Question

Question #7a89b

Answer 1

The moment of inertia is `=1/2ma^2`

Explanation:

The mass of the cylinder is `=m`

The radius of the cylinder is `=a`

The height of the cylinder is `=h`

The volume of the cylinder is `V=pia^2xxh=pia^2h`

The density of the cylinder is `rho=m/V=m/(pia^2h)`

Take a a concentric ring of radius `a` and thickness `=dr`

The area of the ring is `A=2pirdr`

The volume of the ring is `v=2pirhdr`

The mass of the ring is

`dm=v*rho=2pirhdr*m/(pia^2h)=(2m)/a^2rdr`

The moment of inertia of this ring is

`dI=r^2dm=2m/a^2r^3dr`

Integrating

`int_0 ^IdI=2m/a^2int_0^ar^3dr=2m/a^2[r^4/4]_0^a`

`=2m/a^2*a^4/4=1/2ma^2`