The situation is shown fig. As force `F` rotates the sphere, the point of contact has a tendency to slip towards left so that the static friction on the sphere will act towards right. Let `r` the radius of the sphere and a be the linear acceleration of the centre of the sphere the angular acceleration about the centre of the spohere is `alpha=a/r`, as there is no slipping.
For the inear motion of the centre
`F+f=ma`.............i
and for the rotational motion about the centre,
`Fr-fr=ialpha=(2/3mr^(2))a/r`
or `F-f=2/5ma`.............ii
from eqn i andn ii we get
`2F=7/5 ma` or `a (10F)/(7m)`
Alternative method: As the shre is rolling, we can apply torque equation about point of contact.
i.e. `F(2r)=I_(p)alpha=(2/5mr^(2)+mr^(2))alpha`
`impliesF(2r)=(7/5mr^(2))alpha`
Hence angular acceleration `alphas=(10F)/(7mr)`..........iii
As sphere is rolling, acceleration of centre of mass is
`a=alphar=(10F)/(7m)`
