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?

Expert Answers

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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`

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