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Units And Measurement

Question
CBSEENPH11017775

A cylinder of mass m, radius R and moment of inertia I is placed on a rough inclined plane. What should be the minimum coefficient of friction between cylinder and inclined plane so that cylinder rolls down the inclined plane without slipping?

Solution

Consider a cylinder placed on the inclined plane inclined at angle θ the with the horizontal.
The various forces acting on the cylinder are:

(i) Weight mg acting vertically downward.

(ii) The friction F between the cylinder and surface of inclined plane.
                 
On resolving the components of weight mg, along and perpendicular to the inclined plane we have mg sinθand mg cosθ respectively.
Let a be the acceleration with which cylinder rolls down the inclined plane.

Equation of linear motion down the inclined plane is,
ma space equals space mg space sinθ space minus space straight F space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis
Since the cylinder rolls due to the force of friction,
therefore space space space space straight I space straight alpha equals FR space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis
where alpha is angular acceleration of cylinder. 
We know,
straight alpha space equals space straight alpha over straight R space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 3 right parenthesis space

therefore space space space space straight I straight alpha over straight R equals FR space

rightwards double arrow space space space space space space space space straight F equals Iα over straight R squared space space space space space space space space space space space space space space space space space space space space... left parenthesis 4 right parenthesis 
Substituting the value of F in equation (1), we get

space space space space space space space space space space space space ma space equals space mg space sin space straight theta space minus Iα over straight R squared space

rightwards double arrow space space space space space space space space space space space straight a space equals gsinθ minus Iα over mR squared space

rightwards double arrow space space space space space space space space space space space space straight alpha equals fraction numerator straight g space sin space straight theta over denominator 1 plus begin display style straight I over mR squared end style end fraction 
Substituting a  in (4),  we get,
straight F space equals 1 over straight R squared fraction numerator straight g space sin space straight theta over denominator 1 plus begin display style straight I over mR squared end style end fraction space

space space space equals fraction numerator mg space sin space straight theta over denominator 1 plus begin display style mR squared over straight I end style end fraction
There will be no slipping if and only if
Coefficient space of space friction comma space space straight mu greater or equal than straight F over straight R
Therefore minimum coefficient of friction straight mu is, 

space space space space space space space space straight mu space equals straight F over straight R equals fraction numerator mg space sinθ over denominator 1 plus begin display style mR squared over straight I end style end fraction cross times fraction numerator 1 over denominator mg space cosθ end fraction

space space space space space space space space space space space equals fraction numerator tan space straight theta over denominator begin display style mR squared over straight I plus 1 end style end fraction space
space
rightwards double arrow space space space straight mu space equals space fraction numerator tan space straight theta over denominator begin display style mR squared over straight I plus 1 end style end fraction