A disc shaped body (having a hole shown in the figure) of mass `m=10kg` and radius `R=10/9m` is performing pure rolling motion on a rough horizontal surface. In the figure point `O` is geometrical center of the disc and at this instant the centre of mass `C` of the disc is at same horizontal level with `O`. The radius of gyration of the disc about an axis pasing through `C` and perpendicular to the plane of the disc is `R/2` and at the insant shown the angular velocity of the disc is `omega=sqrt(g/R)` rad/sec in clockwise sence. `g` is gravitation acceleration `=10m//s^(2)`. Find angular acceleation `alpha` ( in `rad//s^(2)`) of the disc at this instant.

Answer

6

Explanation

Considering torque of real and pseudo force in the frame of `P`,
`tau=mg R/2+m omega^(2) R R/2`
`=mg R/2 + m g/R R R/2 = mgR`
`I_(P)=m(R/2)^(2)+m(R^(2)+(R^(2))/4)=3/2 mR^(2)`
`:. alpha=(2mgR)/(3mR^(2))=(2g)/(3R)=6 rad//s^(2)`