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Question

A car moves on a circular track of radius r in a horizontal plane. The speed of the car is increasing constantly at a rate of dvdt=a. The coefficient of static friction between the road and the tyre is μ. Find the speed at which the car will skid.

A
[(μ2g2+a2)r2]1/4
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B
μgr
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C
[(μ2g2a2)r2]1/4
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D
ar
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Solution

The correct option is C [(μ2g2a2)r2]1/4
Since the speed of car is increasing continously, it has tangential acceleration given by:
at=a=dvdt
At any time, the centripetal acceleration of car will be directed towards the centre,
ac=v2r
Net acceleration is given by:



aT=a2+(v2r)2 ...(i)
For car to just skid, considering the just slipping condition where friction will act at its maximum value to provide the net force.
f=maT ...(ii)
or f=μN=μmg ...(iii)
(From the equilibrium of car in vertical direction, N=mg)

From Eq (i),(ii) and (iii)
m [a2+(v2r)2]=μmg
Squaring both sides,
m2[a2+(v2r)2]=μ2m2g2
or (v2r)2=μ2g2a2
v=[(μ2g2a2)r2]1/4

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