**A coin is placed on a horizontal turn table, at a distance‘r’ from the axis of rotation of the turn table. Starting from rest, the turn table rotates with a constant angular acceleration ‘****α’. If the coefficient of friction between the coin and the turn table is ‘μ’, the time after which the coin will begin to slip is **

**(a) ****√(μ/α) ****(b) ****√(μg/α ^{2}) **

**(c)**

**√(μg/αr)**

**(d)**

**√(μgα**

^{2}/r)**(e)**

**√(μg/α**

^{2}r)

The coin will start slipping when the centrifugal force begins to exceed the frictional force. In the limiting case therefore, mrω^{2} = μmg where ‘m’ is the mass of the coin and ‘ω’ is the angular velocity of the turn table when the coin begins to slip.

But, ω = ω_{0} + αt = αt since the initial angular velocity (ω_{0}) is zero.

[Note that the above equation is similar to the equation, v = v_{0} + at in the case of linear motion].

Substituting this value of ω in the condition for slipping, we have

mr(αt)^{2} = μmg from which t = √(μg/α^{2}r).

Now, let us modify the above question as follows:

**In a region of space where gravitational force is negligible, a coin is kept in contact with a rod AB as shown, at a distance ‘r’ from the end A. If the rod starts from rest and rotates with a constant angular acceleration ‘****α’ about an axis XX’ perpendicular to the rod and passing through the end A, the time after which the coin will begin to slip is**

**(a) √(****μα)** **(b) √(****μα/r)** **(c) √(****μ/αr)** **(d) √(****μ/α)** **(e) negligibly small **

This problem is very much different since gravity is negligible. In the previous problem, the normal force that produced friction was the result of the weight of the coin where as here it is produced because of the inertial force produced by the acceleration of the rod. When the rod rotates with an angular acceleration α, the coin presses against the rod with an inertial force equal to ma where ‘m’ is the mass and ‘a’ is the tangential acceleration of the mass.

As in the previous problem, the condition for slipping to start is obtained by equating the centrifugal force to the frictional force:

mrω^{2} = μma

Therefore, ω = √(μa/r).

But, ω = ω_{0} + αt = αt (since the initial angular velocity ω_{0} is zero); a = αr.

Substituting these we obtain αt = √(μα) so that t =√(μ/α) .

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