A uniform solid cylindrical roller of mass '`m`' is being pulled on a horizontal surface with force `F` parallel to the surface and applied at its centre. If the acceleration of the cylinder is '`a`' and it is rolling without slipping, then the value of '`F`' is `:-`

Options

A. 2 ma
B. 3 ma
C. `(3 ma)/(2)`
D. `(5 ma)/(2)`

Answer

A

Explanation

By Newton’s law of motion
F-`f_(s) ` =ma

and torque
where `alpha` = tangential component of acceleration
I = moment of inertia
`implies fR = l alpha `
`f R = (MR^(2))/(2) .(a)/( R) [ :. L = (MR^(2))/(2) "and" alpha = (a)/(R ) ]`
`implies f = (ma)/(2)`
`:. F = f + ma `
`= (ma)/(3) + ma = (3ma)/(2)`