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 rad //s ^(2)`
B. `16 rad //s ^(2)`
C. `10 rad //s ^(2)`
D. `20 rad //s ^(2)`

Answer

B

Explanation

If angular acceleration 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)`
From (ii) and (iii)
`40 + F = I _(cm) (a)/(R ^(2)) " "(iv) `
Adding (i) and (iv)
`80= ((I _(cm))/(R ^(2))+m )a `
`a = ((80)/(m+ (m R ^(2))/(R ^(2)))) " " therefore alpha = (80)/(2mR) or alpha = (80)/( 2xx5xx0.5)`
`alpha = 16 rad//s ^(2).`