A disc is purely rolling down an inclined plane of length ‘l’ & angle θ. What is the value of friction acting on it? Let the mass of the disc be M & radius be R.

(a) (Mgsinθ)/3

(b) 0

(c) (4Mgsinθ)/3

(d) (2Mgsinθ)/3

I have been asked this question in homework.

This interesting question is from Rolling Motion topic in portion System of Particles and Rotational Motion of Physics – Class 11

Select the correct answer from above options

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The correct option is (a) (Mgsinθ)/3

To elaborate: Let the angular acceleration of the disc be ‘α’. And the linear acceleration along the incline be ‘a’. For pure rolling, a = Rα. Mgsinθ – f = Ma———-(1), where f is the friction.

fR = Iα————–(2), where I is the moment of inertia = MR^2/2.

fR = Ia/R, we substitute the value of a into the first equation,

Mgsinθ – f = (MfR^2)/( MR^2/2) = 2f

∴ Mgsinθ – f = 2f Or f = (Mgsinθ)/3.