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Thread: [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 $\displaystyle \theta$. 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 $\displaystyle \theta$. 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) $\displaystyle N\sin\theta = \frac{mv^2}{r}= \frac{2500m}{9r}$, where $\displaystyle N$ is normal contact force with road and $\displaystyle r$ = radius of bend

    R(vert) $\displaystyle N\cos\theta = mg$

    Eliminate $\displaystyle N$:

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

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

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

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

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

    Eliminate $\displaystyle N$ from (2) and (3):

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

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

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

    Now solve for $\displaystyle \mu$ and you're done.

    Grandad
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