A car goes on a horizontal circular road of radius R. The speed increases at a constant rate a=dvdt. The friction coefficient between the tyre and road is μ. What is the speed at which the car will skid?
A
(μg−a)R2
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B
[(μ2g2−a2)R2]14
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C
[(μ2g2−a2)R]14
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D
(μg−a)R4
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Solution
The correct option is A[(μ2g2−a2)R2]14 For uniform circular motion, the resultant acceleration is aR=√a2N+a2t aN=v2R,at=dvdt=a From FBD of car on the rough road f=maR μN=m√(v2R)2+(dvdt)2 ⇒μ2N2=m2(v4R2+a2) ⇒v4=μ2N2R2m2−a2R2 =μ2m2g2R2m2−a2R2 =μ2g2R2−a2R2 v=[(μ2g2−a2)R2]14.