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Question: A car is moving on a circular track of radius R. The road is banked at θ. μis the coefficient of fri...

A car is moving on a circular track of radius R. The road is banked at θ. μis the coefficient of friction. Find the maximum speed the car can have.

A

[Rg(sin⁡θ+μcos⁡θ)cos⁡θ+μsin⁡θ]1/2\left\lbrack \frac{Rg\left( \sin\theta + \mu\cos\theta \right)}{\cos\theta + \mu\sin\theta} \right\rbrack^{1/2}

B

[Rg(cos⁡θ+μsin⁡θ)cos⁡θ−μsin⁡θ]1/2\left\lbrack \frac{Rg\left( \cos\theta + \mu\sin\theta \right)}{\cos\theta - \mu\sin\theta} \right\rbrack^{1/2}

C

[Rg(sin⁡θ+μcos⁡θ)cos⁡θ−μsin⁡θ]1/2\left\lbrack \frac{Rg\left( \sin\theta + \mu\cos\theta \right)}{\cos\theta - \mu\sin\theta} \right\rbrack^{1/2}

D

None

Answer

[Rg(sin⁡θ+μcos⁡θ)cos⁡θ−μsin⁡θ]1/2\left\lbrack \frac{Rg\left( \sin\theta + \mu\cos\theta \right)}{\cos\theta - \mu\sin\theta} \right\rbrack^{1/2}

Explanation

Solution

vmax = [Rg(tan⁡θ+μ)1−μtan⁡θ]1/2\left\lbrack \frac{Rg\left( \tan\theta + \mu \right)}{1 - \mu\tan\theta} \right\rbrack^{1/2} = [Rg(sin⁡θ+μcos⁡θ)cos⁡θ−μsin⁡θ]1/2\left\lbrack \frac{Rg\left( \sin\theta + \mu\cos\theta \right)}{\cos\theta - \mu\sin\theta} \right\rbrack^{1/2}