Rotational Mechanics

it is in the direction of the applied force....

FROM ANOTHER VIEW POINT :

FOR THIS F WHICH PRODUCES A TORQUE, ω becomes greater than v . To REDUCE THIS ω an anti-torque needs to be given which is given by the friction in the forward direction.
This direction of friction also increases v as it's in the same direction.

This is the condition for pure rolling.

just asume that the friction is in forward direction...and the make equation
F + f = Ma..where f is frictional force..and (F - f)R = MR²Î±
so from here f comes to be zero..but this is only valid in case of ring..in other cases frictional force wont be 0

ya to finish off....with hep from govind

F + f = Ma........(1)

clockwise torque due to Force F..... = FR

anti-clockwise torque due to friction.. = fR

so net clock wise torque =FR - fR = (F - f)R

net torque = I @ (@ = alpha)

( F - f ) R = M R² @...(2)

solving 1 & 2....

Ma - 2f = M R @

for pure rolling....

a = R @

so f = 0

AN ADDITION TO SANDY'S STATEMENT :
BY f SANDY MEANS KINETIC FRICTION.........

ROLLING FRICTION IS NOT ZERO.....WITHOUT FRICTION HOW CAN A TORQUE BE GENERATED TO KEEP IT ROLLING YAAR ???

ya...

so ....

f_kinetic = 0....(ans)

N.B. f_rolling ≠0

@aveek now thats a complete answer... rite??