A thin uniform spherical shell of mass `m = 2 kg` and radius `R = 10 cm` is placed on a rough horizontal surface with a coefficient of friction `mu = 0.4`. A horizontal force 'F' is applied to the spherical shell at a height `R//3` ablove the centre as shown. The maximum magnitude of the applied force 'F' for which the spherical shell will be rolling without slipping is (taking `g = 10 m//a^(2)` )

Options

A. `20 N`
B. `30 N`
C. `40 N`
D. `60 N`

Answer

C

Explanation

`F - f_(s) = ma` …..(i)
`(FR)/(3)+ f_(s)R = (2)/(3) mR^(2)alpha`
`(F)/(3) + f_(s) = (2)/(3)ma` [since, `a = Ralpha`] …..(i)
Solving (i) and (ii), we get
`(4F)/(3) = (5)/(3)ma`
`a = (4F)/(5m)`
Now, from equation (i)
`f_(s) = F - ma = F - (4F)/(5) = (F)/(5)`
Hence, `f_(s) le muN`
`F//5 le muN`
`F le 5 mumg`
`therefore F_(max) = 5 xx 0.4 xx 20 = 40 N`