In this Phys.SE question a ball rolls down a circular incline. We determine the point of zero net force using equations of kinetic energy for final velocity. However, I'm wondering how final velocity of a ball rolling down a concave circular incline can be derived using equations of motion.
I've tried doing this by starting with acceleration along the direction perpendicular to the line from the ball to the center of the circle. With mass $m$, the acceleration is $a=mg\sin(\theta)$. Now I'm not sure how to approach deriving a velocity equation as it seems velocity is dependent on both time and initial angle.
Is it true that in this situation, we can obtain velocity by integrating acceleration with respect to time to obtain: $v-v_0=mg\sin(\theta)t$? This doesn't seem correct because the acceleration is non-constant. Please some help with this?
Best Answer
You have to remember that $\theta$ will depend on time as well so that you have $ma=mg\sin(\theta(t))$ that this will not be a simple think to integrate.
In order to proceed, you first need to relate $v$ to $\frac{d\theta}{dt}$. As the paricle is trapped on the surface of a sphere you then have $v=r\frac{d\theta}{dt}$ so the equation that you need to integrate will be
$$ r\frac{d^2\theta}{dt^2} = mg\sin(\theta),$$
as $a=\frac{dv}{dt}$. This is most easily done by separation of variables giving
$$ \frac{d^2\theta}{\sin(\theta)} = \frac{g}{r} dt^2.$$
Next, you integrate the two sides individually and get
$$ \log\left(\tan\left(\frac\theta2\right)\right)d\theta = (\frac{g}{r}t+C )dt$$ where $C$ is an integration constant. From this you can work our the velocity $\frac{d\theta}{dt}$ as a function of $\theta(t)$ and $t$.
Integrating both sides again gives an answer that is to ugly to reproduce here, which show that $\theta(t)$ is a complicated function.