# What's wrong with this derivation for rotational energy of a sphere?

1. Mar 9, 2013

### henpen

First, energy of a disk:

$$\int \frac{dm}{2}r^2 \omega^2 =\frac{\omega^2}{2}\int_0^R m\frac{2 \pi r dr}{\pi R^2}r^2 =\frac{m\omega^2}{ R^2}\int_0^R r^3 dr=\frac{m\omega^2 R^2}{4 }$$
Which agrees with other sources. However, in the following lies my problem:

The equation for a circle: $r^2+x^2=R^2 \Rightarrow r^2=R^2-x^2$

Energy of a sphere- integrate infinitesimal disks's rotational kinetic energy (assuming rotational energy is additive, which makes sense physically), all of which have their centre through the x-axis:

$$\int_0^R \frac{m\omega^2 r^2}{4 }dx=\frac{m\omega^2}{4 }\int_0^R(R^2-x^2)dx=\frac{m\omega^2}{4 }\frac{2R^3}{3}=\frac{m\omega^2 R^3}{6 }$$

This differs from the result here of $$\frac{I \omega^2}{2}=\frac{m\omega^2 R^2}{5 }$$ quite substantially. Where have I gone wrong?

2. Mar 9, 2013

### Staff: Mentor

In setting up your second integral you have treated m as a constant.

3. Mar 9, 2013

### henpen

Gah! Thanks for that.

4. Mar 9, 2013

### henpen

$$\int_0^R \frac{m(r)\omega^2 r^2}{4 }dx=\frac{\omega^2 }{4 }\int_0^R m(r)r^2 dx$$
$$m(r)= \rho \pi r^2 = k r^2$$
$$\frac{k\omega^2 }{4 }\int_0^R r^4 dx=\frac{k\omega^2 }{4 }\int_0^R (R^2-x^2)^2 dx=\frac{k\omega^2 }{4 }\int_0^R R^4+x^4-2R^2x^2 dx$$
$$=\frac{k\omega^2 }{4 }(R^5+\frac{1}{5}R^5-\frac{1}{3}2R^5 )=\frac{k\omega^2 }{4 }\frac{1}{15}(15R^5+3R^5-10R^5 )=\frac{k\omega^2 }{4 }\frac{1}{15}(8R^5 )$$
$$=\frac{2k\omega^2 R^5 }{15 }$$

$$k=\pi \rho = \pi \frac{M}{\frac{4}{3}\pi R^3}=\frac{3M}{4R^3}$$

$$\frac{2k\omega^2 R^5 }{15 }=\frac{3M}{4R^3}\frac{2\omega^2 R^5 }{15 }=\frac{M \omega^2 R^2}{10}$$

That's correct: excellent!

5. Mar 9, 2013

### Philip Wood

Since $\frac{1}{2}\omega^2$ is in common to all discs, it’s (arguably) nicer to evaluate I, and, having found it, to multiply it by $\frac{1}{2}\omega^2$ afterwards.

6. Mar 9, 2013

### Staff: Mentor

That's how I would do it.