Is the line integral of tangential force in pendulum equal to the negative of change in potential energy?

[zenterix]

Homework Statement

In doing some seemingly simple calculations related to a simple pendulum, I ran into some doubts.

Relevant Equations

I am wondering if the line integral of the tangential component of gravitational force along the pendulum trajectory is equal to the negative of change in potential energy?

Consider a simple pendulum as depicted below

Consider the integral

$$\int \vec{F_g}\cdot d\vec{r}$$

My question is if we can equate this to the negative of a change in a potential energy function, ie $-\Delta U$?

Since $F_g$ is conservative, by the 2nd fundamental theorem of calculus for line integrals, the line integral above is indeed the change in some potential function.

My question is if this potential function is the same as whatever the commonly used potential energy function is for this problem?

In the case of a particle moving straight along an $x$-axis we have $\int \vec{F}\cdot d\vec{r}=\int_{x_i}^{x_f} Fdx$ and if $F$ is conservative then we consider the potential function of $F$ to be a function $-U(x)$.

In the case of the pendulum we have

$$\vec{F}_g=-mg\sin{\theta}\hat{\theta}+mg\cos{\theta}\hat{r}\tag{1}$$

$$d\vec{r}=Ld\theta\hat{\theta}\tag{2}$$

Then

$$\int\vec{F_g}\cdot d\vec{r}=\int_\theta^0 (-mgL\sin{\theta}) d\theta\tag{3}$$

$$=mgL(1-\cos{\theta})\tag{4}$$

One option is to equate this to $-\Delta U = U(\theta)-U(0)$ and define $U(0)=0$. Then

$$mgL(1-\cos{\theta})=U(\theta)\tag{5}$$

On the other hand, in (3) we could have written

$$\int\vec{F_g}\cdot d\vec{r}=\int_\theta^0 (-mgL\sin{\theta}) d\theta=L\int_\theta^0 (-mg\sin{\theta})d\theta\tag{5}$$

$$=L\cdot (mg(1-\cos{\theta})\tag{6}$$

$$=LU(\theta)\tag{7}$$

Consider a different calculation

$$\int \vec{F}\cdot d\vec{r}=\int \frac{d}{d\theta} \left (-U(\theta)\right )\hat{\theta}\cdot Ld\theta \hat{\theta}$$

$$=-L\int dU$$

$$=-L\Delta U$$

It seems that the integral $\int \vec{F}\cdot d\vec{r}$ is not equal to $-\Delta U$ anymore, which at first glance seems it should be since I am integrating $\frac{d}{d\theta} (-U(\theta))$.

However, I am integrating with respect to $d\vec{r}=Ld\theta$ not $d\theta$.

 

[zenterix]

I think the mistake is thinking I can write $\vec{F}=-\frac{d}{d\theta}U(\theta)$ though I am not sure exactly why yet.

 

[kuruman]

I think the mistake is thinking I can write $\vec{F}=-\frac{d}{d\theta}U(\theta)$ though I am not sure exactly why yet.

It's dimensionally incorrect. The correct expression is $\mathbf F=-\mathbf{\nabla}U.$ You can then write the gradient in Cartesian, cylindrical etc. coordinates.

 

[zenterix]

The correct expression is F=−∇U. You can then write the gradient in Cartesian, cylindrical etc. coordinates.

I am aware of this but in the case of the pendulum it seems we have only one dimension, $\theta$.

At this point I think I understand what is happening.

We have $\vec{F}_T(\theta)=-mg\sin{\theta}\hat{\theta}$, the tangential component of the gravitational force. We see it is a function only of $\theta$. We know it is conservative.

If we integrate this function with respect to $\theta$ then we get the difference in a potential function. We define the (potential energy) function $U(\theta)$ to be the negative of the potential function of $\vec{F}(\theta)$.

When we want to compute the work done by $\vec{F}_T$ along the trajectory of the pendulum we need to compute

$$\int\vec{F_T}\cdot d\vec{r}$$

We are not integrating $\vec{F}_T$ with respect to $\theta$ but rather with respect to $L\theta$.

All this entails is multiplying the integral of $\vec{F}_T$ with respect to $\theta$ by the constant $L$.

Thus,

$$\int\vec{F}_T\cdot d\vec{r}=L\int\vec{F}_T\cdot d\theta\hat{\theta}=L\int F_Td\theta=L\Delta U(\theta)$$