__Question 7.31__:*µ*

_{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

*θ*

_{ }of the plane is increased, at what value of

*θ*does the cylinder begin to skid, and not roll perfectly?

__Solution__:*m*= 10 kg

Radius of the cylinder,

*r*= 15 cm = 0.15 m

Co-efficient of kinetic friction,

*µ*

_{k }= 0.25

Angle of inclination,

*θ*= 30°

Moment of inertia of a solid cylinder about its geometric axis, I = (1/2)mr

^{2}

The various forces acting on the cylinder are shown in the following figure:

a = mg Sinθ / [m + (I/r

^{2}) ]

= mg Sinθ / [m + {(1/2)mr

^{2}/ r

^{2}} ]

= (2/3) g Sin 30

^{0}

= (2/3) X 9.8 X 0.5 = 3.27 ms

^{-2}

^{}

**(a)**Using Newton’s second law of motion, we can write net force as:

*f*

_{net}=

*ma*

mg Sin 30

^{0}- f = ma

f = mg Sin 30

^{0}- ma

= 10 X 9.8 X 0.5 - 10 X 3.27

49 - 32.7 = 16.3 N

**(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 skid, we have the relation:

μ = (1/3) tan θ

tan θ = 3μ = 3 X 0.25

∴ θ = tan

^{-1}(0.75) = 36.87

^{0}

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