If angular velocity $\omega=\sqrt{\frac{3g\sin\theta}{2a}}$ can I find angular acceleration $\alpha$ by differentiating $\omega$? It was my understanding that angular acceleration is the derivative of angular velocity.
The reason I ask is 
Thanks.
 A: Yeah, one can write:
$$\alpha=\dfrac{d\omega}{dt}$$
If $\theta=\theta(t)$ and $a=a(t)$, then:
$$\alpha=\dfrac{d}{dt}\left(\sqrt{\dfrac{3g\sin\theta(t)}{2a(t)}}\right)=\sqrt{\dfrac{3g}{2}}\dfrac{d}{dt}\left(\sqrt{\dfrac{\sin\theta(t)}{a(t)}}\right)$$
Not, you may just use the chain rule and the quotient rule to evaluate this differential.
A: Since $\omega$ is a vector, it also has a direction. Let it be such that:
$$\vec{\omega}=\sqrt{\frac{3gsin\theta}{2a}} \vec{e}$$($\vec{e}$ is the unit vector in direction of $\omega$)
$$\vec{\alpha}=\frac{d\vec{\omega}}{dt}$$
So take care of both while differentiating: the magnitude and direction $\vec{e}$. Hint: you can use product rule to differentiate magnitude and direction separately
A: Yes , you can find the angular acceleration by simply differentiating your angular velocity equation , given that your equation is a function of 't' i.e time(bcz differentiating velocity wrt time will always give you the acceleration.) If there are few coefficients in the equation just plug there value in n ur done.
A: Note that you can also solve the problem using $\tau = I\alpha$: $\tau$ is torque, $I$ is moment of inertia, $\alpha$ is rotational acceleration. If the center of rotation is the contact point of the rod and the table, the only force contributing to the torque is gravity.
