# Why is this derivation of rotational kinetic energy wrong?

I was trying to derive the rotational kinetic energy of a solid sphere with radius $$R$$ using the following idea.

Each infinitesimally small shell volume $$4\pi r^2 \cdot dr$$ at radius $$r$$ within the sphere has a mass $$m$$ of $$m(r) = \frac{4\pi r^2 \cdot dr}{V} \cdot M$$ where $$V$$ and $$M$$ are respectively the total volume and total mass of the sphere.

That piece of shell mass is rotating at a velocity of $$r\omega$$ where $$\omega$$ is the angular velocity. Such that the rotational kinetic energy of that piece of shellmass, according to $$\frac{1}{2}mv^2$$, is: $$E_{rot}(r\geq r+dr)=\frac{1}{2} \cdot \frac{4\pi r^2 \cdot dr}{V} \cdot M \cdot (r \cdot w)^2$$ Integrating this equation all up to the total radius $$R$$ of the sphere would give the total rotational kinetic energy of the sphere. The integration would give: $$E_{rot}=\int_0^R \frac{1}{2} \cdot \frac{4\pi r^2 }{V} \cdot M \cdot (r \cdot w)^2 \cdot dr = \frac{3}{10} M (\omega \cdot R)^2$$ However, the correct rotational kinetic energy formula for a solid sphere has a factor of $$\frac{1}{5}$$ instead of $$\frac{3}{10}$$ and integrates over mass increments $$dm$$ instead of over radius increments $$dr$$.

Why is my idea for the derivation wrong?

## 1 Answer

It’s wrong because you assumed that all of the shell is rotating with speed $$r\omega$$. Only the equator of the shell has that speed. The poles aren’t moving at all. You have to use the distance from the axis of rotation. The spherical coordinate $$r$$ is the distance from the origin, not the distance from the axis of rotation.

By the way, you are also misusing the word “velocity” when you mean “speed”.