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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,
$$\mu=\dfrac{1}{3}tan\theta$$
$$\implies tan\theta=3\mu=0.75$$
$$\implies\theta=36.87^{\circ}$$

477002_458337_ans_61ccdfa33e154f6fa0902fbd25fde861.png

Physics
NCERT
Standard XI

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