A uniform solid cylinder is placed with its axis horizontal on a plane, whose inclination to the horizon is α. Show that the least coefficient of the friction between it and the plane, so that it may roll and not slide, is 1/3 tanα. If the cylinder be hollow and of small thickness, the least value is 1/2 tanα.

Solution: In this figure 1 ,

 a uniform solid cylinder is placed horizontally on the plane BOCD Which makes an angle α with the horizon. Now we make another cross sectional figure from side view only to make our calculation easy.
we get figure 2.

 Let O be the initial point from which the cylinder rolls down. The supposition related to the cylinder are :
                                                       C→centre, M→mass and a→radius
 The plane OD makes an angle α with horizon then we get ∡PCT=α from this following figure and reasons.

 The weight mg act downwardly and it makes an angle α with ck and when we decompose it we get the force along ck is Mgcosα and the force along ct is Mgsinα.
Let CA be the line fixed in the body and the normal line ck in the space and let ∡Ack=θ and F& R be the friction and normal reaction acting on a plane respectively.
 Now, after time t, let the cylinder rolls in x-axis(it is supposed) so that F & Mgsinα act along x axis and R & Mgcosα act along y-axis and covers a distance x from initial point O at k.
 Then arc Ak = Ok = x
                    aθ = x                      [ ∵θ=arc/radius ]
differentiating w.r.to t upto second order
we get (d^2 x)/(dt^2 ) = a(d^2 θ)/(dt^2 )………………..(1)
 We have the equation of motion are M (d^2 x)/(dt^2 ) = Sx …(2)
                                                           M (d^2 y)/(dt^2 ) = Sy …(3)
            and also
 we have
 Mk^2 (d^2 θ)/(dt^2 ) = the forces along x axis in opposite direction × radius = F.a …(4)
From (1) & (2)
Or M a(d^2 θ)/(dt^2 ) = Mgcosα + (-F) [∵ the forces along x axis are mgcosα and (-f) due to acting                                                                    opposite direction]
Multiplying both sides by k^2
                                  Or M a(d^2 θ)/(dt^2 ) k^2 = k^2Mgcosα - k^2 F
                                 Or F(a^2+k^2) = k^2Mgcosα
                                  Or F= (k^2/(k^2+a^2 ))Mgcosα...(5)
From (3)
                         M (d^2 y)/(dt^2 ) = Mgcosα + (-R)    [∵ the forces along x axis are Mgcosα and (-R)                                                                                             due to acting opposite direction]
                    or, 0=Mgcosα -R                      [∵ the body moves only in one direction]
                    or, R=Mgcosα...(6)
Hence from (5) and (6)
        F/R=(k^2/(k^2+a^2 ))tanα
If the cylinder rolls then,
    F/R<m(coefficient of friction)
i.e (k^2/(k^2+a^2 ))tanα<m
here we have the cylinder is solid so k^2=a^2/2  so {(a^2/2)/  (a^2/2+a^2)}tanα<mi.e
                                                        (1/3)tanα<m
and if the cylinder is hollow then k^2=a^2 so, (a^2/a^2+a^2)tanα<m i.e 
                                                        (1/2)tanα<m