Radial & Tangential Acceleration for Non Uniform Circular Motion
A car moves o...
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
[(μ2g2−a2)r2]1/4
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D
√ar
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Solution
The correct option is C[(μ2g2−a2)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=μ2g2−a2 ∴v=[(μ2g2−a2)r2]1/4