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Question

# A cylinder of mass 10 kg and radius 15 cm is rolling perfectly on a plane of inclination 30o. The coefficient of static friction μs = 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 q of the plane is increased, at what value of q does the cylinder begin to skid, and not roll perfectly ?

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Solution

## Given: The mass of the cylinder is 10 kg, radius of cylinder is 15 cm, the angle of inclination is 30° and the coefficient of static friction is 0.25. a) The free body diagram of the cylinder is shown as, Figure (1) Where, m is the mass of the cylinder, g is the gravitational acceleration, f is the friction force acting on the cylinder, mgsin30° is the weight component along the inclined plane and mgcos30° is the weight component along the direction perpendicular to the inclined plane. The angular acceleration of cylinder is given as, a=Rα α= a R The net torque acting on the cylinder is, τ net =f×R The net torque also given as, τ net =Iα Therefore, f×R=Iα f= Iα R By substitute the value of α, we get f= I( a R ) R = Ia R 2 …… (1) The equilibrium equation of force along the inclined plane is given as, F net =mgsin30°−f By substituting the value of F net and f in the above equation, we get ma=mgsin30°− Ia R 2 ma=mgsin30°− ( 1 2 m R 2 )a R 2 a= 2 3 gsin30° By substituting the value of a in the equation (1), we get f= ( 1 2 m R 2 )( 2 3 gsin30° ) R 2 = 1 3 mgsin30° …… (2) By substituting the values in the above equation, we get f= 1 3 ×10×9.81×sin30° =16.35 N ≈16.4 N Thus, the force of friction acting on the cylinder is 16.4 N. b) During pure rolling motion, the net velocity at the point of contact is zero. Therefore, work done against the friction is zero as the friction force acting at the point of contact is zero at each instant. c) The friction force acting on the cylinder is, f= μ s mgcosθ From the equation (2), the friction force for the given angle of inclination θ is, f= 1 3 mgsinθ By substituting the value of f in the above equation, we get μ s mgcosθ= 1 3 mgsinθ μ s cosθ= 1 3 sinθ tanθ=3 μ s θ= tan −1 ( 3 μ s ) By substituting the value in the above equation, we get θ= tan −1 ( 3×0.25 ) =36.87° ≈37° Thus, the angle of inclination at which the cylinder begin to skid is 37°.

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