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 How much is the force of friction acting on the cylinder ?

A cylinder of mass 10kg and radius 0.15m is rolling perfectly on a pla

1.	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 How much is the force of friction acting on the cylinder ?
Answer» Solution : 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^(@)`