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.

  • 1
    $\begingroup$ The notation here is all over the place - can you say precisely what it is you're integrating, so we could fix it? $\endgroup$
    – J. Murray
    Oct 5, 2017 at 5:38
  • $\begingroup$ Trying to solve the first equation for v. I'm just not exactly sure about the one step of changing the variable you are integrating with respect to. From the first question to the second. $\endgroup$
    – Jmcnamara
    Oct 5, 2017 at 5:39
  • 1
    $\begingroup$ What I mean is that the equation you wrote there does not make mathematical sense - that's not something that can be "solved for v." If you can state the actual problem which precedes the integral, we can fix that. $\endgroup$
    – J. Murray
    Oct 5, 2017 at 5:42
  • $\begingroup$ Sorry I should have said, an expression for v in terms of r and theta. Basically I just need to integrate both sides of the equation and get v by itself. The equation comes from the kinematics equation ads = vdv. However this means that the integral is in terms of s. The aim is to change variables to an integral in terms of theta and have the correct corresponding limits. In the end I need an equation v= ... Thanks. $\endgroup$
    – Jmcnamara
    Oct 5, 2017 at 6:12

1 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.

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, $$ gr \sin(\alpha) = \Delta \left(\frac{1}{2}v^2\right)$$

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. $\alpha$).

  • $\begingroup$ Thank you for such a complete answer. That explains it perfectly. I was not thinking about the integral as the path of the particle through an angle theta. Sorry for the awkwardly worded question. $\endgroup$
    – Jmcnamara
    Oct 5, 2017 at 7:26
  • $\begingroup$ No problem. I just re-worded my answer to be more consistent with your question, so you may want to make sure it still makes sense. $\endgroup$
    – J. Murray
    Oct 5, 2017 at 7:27

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