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.

 

[PJC01]

Dear Marius,

Yes, you can find velocity by integrating acceleration with respect to position. This is known as the fundamental theorem of calculus, which states that the derivative of an integral is equal to the original function. In this case, the integral of acceleration with respect to position would give you the velocity.

However, it is important to note that this only works if the acceleration is constant or can be expressed as a function of position. If the acceleration is changing, then you would need to use more advanced techniques such as differential equations to find the velocity.

Regarding your question about dimensional analysis, it is a useful tool for checking the validity of equations and ensuring that all units are consistent. In this case, if you are integrating acceleration with respect to position, the units would be (m/s^2) * m = m/s, which is indeed the unit for velocity.

As for the question about work, you can use the equation tau = Ia to find the work done by a pendulum. However, the integration would be with respect to time, not angle, as work is defined as the integral of force with respect to displacement. So, the correct equation would be W = ∫tau * dθ, where dθ represents the change in angle.

I hope this helps clarify your doubts. Keep exploring and asking questions!

Best regards,