# Consider a cylinder of radius `R` , mass `M` , length `z` , and density `rho(r)=Ar` that rolls without slipping down an inclined plane of height `h` at an angle `theta` . What is the velocity of the cylinder at the bottom of the inclined plane?

We will use conservation of energy to solve this problem. We need to consider the rotational energy of the cylinder and the translational energy of the center of mass.

`E_i=E_f`

`U(h)=K_(trans)+K_(rot)`

`Mgh=1/2 Mv^2+1/2 I omega^2`

`Mgh=1/2 Mv^2+1/2 I (v/R)^2`

We need to find the moment of inertia.

`I= int r^2 dm=int r^2 rho(r) dv`

`I=int_0^R r^2 rho(r) z (2pi r) dr`

`I=2pi zA int _0^R r^4 dr`

`I=(2pi zAR^5)/5`

Now to get A in terms of `M` .

`M=int dm=int_0^R rho(r) z(2pi r) dr`

`M=2A z pi int_0^R r^2 dr=2A z pi (1/3)R^3`

`A=(3M)/(2z pi R^3)`

`I=(2pi zAR^5)/5=(2pi z)(3M)/(2z pi R^3)*(R^5/5)=3/5MR^2`

Now solve the energy equation for `v` .

`Mgh=1/2 Mv^2+1/2 I (v/R)^2`

`2Mgh=Mv^2+(3/5MR^2)*(v/R)^2`

`2gh=v^2+(3/5)v^2`

`2gh=(8/5)v^2`

`5/4 gh=v^2`

`sqrt(5gh)/2=v`