A cylinder is rolling without sliding over two horizontal planks (surfaces) `1` and `2`. If the velocities of the surfaces `A` and `B` are `-vhati` and `2vhati` respectively, find the:
a. Position of instantaneous axis of rotation.
b. Angular velocity of the cylinder.

Answer

`3v//2R`

Explanation

a. Since there is no relative sliding between the cylinder and the planks `1` and `2`, the points, `A` and `B` of the disc will move with velocities equal to the velocities of the respective surface.
`vecvA=-vhati` and `vecvB=-2vhati`
respectively. Joining `A` and `B` and `A'` and `B'` we find the point `P` as `IAR`. Then we have the similar triangles `"PAA"'` and `"PBB"'`. Using the properties of similar triangles we have

`omega=("BB"')/("AA"')=("B"P)/("AP")`
`"AA"'=v_(A)=v, "BB"'=v_(B)=2v, AP=2R-x` and `BP=x.`
Then we have `(2v)/v=x/(2R-x)`
This gives, `x=(3R)/3`
hence `IAR` is located at a distance of `4R//3` below the top of the disc.
b. As we know `omega=v_(AB)/(AB)`
`v_(AB)=(vecv_(A)-vecv_(B)|=|-vecvhati-2vhati|=3v` and `AB=2R`
This gives `omega=(3v)/(2R)` clockwise