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Please consider the below image which is from Rana and Joag, Classical Mechanics. They build on a proof, which I reiterate below and through it they show that angular velocity of any point B in the rigid body is the same about a point $B_0$ and then state that the angular velocity of B is same about any other point too.

enter image description here

From the image above and considering the rigid body constraints we get,

$\pmb{{\rho_1}\times (\dot{\rho_1}\times \dot{\rho_2} )} = \pmb{(\rho_1 . \dot{\rho_2})\dot{\rho_1}} - \pmb{(\rho_1 . \dot{\rho_1})\dot{\rho_2}} = -\pmb{(\rho_2 . \dot{\rho_1})\dot{\rho_1}}$

which implies that

$\pmb{ \dot{\rho_1} = (\frac{\dot{\rho_1}\times\dot{\rho_2}}{\dot{\rho_1} . \rho_2})}\times \rho_1 $

and similarly we can get

$\pmb{ \dot{\rho_2} = (\frac{\dot{\rho_1}\times\dot{\rho_2}}{\dot{\rho_1} . \rho_2})}\times \rho_2 $

The book says " that the above equations indicate that the vector

$\pmb{ \omega = (\frac{\dot{\rho_1}\times\dot{\rho_2}}{\dot{\rho_1} . \rho_2})}$

behaves as an angular velocity vector for $\pmb{\rho_1}$ and $\pmb{\rho_2}$ about $B_0$. The book proves that the any given point in the body say B has the same angular velocity as given by the expression but doesn't prove that why any point in the body should have the same angular velocity about any other point say $B_0'$.

This is what I want to prove mathematically rigorously. To do that I begin writing

$\pmb{ \rho_1 = \rho_1' + a}$ and similarly $\pmb{ \rho_2 = \rho_2' + a}$ and I try to find their time derivative and substitute the result into the expression for $\pmb{\omega}$ that has been calculated above. Now I will get my correct answer only if

$\pmb{\dot{a}} = 0$

Now I don't think that $\pmb{\dot{a}} = 0$ should be true because the direction of a will change even if its magnitude is constant and hence I fail to prove the result.

Could anyone help me how to proceed forward using the above expression for $\pmb{\omega}$ to prove that it is indeed the same about any point in the body. Should $\pmb{\dot{a}} = 0$ be true because if it is false then the result is false itself. But then why should be $\pmb{\dot{a}} = 0$ be true.

Or is there another way out.

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  • $\begingroup$ I could write the Landau-Lifshitz argument, but it's so short, and in general you would benefit much more if you directly read/work out the book. I recommend the latter. If you do that, you'd also see that for linear rigid bodies the proof doesn't go through... $\endgroup$
    – Vivek
    Jun 13 '20 at 21:15
  • $\begingroup$ @Vivek ok. Any fault you could see in this one. Or how should I proceed in this proof along the lines of my argument $\endgroup$
    – Shashaank
    Jun 14 '20 at 4:33
  • $\begingroup$ @Vivek that is, will $\dot{a}$ be 0 or not... $\endgroup$
    – Shashaank
    Jun 14 '20 at 4:42
  • $\begingroup$ "The book says " that the above equations indicate that the vector" I think this is wrong: $\omega=\frac{v\times R}{R\cdot R}$ check the units for omega $\frac{m/s\,m}{m^2}$ $\endgroup$
    – Eli
    Jun 26 '20 at 16:53
  • $\begingroup$ @Eli yes you can have a look at that equation for omega, it gives the units of omega as $\frac{m/s}{m}$, the same as which your equation for omega gives. It gives the same units, isn't it... Have a look at the equation and let me know please. $\endgroup$
    – Shashaank
    Jun 26 '20 at 17:23
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From Euler's theorem we know that a rigid motion in odd number of dimensions with one point fixed is a pure rotation about an axis (cf. Goldstein). This allows one to write,

$$\mathbf{v}(\mathbf{r}) = \mathbf{v}_A + \omega_A \times(\mathbf{r}-\mathbf{r}_A), \quad \quad (1) $$

$$\mathbf{v}(\mathbf{r}) = \mathbf{v}_B + \omega_B \times(\mathbf{r}-\mathbf{r}_B). \quad \quad (2)$$

Now the velocity fields described by equation (1) and (2) must be the same. This means, in particular, that

$$ \omega_A \times \mathbf{r} = \omega_B \times \mathbf{r}.$$

The only way this is possible is if the vectors of angular velocity are equal, i.e. $\omega_A = \omega_B$, unless the rigid body is linear, in which case there is no way to uniquely define the angular velocity since rotations along the line which contains the rigid body do not contribute to physical motion.

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  • $\begingroup$ So will $\dot{a}$ be 0 in my proof above or not. $\endgroup$
    – Shashaank
    Jun 14 '20 at 10:05
  • $\begingroup$ No, why would it be? It's the relative velocity of $B'_0$ wrt $B_0$. $\endgroup$
    – Vivek
    Jun 14 '20 at 10:44
  • $\begingroup$ because if it isn't then I will get the wrong result. See if $\omega$ has to be same then $\dot{a}$ should be zero and my proof is exactly along the lines of landau. It's just a translation. We are unable to see why it should be 0...but if the result has to be true then it should be $\endgroup$
    – Shashaank
    Jun 14 '20 at 11:13
  • $\begingroup$ Then that indicates a flaw in your reasoning or calculation somewhere... $\endgroup$
    – Vivek
    Jun 14 '20 at 16:12
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    $\begingroup$ Yes, that's correct. You can translate with any point on the rigid body and only a residual rotation of the rigid body will remain. This rotation is the same no matter what choice of point you make to travel alongside (hence the concept of "angular velocity" without reference to a specific point on the rigid body). $\endgroup$
    – Vivek
    Jul 3 '20 at 20:31

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