# How to use Calculus to give an explanation for a video by Lewin [closed]

In the video by the well-known physicist Water Lewin Link Here, in which he demonstated a series of experiments of rolling objects. Many are quite surprising. At the end of the video, the attended students asked him to give some explanation for the phenomenon. But he just said that it is not easy! and needs "Calculus" and then only gave a very brief idea.

I am not a physics student but I am very interested to learn how Calculus comes into play here.

Could anyone give some explanations based on more the Mathematics aspect (like Lewin suggested) for his experiments!

Thanks for any help.

• This question should be posted to the Physical StackExchange. – Matthew Daly Jan 31 '20 at 17:04

To understand rolling without slipping we need to understand the kinetic energy falls into two modes; linear kinetic energy and rotational kinetic energy. For a rigid body if the velocity of the center of mass $$m$$ is $$v$$ then $$\frac{1}{2}mv^2$$ gives the linear KE. On the other hand, analyzing the motion of the little bits of mass rotating around the center of mass allows us to capture that collective energy by $$\frac{1}{2}I \omega^2$$ where $$I$$ is the moment of inertia and $$\omega$$ is the angular velocity about the center of mass. Generally it is more complicated and you need an inertia tensor to capture the story, but when the rotational motion is just around one axis then we get the usual Freshman stuff.
Ok, so, $$KE = \frac{1}{2}mv^2+ \frac{1}{2}I \omega^2$$ Conservation of energy for a rigid body starting from rest gives us that the initial gravitational potential energy $$PE = mgh$$ is matched by the final $$KE$$.
Rolling without slipping means that $$\omega$$ and $$v$$ are connected; $$v = \omega R$$ hence we find: $$mgh = \frac{1}{2}mv^2+ \frac{1}{2}I (v/R)^2$$ hence $$mgh = \frac{1}{2}(m+ I/R^2)v^2$$
and so, $$v = \sqrt{\frac{2mgh}{m+I/R^2}}$$ We find larger $$I$$ iimplies smaller $$v$$. Calculus shows $$I = \frac{1}{2}mR^2$$ for a solid cylinder whereas $$I = mR^2$$ for a cylindrical shell. Another common example somewhere between these extremes is the solid sphere with $$I = \frac{2}{5}mR^2$$.
• Also, it is noteworthy that $I$ typically has an $R^2$ dependence independent of $m$. This fortunate cancellation is critical to his analysis. I am unaware if there exist exotic shapes where $I$ is not dependent on $R^2$, or how to even make sense of this question I raise... – James S. Cook Feb 1 '20 at 2:35