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

1. May 1, 2014

### 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

2. May 1, 2014

### Staff: Mentor

Yes, to within a constant of integration.

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

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

Last edited: May 1, 2014
3. May 1, 2014

### 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?

Kind regards,
Marius

Last edited: May 1, 2014
4. May 1, 2014

### 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.

Last edited by a moderator: May 1, 2014