We know that W(nc) = (Delta)E (Where W(nc) is the non-conservative work done on the system.)
Now, we are rolling without slipping so there IS friction in the problem. However this work is already taken into account by the rotational Kinetic energy term and thus we can still take W(nc) = 0 J.
0 = (Delta)K + (Delta)K(rot) + (Delta)P
(Delta)P = Mg(h - h0) where h is the final height of the CM and h0 is the initial height of the CM.
(Delta)K = (1/2)M(v^2 - v0^2)
(Delta)K(rot) = (1/2)I(w^2 - w0^2) where "w" is (omega) the angular speed, and I is the moment of inertia of the object in question.
We know that there is rolling without slipping. Thus we know that v = Rw.
So plug all this into the energy equation, and use v = Rw to put the equation in terms of v. (Not w, I'll explain why in a minute, though in the long run it doesn't make much difference.) The only "unknown" in this equation will be v. (Specifically I'm assuming values for M, R, h, and h0 are "known.")
The thin walled tube for example: I = MR^2
I'll assume a coordinate system for the gravitational potential energy such that the CM of the tube at the bottom of the incline is at y = h = 0. Thus the initial height of the tube can be considered to be y = h0, where h is the height of the incline. (Upward is, of course, positive.)
0 = (1/2)M(v^2 - 0^2) + (1/2)*MR^2*(w^2 - 0^2) + Mg(0 - h0)
0 = (1/2)Mv^2 + (1/2)MR^2*w^2 - Mgh0
and v = Rw, so w = (v/R):
0 = (1/2)Mv^2 + (1/2)MR^2*(v^2/R^2) - Mgh0
0 = (1/2)Mv^2 + (1/2)Mv^2 - Mgh0
0 = Mv^2 - Mgh0
v = sqrt(gh).
Now find this for the disk and sphere.
This doesn't have to do with acceleration directly, but we know that the acceleration will be constant and (presumably) we know they all rolled the same distance. So by using the distance travelled by the CM and the initial and final speeds we can figure out the accelerations. (Though we can simply compare the final v and get the same answer.)