A cylinder of mass 10kg and radius 0.15m is rolling perfectly on a plane of inclination `30^(@)`. The coefficient of static friction `mu_s=0.25`
What is the work done against friction during rolling ?

Explanation

Given, `m=10kg,r=0.15m,theta=30^(@),mu_3=0.25`
`I=(MR^2)/2=(10xx(0.15)^2)/2=0.1125kgm^2`
From the figure,
`((mgsintheta-mu_kmgcostheta)/(1+(K^2)/(R^2))=ma`
i.e, `[(gsintheta-mu_kgcostheta)/(1+K^2)/(R^2))]=a`
Taking `mu_k=0` (not given)
`a=(gsintheta)/(1+(K^2)/(R))` where `(K^2)/(R^2)`=1/2` for a cylinder `:. a=`(gsintheta)/(3/2)=2/3gsintheta`
We know that `tau=Iprop=fR`
i.e `f=(Iprop)/(R)` i.e `f=(mR^2/(2R)xxa/R" ":.f=1/3mgsintheta`
`f=1/3xx10xx9.8xxsin30^(@)=5xx3.267=16.335`
`:.` Force of kinetic friction=16.34N
Work done against friction while in motion=0
For the cylinder to kid `mu=1/3tantheta`
i.e `tantheta=3xx0.25=0.75`
`:.theta~~37^(@)`