laws of motion

well here are the steps . .

first step find the velocity of m as a function of θ (angle to horizontal) (by conservation of energy)

the normal reaction in their interface will increase compensating mv²/R + mgsinθ

the force on M along horizontal direction will be backwards = Ncosθ

the normal force on ground= Nsinθ+Mg

so we have for the wedge to not slip

Ncosθ<=μ(Nsinθ+Mg)

from here we will get μ as a inequality in θ (as we know N and v in terms of θ)

for the wedge to not slip at each moment μ should be greater than the maximum value of the corresponding left expression

.. thus you will arrive at the answer

rohan i know the steps, but can u help me out in the last step after we have reached the inequality ..

ok , i hope this is the inequality you got =

μ>(3mgsinθcosθ)/(3mgsin²Î¸+Mg)

divide both num. and denominator by cos²Î¸

now express sec²Î¸ as 1+tan²Î¸ and divide again by tanθ

u will get

3m/(3m+M)tanθ+M/tanθ

apply A.M >=G.M to the bottom :)

thx again rohan [1]
yeah that was the inequality i got