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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?

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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.

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  • $\begingroup$ That's what I thought – that this can be formulated in terms for rotational kinematics. However, I had just learned about them and didn't see to apply them here. Thanks so much. I havn't learned integral calculus and was hoping you'd be able to complete the integration step as well. $\endgroup$ – theideasmith Oct 31 '16 at 12:55
  • $\begingroup$ @theideasmith Added solution of integral, and some comment on why you genreally don't want to proceed further down this road. $\endgroup$ – Mikael Fremling Oct 31 '16 at 13:15
  • $\begingroup$ Additionally – can you explain how you got from $v=r\frac{d\theta}{dt}$ to $mg\sin\theta=r^2\frac{d^2\theta}{dt^2}$? $\endgroup$ – theideasmith Oct 31 '16 at 14:01
  • $\begingroup$ @theideasmith Added that $a=\frac{dv}{dt}$ and changed $r^2$ to $r$ in the equation. $\endgroup$ – Mikael Fremling Oct 31 '16 at 14:20
  • $\begingroup$ Thank you so much for this answer. So what you are saying is that while the problem could be solved with kinetic equations it is much much simpler with energy equations $\endgroup$ – theideasmith Oct 31 '16 at 15:34
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Getting the velocity as a function of the angle is very simple. If the equation of motion is $$mr\frac{d^2\theta}{dt^2}=-mg\sin{\theta}$$where $\theta$ is the angle with the vertical, multiply both sides by $d\theta /dt$ and then integrate with respect to time, or, equivalently, set the change in kinetic energy equal to the change in potential energy. Either way, you get $$\frac{v^2}{2}=gr(\cos{\theta}-\cos{\theta_0})$$where $\theta_0$ is the initial angle. This neglects the rotational kinetic energy of the ball. Including that involves adding only one additional term to the kinetic energy.

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  • $\begingroup$ If the initial angle $\theta_{\rm o}$ is smaller than the final angle $\theta$ the right hand side of your second equation is negative. I think that in your equation for the angular acceleration the right hand side should be positive? $\endgroup$ – Farcher Nov 1 '16 at 9:23
  • $\begingroup$ @Farcher Thanks for your comment. In my analysis, $\theta$ is the angle of the ball relative to the vertical, and I envisioned $\theta_0$ as being close to $\pi /2$, so that, as the ball descends, the angle $\theta$ gets smaller. Maybe I should have drawn a diagram. $\endgroup$ – Chet Miller Nov 1 '16 at 11:11
  • $\begingroup$ I now understand. $\endgroup$ – Farcher Nov 1 '16 at 11:17
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As you noted at the end in doing the integration to obtain $v-v_0=mg\sin(\theta)t$ you have assumed that $\sin(\theta)$ is constant which it is not.

The analysis is easier if the ball does not roll and the frictional force is zero.
The tangential force at any time $t$ is your $mg\sin(\theta)$.

Also the linear speed $v$ is connected to the rate of change of angle $\omega =\frac{d\theta}{dt}=\dot \theta$

$v=(R+r)\omega$ where $R$ is the radius of the track and $r$ is the radius of the ball.

So the linear acceleration $a=\frac{dv }{dt } = (R+r) \dot \omega = (R+r) \ddot \theta \approx R \ddot \theta$ if $R>>r$

So you have to solve $g\sin \theta=R \ddot \theta$.

Multiply both sides by the linear speed $v=R\dot \theta$

$v R \ddot \theta = v \dfrac{dv}{dt}$ and $vg \sin \theta = gR \sin \theta \dfrac {d\theta}{dt}$

and integrate

$\displaystyle \int ^v _0 v \; dv= gR \int ^\theta _0 \sin \theta \; d \theta \Rightarrow \dfrac 12 v^2 = gr (1 -\cos \theta)$

which is a long-winded way of applying conservation of energy directly.

Gain in kinetic energy = loss in potential energy $\Rightarrow \dfrac 12 mv^2 = mgr (1 -\cos \theta)$

To find the angle $\theta$ as a function of time ie solve $g\sin \theta=R \ddot \theta$ for $\theta(t)$ is much more difficult and WolframAlpha with the input of solve[theta''[t]=sin( theta[t])] tells me it requires the Jacobi amplitude function.


With the ball rolling without slipping requires there to be a static frictional force $F$ on the ball at the point of contact between the ball and the track which will make the analysis more difficult.

If the frictional force is $F$ it will act at the point of contact between the ball and the track and will lie alone a tangent to the track.

So you now have $mg\sin \theta - F=mR \ddot \theta$ and $Fr = I_{\rm c} \ddot \phi$ where $\ddot \phi$ is the angular acceleration of the ball, $r$ is the radius of the ball and $I_{\rm c}$ is the moment of inertial of the ball about its centre of mass.

$R\ddot \theta = r \ddot \phi \Rightarrow F= \dfrac{I_{\rm c}R \ddot \theta}{r^2} $

Substituting for $F$ gives

$mg\sin \theta = \left (m+\dfrac {I_{\rm c}}{ r^2} \right ) R \ddot \theta$

with the bracket having an extra term from the equation when it was assumed that the ball was not spinning $\dfrac {I_{\rm c}}{ r^2} $ which is the extra inertia due to the ball spinning.

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