The following question appeared in the Physics paper of A.I.E.E.E.-2005:

**The upper half of an inclined plane with inclination Ф is perfectly smooth while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom if the coefficient of friction for lower half is**

(a) tanФ (b) 2tanФ (c) 2cosФ (d) 2sinФ

(a) tanФ (b) 2tanФ (c) 2cosФ (d) 2sinФ

You have to use the equation of linear motion, v

0 = 2gsinФ×s + 2(gsinФ - μgcosФ)×s, from which 2sinФ = μcosФ

Therefore, μ = 2tanФ [Option (b)]

Consider the following MCQ:

^{2}= u^{2}+ 2as for both parts of the motion. For the smooth half of the incline, we have v^{2}= 0 + 2gsinФ×s and for the rough half we have 0 = v^{2}+2(gsinФ - μgcosФ)×s. Combining these two equations, we get0 = 2gsinФ×s + 2(gsinФ - μgcosФ)×s, from which 2sinФ = μcosФ

Therefore, μ = 2tanФ [Option (b)]

Consider the following MCQ:

**A car of mass 1000kg moves on a circular track of radius 20m. If the coefficient of friction is 0.64, then the maximum velocity with which the car can move is**

(a) 15m/s (b) 11.2m/s (c) 20m/s (d) 18m/s (e) 22.4m/s

This simple question appeared in the Kerala Engineering Entrance Test paper of 2006. As the frictional force supplies the centripetal force required for the circular motion, we have μmg = mv

(a) 15m/s (b) 11.2m/s (c) 20m/s (d) 18m/s (e) 22.4m/s

^{2}/r so that v = √(μrg) = √(0.64×20×9.8) = 11.2m/s

Let us discuss two more questions involving friction:

Suppose you are not asked to find the maximum value of ω in the above problem. Instead, you are asked to find the frictional force at the angular velocity ω, as in the modified question below:

. The correct option therefore is (d).

**A coin of mass ‘m’ is placed on a rough horizontal turn table with coefficient of limiting friction ‘μ’. The distance of the coin from the axis of the turn table is ‘r’ and the turn table is turning with angular velocity ‘ω’. The maximum value of ω so as to keep the coin intact on the turn table is**

(a) √(μg/r) (b) √(μgr) (c) μgr (d) √(μr/g) (e) μgr

It is the frictional force that prevents the coin from sliding outwards under the action of the centrifugal force. Therefore we equate the limiting value of the frictional force to the centrifugal force (magnitudes): μmg = mrω(a) √(μg/r) (b) √(μgr) (c) μgr (d) √(μr/g) (e) μgr

^{2}^{2}from which ω = √(μg/r). The correct option is therefore (a).Suppose you are not asked to find the maximum value of ω in the above problem. Instead, you are asked to find the frictional force at the angular velocity ω, as in the modified question below:

**A coin of mass ‘m’ is placed on a rough horizontal turn table with coefficient of limiting friction ‘μ’. The distance of the coin from the axis of the turn table is ‘r’ and the turn table is turning with angular velocity ‘ω’. If the coin remains intact on the turn table, the frictional force between the coin and the surface of the turn table is**

(a) μmg (b) zero (c) mg (d) mrω

Nothing is mentioned about the limiting equilibrium of the coin and therefore, the self adjusting frictional force just balances the centrifugal force mrω(a) μmg (b) zero (c) mg (d) mrω

^{2}(e) mrω^{2}. The correct option therefore is (d).

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