A string is wound around a hollow cylinder of mass 5 kg and radius 0.5 m. If the string is now pulled with a horizontal force of 40 N, and the cylinder is rolling without slipping on a horizontal surface (see figure) then the angular acceleration (in `"rad"//s^(2)`) of the cylinder will be (Neglect the mass and thickness of the string),

Explanation

From newton.s second law
`40+f=m(R alpha)`………I
Taking torque about 0 we get
`40xxR-fxxR=I alpha`
`40xxR-fxxR=mR^(2)alpha`
`40-f=mR alpha`………..ii
Solving equation (i) and (ii)
`alpha=40/(mR)=16 "rad"//s^(2)`