Changing limits when changing integration variable

Question

In class we were given a circular motion problem with a particle travelling in a circle of radius r. The distance travelled by the particle is given by s. To solve the question we needed to solve:

$\int_0^v g \cos \theta \ dv$ = $\int_0^s g \cos \theta \ ds$ for v
where s = r $\theta \ $

I'm trying to find an expression for v in terms of r and $\theta$.
Our teacher's solution said this equation could be simplified to:

$\frac{1}{2} v^2$ = $\int_0^\theta g \cos \theta \ rd\theta$

I understand how $ds$ becomes $rd\theta$ however I don't understand why the upper limit of integration is not also changed to $r\theta$ when changing the variable you are integrating with respect to. For example

$\frac{1}{2} v^2$ = $\int_0^{r\theta} g \cos \theta \ rd\theta$

Any help would be appeciated.

Answer 1

The actual answer to this comes at the bottom, but here is the preliminary part:

The infinitesimal work done on a particle by a force $\vec F$ as it undergoes an infinitesimal displacement $d\vec s$ is given by $\vec F \cdot d \vec s$. Therefore, the total work done by that force as the particle moves between points $A$ and $B$ is

$$ W = \int_A^B \vec F \cdot d\vec s$$

The position of a particle is a function of time - $\vec s = \vec s(t)$. Therefore, since $d\vec s = \frac{d\vec s}{dt} dt$, we can say that

$$ W = \int_{t_A}^{t_B} \vec F \cdot \frac{d\vec s}{dt} dt =\int_{t_A}^{t_B} \vec F \cdot \vec v dt$$

In general this would depend on the path you take from point $A$ to point $B$. However, we can worry about that later.

Now, Newton's 2nd law says that if $\vec F$ is the only force acting on the particle, then $$ \vec F = m \vec a = m \frac{d\vec v}{dt} = m\frac{d^2 \vec s}{dt^2}$$ and so $$ \vec F \cdot \vec v = m \frac{d\vec v}{dt} \cdot \vec v = \frac{1}{2} m \frac{d}{dt} (\vec v \cdot \vec v) =\frac{d}{dt} \left( \frac{1}{2} m v^2\right)$$

Therefore, we have that

$$ W = \int_A^B \vec F \cdot d\vec s = \int_{t_A}^{t_B} \frac{d}{dt}\left(\frac{1}{2} m v^2\right) dt$$

But that last integral is simply the integral of a derivative, which is trivial to solve. The result is that

$$ W = \int_A^B \vec F \cdot d\vec s = \frac{1}{2}mv_B^2 - \frac{1}{2}mv_A^2 \equiv \Delta\left(\frac{1}{2} m v^2\right)$$

which is the work-energy theorem.

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Now, your question essentially relates to the actual evaluation of the integral $$W= \int_A^B \vec F \cdot d\vec s$$ For a particular $\vec F$ and a particular path from point $A$ to point $B$. Based on what you've written, I have to assume that:

• The path is along a section of a circle of radius $r$ from $\theta = 0$ to some angle $\alpha$, so $ds = rd\theta$, and
• The force points in the +$y$ direction, so $\vec F \cdot d\vec s = F\cos(\theta) ds$, and its magnitude is the weight of the particle ($mg$)

Under those two assumptions, the integral becomes $$ W = \int_0^\alpha mg\cos(\theta)rd\theta = mgr \int_0^\alpha \cos(\theta) d\theta = mgr \left.\sin(\theta)\right|^\alpha_0 = mgr\sin(\alpha)$$

As we saw above, this is equal to $\Delta\left(\frac{1}{2} mv^2\right)$, so

$$ mgr \sin(\alpha) = \Delta\left(\frac{1}{2} mv^2\right)$$ or, if you prefer, $$ gr \sin(\alpha) = \Delta \left(\frac{1}{2}v^2\right)$$

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I think your confusion relates to the fact that you're integrating between points in the plane rather than points on the real line. Think about the integration bounds as being the values of the integration variable which correspond to the beginning and end of the path, respectively. In this case, we are integrating over $\theta$, and the path begins when $\theta=0$ and ends when $\theta=\alpha$.

A lot of confusion stems from writing something like this:

$$ \int_0^s g\cos(\theta) ds$$

I hope that's not actually what your teacher wrote. The notation is very bad. That expression says

Integrate $g\cos(\theta)$ from $s=0$ to $s=s$

You can see why this makes no sense. $s=s$ is always true - you can't put the integration variable in the integration bounds!

If you wanted, you could put something like $$\int_0^{s_{max}} g \cos(\theta) ds$$ where $s_{max}=r\alpha$ but you should also be careful to remember that $\theta=\theta(s)$ would be considered a function of $s$ in that context.

Making the transformation $ds \rightarrow rd\theta$ we get the integral I wrote above. Remember - the integration bounds reflect the values of the integration variable, so when we switch to $\theta$, the lower bound is the initial value of $\theta$ (i.e. 0) and the upper bound is the final value of $\theta$ (i.e. $\alpha$).