \mathbf{p} \right)$$ Since $\mathbf{r}_0$ is constant, the derivative only acts on $\mathbf{p}$, giving $\sum \mathbf{r}_0 \times \mathbf{F}$ on the right hand side. So the equation is true. Physics works!

$\endgroup$
  • $\begingroup$ were the last two words inspired by Prof Lewin? :D $\endgroup$ – Saurabh Raje Jul 28 '15 at 7:58
  • 1
    $\begingroup$ yes! it's a great way to end a derivation. $\endgroup$ – knzhou Jul 28 '15 at 7:59
  • $\begingroup$ ... the physics equivalent of QED in mathematics, perhaps. $\endgroup$ – sammy gerbil Feb 21 '17 at 19:50
0
$\begingroup$

First of all, how could you be sure that the same force $F$ will produce rotation about different axes simultaneously?

When you apply some force on the body, calculating from your body fixed frame, it will have certain torque about the origin. If you change the system, then you will get different values of the torque, but here torque has all the 3 components, so you cannot apply $\tau=I\alpha$ formulae.

What you can do is to apply Euler's equation of dynamics and find $\omega_x,\,\omega_y,\,\omega_z$, at some instant $t = t_1$ this values will give you rotation about some instantaneous axes

Now changing the coordinate system will give you different values of the $\omega$ vector's components. But to your surprise, at that very instant $t = t_1$, $\omega_x',\,\omega_y',\, \omega_z'$ will have such values that will indicate the same rotation axis (instantaneous) as we got before ,

For your convenience, you can go through Euler's Equation of dynamics, David Morin's Classical Mechanics textbook.

$\endgroup$
  • $\begingroup$ I am talking only about pure rotation. And the new frame will also be inertial. $\endgroup$ – Tripti Khulbe Jul 28 '15 at 7:33
  • $\begingroup$ i never talked about translation + rotation , u have to specify rotation about fixed principal axis , then only T=Ia relation is valid , but then for a fixed force torque will have some fixed direction about any of the principal axes , for the other axes it will have 0 value , and the frames are body fixed frame always ,if the body is under acceleration it is always some non-inertial frame $\endgroup$ – Priyadarshi Paul Jul 28 '15 at 7:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.