1
$\begingroup$

The problem was posed as follows. Given a pendulum of length $L$ with a mass $m$ attached to it, which forms an angle $\theta$ from the y-axis to the direction of swinging.

First we had to find the potential energy as a function of the angle, which was trivial enough.

$$U = mgL(1 - cos(\theta))$$

But the next question was to develop the gradient for this potential energy function via the arc length $s = L\cdot\theta$, where $s$ is the arc length.

I calculated the gradient by using the trigonometric definition of $\cos(\theta)$, but I developed the gradient by changing the variable of my function $U(\theta)$ to $U(x,y)$ after which I calculated it further, and found that $F_{tan} = -mg\sin(\theta)$, as desired.

But is there an alternative way to calculate it, by using the gradient, and by utilizing the arc length formula above? The question explicitly asks to develop the gradient via the arc length rather than $x$ and $y$ components.

$\endgroup$
  • $\begingroup$ Please try to use math formatting for better readability. $\endgroup$ – ja72 Oct 4 '13 at 16:24
  • 1
    $\begingroup$ Do you know about the gradient in polar coordinates? $\endgroup$ – ja72 Oct 4 '13 at 16:26
  • $\begingroup$ I am not familiar with it, but after looking at some pages online, I am not quite sure how to proceed. In polar coordinates would my equation be U(L, a)? Doesn't this pose a problem as the radius is constant, and if I understood what I am read, then the entire gradient would reduce to (1/L)*(partial derivative of U to a), which would just give me (1/L)(mgLsin(a)), which leaves me with my original equation mgsin(a), which still does not include the arc length. $\endgroup$ – Kayle of the Creeks Oct 4 '13 at 17:54
1
$\begingroup$

Writing $U$ as a function of $x$ and $y$ doesn't really make sense because the length of the pendulum is fixed at $L$ so that the system is really one dimensional.

I think what you were supposed to do is change variables from $\theta$ to $s$. Notice $\theta=s / L$. Plugging this into the expression for $U$, we get $$U(s) = mgL(1- \cos(s/L)).$$ Next I think you are supposed to calculate the derivative of the energy in this coordinate system. You should get $$\frac{d}{ds} U = mg \sin(s/L).$$

$\endgroup$
  • $\begingroup$ I think this might have been the approach they were looking for. My original idea was that if I swapped the variables to x and y, I could find the components of the tangential force, so that I could find it's magnitude. It worked, and I got the above answer as well, it just wasn't what was asked. In the above equation, why are we taking the partial derivative, rather than just a normal derivative with ds? $\endgroup$ – Kayle of the Creeks Oct 4 '13 at 18:05
  • $\begingroup$ oh I think it should be a total derivative, but they are the same here. I don't really think about things like that. I will change it $\endgroup$ – Brian Moths Oct 4 '13 at 18:07
  • $\begingroup$ Alright, I figured that they were the same here. I understand this approach, but it still confuses me why the question explicitly asked for me to use the gradient. $\endgroup$ – Kayle of the Creeks Oct 4 '13 at 18:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.