A car moves on a horizontal circular road of radius R while its speed increases at constant tangential acceleration a=dv/dt. If the tyre-road friction coefficient is µ, find the speed at which it skids.
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Question

A car moves on a horizontal circular road of radius R while its speed increases at constant tangential acceleration a=dv/dt. If the tyre-road friction coefficient is µ, find the speed at which it skids.

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G1 Physics · Accepted Answer

At skidding, friction acceleration µg is the vector resultant of tangential acceleration a and centripetal acceleration v²/R: µ²g²=a²+v⁴/R². Thus v=[(µ²g²-a²)R²]^(1/4).