Know acceleration as a function of position, can I find velocity?

[Jonsson]

Hello I know acceleration of a pendulum as a function of angle. Can I find velocity by integration with respect to theta? My intuition says yes, but dimensional analysis tells me otherwise. What is right and wrong?

And what about work? Can I integrate tau = Ia with respect to theta and find work?

Thanks!

Kind regards,
Marius

 

[Nugatory]

Hello I know acceleration of a pendulum as a function of angle. Can I find velocity by integration with respect to theta?

Yes, to within a constant of integration.

My intuition says yes, but dimensional analysis tells me otherwise. What is right and wrong?

Show us your dimensional analysis and we'll be more able to help.

And what about work? Can I integrate tau = Ia with respect to theta and find work?

yes (assuming that $\tau$ is a torque and $a$ is the angular acceleration).

 

[Jonsson]

I have acceleration, $\vec{\alpha}(\theta)$ for the pendulum given by:

$$
\vec{\alpha}(\theta) = -\hat{k}\left( \frac{L\,m\,g}{I} \right) \sin\theta
$$If I could use integration to find $\omega$:

$$
\vec{\omega}(\theta) = \hat{k}\left( \frac{L\,m\,g}{I} \right) \cos\theta + \vec{C}
$$

But If could integrate again:

$$
\vec{\theta}(\theta) = -\vec{\alpha} + \vec{C}\theta + \vec{D}
$$

Question 1: Surely that cannot make sense? $\vec{\theta}$ as a function of itself?

Question 2: And looking at units, $\alpha$ is angular acceleration, so it must have units $\rm s^{-2}$, if I integrate twice, i basically multiply by $\theta$ twice, which is dimensionless so I get $\vec{\theta}(\theta) : \rm s^{-2}$. #\vec{\theta}## should have no dimensions. So think I have a contradiction. What is correct?

What have I misunderstood?

Thank you for your time.

Kind regards,
Marius

 

[HallsofIvy]

Your first integration is wrong. \alpha is d\omega/dt, not d\omega/d\theta so you cannot integrate the right side with respect to \theta as it stands. You can use the chain rule to write d\omega/dt= (d\omega/d\theta)(d\theta/dt)= \omega d\omega/d\theta. So you can write your equation as
\omega d\omega/d\theta= -k\left(\frac{Lmg}{I}\right)sin(\theta)
which, when you separate variables, becomes
\omega d\omega= -k\left(\frac{Lmg}{I}\right)sin(\theta)d\theta

and integrate both sides.