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

Step-by-step solution

By Newton's law of motion F - f s = m a ...(i) and torque, τ = l α where, α = tangential component of acceleration I =  moment of inertia  ⇒    f R = l α f R = M R 2 2 · a R ∴ I = M R 2 2 and , α = a R ⇒    f = m a 2 ∴     F = f + m a = m a 2 + m a = 3 m a 2