A cylinder of mass 10 kg and radius 15 cm is rolling perfectly on a plane of inclination 30°. The coefficient of static friction µ_s = 0.25. 

(a) How much is the force of friction acting on the cylinder? 

(b) What is the work done against friction during rolling? 

(c) If the inclination θ of the plane is increased, at what value of θ does the cylinder begin to skid, and not roll perfectly?

Mass of the cylinder, m = 10 kg 
Radius of the cylinder, r = 15 cm = 0.15 m 
Co-efficient of kinetic friction, µ_k = 0.25 
Angle of inclination, θ = 30° 
Moment of inertia of a solid cylinder about its geometric axis, I = mr²
The various forces acting on the cylinder are shown in the following figure, 


    a = mg Sinθ / [m + (I/r²) ] 
      = mg Sinθ / [m + {(1/2)mr²/ r²} ] 
      =  g Sin 30° 
      = × 9.8 × 0.5 
      =  3.27 ms^-2
(a) Using Newton’s second law of motion,
  Net force, f_net = ma 
 
 mg Sin 30° - f = ma 

                     f = mg Sin 30° - ma 

                       = 10 × 9.8 × 0.5 - 10 × 3.27 
       49 - 32.7 = 16.3 N
(b) During rolling, the instantaneous point of contact with the plane comes to rest.
Hence, the work done against frictional force is zero.
(c) For rolling without skid,
Using the relation, 
                    μ = (1/3) tan θ 
              tan θ = 3μ
                       = 3 × 0.25 
∴ θ = tan^-1 (0.75)
      = 36.87°.