AL APPLIED

Moment of inertia of a solid disc, I = 1/2 ma^2

Let N be the normal reaction on the disc.

Let the potential energy be zero at the horizontal radius of the cylinder.

Since the disc rolls without slipping, hence there is no energy loss in the process.

And the rolling without slipping requirement: (a + b)x = a@, where @ is the angle subtended by the disc.

So, d@/dt = (a + b)(dx/dt) / a = v / a

Initial P.E. = P.E. + translational K.E. + rotational K.E.

mg(a + b) = mg(a + b)cosx + 1/2 mv^2 + 1/2 I(d@/dt)^2

mg(a + b)(1 - cosx) = 1/2 mv^2 + 1/2 (1/2 ma^2)(v/a)^2

mg(a + b)(1 - cosx) = 3/4 mv^2 ... (1)

By equation of motion in the radial direction,

mgcosx - N = mv^2 / (a + b) ... (2)

For the disc to leave the cylinder, the normal reaction is zero, N = 0.

Sub (1) into (2) and N = 0,

mgcosx = (4/3)mg(1 - cosx)

(7/3)cosx = 4/3

cosx = 4/7 ###