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, the cylinder is rolling without slipping on a horizontal surface (see figure), then the angular acceleration of the cylinder will be (Neglect the mass and thickness of the string).

Options

A. `12 md //s^(2)`
B. `16 md//s^(2)`
C. `10 md//s^(2)`
D. `20 md//s^(2)`

Answer

B

Explanation

If angular accleration of the cylinder is `alpha` and acceleration of C.O.M. is a, then
`40 -F=ma`……….(i)
`(40 +F)R = I_(cm)alpha`……..(ii)
`a = R alpha`….(iii)

`40 + F = I_(CM) a/R^(2)`........(iv)
Adding (i) and (iv)
`80 = (I_(CM)/R^(2) + m)a, a=80/(m+(mR^(2))/R^(2)) therefore alpha = 80/(2mR)` or `alpha = 80/(2 xx 5 xx 0.5)`,