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Systems of Particles and Rotational Motion

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Question : 31 of 33
Marks: +1, -0
A cylinder of mass 10 kg and radius 15 cm is rolling perfectly on a plane of inclination 30°. The coefficient of static friction μs=0.25\mu_s = 0.25.
(a) How much is the force of friction acting on the cylinder?
(b) What is the work done against frictionduring 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?
Solution:  
Mass of cylinder M=10 kgM = 10\,\text{kg}, Radius of cylinder, R=0.15 mR = 0.15\,\text{m}Angle of inclination, θ=30∘\theta = 30^{\circ}, Coefficient of static friction, μs=0.25\mu_s = 0.25
(a) Force of friction on the cylinder is given by
F=13mgsin⁡θF = \frac{1}{3} m g \sin \theta
=13×10×9.8×sin⁡30∘= \frac{1}{3} \times 10 \times 9.8 \times \sin 30^{\circ}
=13×10×9.8×12=16.3 N= \frac{1}{3} \times 10 \times 9.8 \times \frac{1}{2} = 16.3\,\text{N}
(b) During rolling the point of contact is at rest. Therefore, work done against friction is zero.
(c) The cylinder skids (does not roll) when
μs=13tan⁡θ\mu_{s} = \frac{1}{3} \tan \theta
tan⁡θ=3μs\tan \theta = 3 \mu_{s} =3×0.25=0.75= 3 \times 0.25 = 0.75
⇒θ=36.87∘≈37∘\Rightarrow \theta = 36.87^{\circ} \approx 37^{\circ}
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