  Question

A cylinder of mass 10 kg and radius 15 cm is rolling perfectly on a plane of inclination $$30^o$$. The coefficient of static friction $$\mu_s = 0.25$$. (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 $$\theta$$ of the plane is increased, at what value of $$\theta$$ does the cylinder begin to skid, and not roll perfectly ?

Solution

Moment of inertia of the cylinder about its geometric axis=$$\dfrac{1}{2}mr^2$$Acceleration of the cylinder is given as $$a=\dfrac{mgsin\theta}{m+\dfrac{I}{r^2}}=\dfrac{2}{3}gsin30^{\circ}=3.27m/s^2$$(a) Using Newton's Second Law, we can write, $$f_{net}=ma$$$$\implies mgsin30^{\circ}-f=ma$$$$\implies f=49N-32.7N=16.3N$$(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 skidding, we have the relation,$$\mu=\dfrac{1}{3}tan\theta$$$$\implies tan\theta=3\mu=0.75$$$$\implies\theta=36.87^{\circ}$$ PhysicsNCERTStandard XI

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