# Why is it that angular acceleration is constant in different instantaneous reference frames?

Take the following example: A rod (of length L and mass m) is held horizontally at both ends by supports. One is instantaneously removed.

The specific problem is to prove that the force on the other support drops from mg/2 to mg/4, which I proved by first considering the centre of mass as the instantaneous reference frame, and thus considering a rotation around the support.

Resolving angular forces: (F = Force at pivot, I = Moment of Inertia = m(L^2)/12, ω = angular velocity)

 FL/2 = I * dω/dt
FL/2 = m(L^2)/12 * dω/dt
F = mL/6 * dw/dt   (1)


Now taking the instantaneous reference frame around the pivot: (I = M(L^2)/3, ω' = angular velocity, force at CoM = mg)

mg * L/2 = I * dω'/dt
mgL/2 = mL2/3 * dw'/dt
dw'/dt = 3g/2L    (2)


The desired solution can be found by substituting (2) into (1), i.e. by equating dw'/dt and dw/dt. Why is it that this can be done?

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What is angular speed? Clearly it is $\frac {v_\perp} {r}$ where symbols have their usual meanings.

Rod rotates about its, say, rightmost point, say $O$. We will consider left side as positive $x$-axis.

Now consider a point $A$ at distance $r_1$ from it. Let the rod have instantaneous angular speed $\omega$. All points on the rod will have this $\omega$ wrt $O$.

Consider a point B at a distance $r_2$ from it, clearly with same $\omega$ wrt $O$. This can be seen by the fact that rate of change of angular displacement is same for all points as $\omega =\frac {d\theta}{dt}$

Assume $r_2 > r_1$

Now consider the point A as frame of reference and let us calculate $\omega$ $'$ which is angular speed of $B$ wrt $A$. Clearly, $v_A=\omega r_1$ wrt ground and that of $B$ is $\omega r_2$. Now calculate $v_\perp$ of $B$ wrt $A$.

Clearly, it is $v_b-v_a =\omega(r_2-r_1)$ And distance between $A$ and $B$ is $r_2-r_1$.

So, what do you get $\omega$ $'$ ?

$\omega$ $' =\frac {\omega(r_2-r_1)} {r_2-r_!} =\omega$

As instantaneous angular speed is same, its rate of change $\alpha$ will also be same. Get it?

This is a bit of general answer but applies to your question also. Also you can solve your question by using newton's second law and using $a=\alpha \frac L 2$

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