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Math Help - [SOLVED] Centripetal Force

  1. #1
    Senior Member topher0805's Avatar
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    [SOLVED] Centripetal Force

    A car navigates a turn in the road at 60 km/h without sliding. The road is inclined at an angle \theta. What does the coefficient of static friction need to be so that the car can take the turn at 90 km/h without skidding?
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  2. #2
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    Motion in a circle

    Hello topher0805
    Quote Originally Posted by topher0805 View Post
    A car navigates a turn in the road at 60 km/h without sliding. The road is inclined at an angle \theta. What does the coefficient of static friction need to be so that the car can take the turn at 90 km/h without skidding?
    When the car is travelling at 60 km/h (v = 50/3 m/sec), assuming there is no tendency to slide:

    R(horiz) N\sin\theta = \frac{mv^2}{r}= \frac{2500m}{9r}, where N is normal contact force with road and r = radius of bend

    R(vert) N\cos\theta = mg

    Eliminate N:

    \Rightarrow \tan\theta = \frac{2500}{9gr} (1)

    When speed = 90 km/h (v = 25 m/sec) and car is on point of sliding:

    R(horiz) \mu N\cos\theta + N\sin\theta = \frac{mv^2}{r} = \frac{625m}{r} (2)

    R(vert) N\cos\theta = mg + \mu N\sin\theta

    \Rightarrow N (\cos\theta -\mu\sin\theta) = mg (3)

    Eliminate N from (2) and (3):

    \frac{\mu\cos\theta + \sin\theta}{\cos\theta -\mu\sin\theta} = \frac{625}{gr}

    \Rightarrow \frac{\mu + \tan\theta}{1 - \mu\tan\theta} =\frac{625}{gr}

    = \frac{9}{4}\tan\theta, from (1)

    Now solve for \mu and you're done.

    Grandad
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