There are too many problems put together, which looks very troublesome.

Here is the solution to the seventh problem:

Assume that the angular velocity of the roller is ω and the angular acceleration is α.

The speed of the drum in horizontal direction can be divided into: the speed in AB direction and the speed in vertical AB direction.

Starting from OA⊥AB, the polar velocity in AB direction is: Vab=ω0r.

Therefore, the translation speed of the center of mass is: Vc=Vab/sin60, ω=2ω0r√3/3, because the roller only rolls and does not slide,

Then: ω=Vc/R,

Solution: ω=2ω0r√3/3R.

The acceleration of point b relative to point a is: a = (vabtan 30) 2/ab = (vab) 2 √ 3/9r = ω 02r √ 3/9, and the direction points to a along b,

Then: the absolute acceleration ab of the centroid point b is projected along BA to be equal to a.

If ABS in60 = ω 0 2r √ 3/9, then the acceleration of the center of mass is: AB = 2ω 0 2r/9.

By: α = AB/R = 2ω 0 2r/9r,

8. Angular velocity of 8.BC rod: ω=ω0*AB/BD. According to the geometric relationship, BD=BC=CD=2m.

Therefore: ω=ω0/2=5 radians/second.

Absolute acceleration of point C: the component of ac along the vertical BC direction is equal to the projection of the absolute acceleration of point B to the vertical BC direction.

So the angular acceleration is α, α * BC = ω 0 2 * abcos 30.

Solution: α = (10 2) *1√ 3/(2 * 2) = 25 √ 3 rad/S2.