Elementary but from a little different source

ya true :) , I loved it .

Let the aspirants try , it should be some good practice for them before the exams .

thanks rohan got it

ok so heres the solution !

consider friction at the interface between the two cylinders ..
let it be f

let the normal reaction at that surface be N

at each moment the smaller mass has to remain in equillibrium because if it isnt there will be slippin

let the friction at its ground be F to the left

then equating torques we have f=F

making the resultant of foces along horizontal 0

F+Fsinθ=Ncosθ

where θ is the angle made by the line joining the centres at any moment

but F<=μN

so μ>=cosθ/(1+sinθ)

for no slipping μ should be greater than the maximum value of the exp. right .. (so that it reamins greater throughout)

there will be different values of the right exp. for different θ but the expression is tan(π/4-θ)

so it will maximum for minimum θ which is at the intial stage ..

sinθ=R-r/(R+r) from figure
find cosθ ,..
substituting we get

μ>=√(d/D)

were is friction at ground shown in the diagram ??????????

can u tell wats rong with my argument abv ;?

Sorry , I was a little busy

Why don't you write the eqns of torque ?

diptosh find mistake in my sol or give ur sol
no one else trying i guess

ne one

reply wats rong with my exp

till now

by considering that small roller exerts normal rn along line joining centres , got that

Ncosθ + μNsinθ ≤ P , θ angle made by line joining centre with horizontal

Nsinθ - μNcosθ ≥ Mg , M - mass of big roller

i ve assumed that smaller sphere doent move as its very heavy

i m unable to proceed further , do i need xtra info or wat concept am i missing

ya mass is not required take frictional forces, thier torques and proceeding. in the later stage they will get eliminated.

but i am getting the answer with reverse of inequality.