Verified answer

we need to find the linear acceleration of the cylinder

newton's second law tells us that

mg sin (theta) - friction = ma where a is the linear acceleration

the friction will cause a torque on the cylinder such that

friction x R = I alpha where I is the moment of inertia of the cylinder and alpha is the angular acceleration

the moment of inertia of a solid cylinder is 1/2 mR^2 and the angular acceleration is a/R

so we have:

friction x R = 1/2 mR^2(a/R) or friction = 1/2 ma

substitute back into newton's second law

mgsin(theta) - friction = ma

mgsin(theta)-1/2ma = ma or a = 2/3 g sin(theta)

the time to travel a distance L starting fromr rest and with acceleration a is

L=1/2 at^2

so the time to reach the bottom is

sqrt[2L/a] where a=2/3 g sin(theta) and L=h/sin(theta)

substitute values and solve for t

i will positioned my Arabic in to apply now i do no longer you it alot and that i do no longer understand very lots yet i think of you deserve it. Ana wahashany kitear!!!!!!!!!!!!!!!!!! Ana bahbek ya Farah It had in effortless terms been some days with out you in spite of the incontrovertible fact that it became into no longer the comparable the Egypt type isn't the comparable with out you no longer even for a million day.