Maximal speed for taking a turn

Classical Mechanics Level 3

A car is taking a circular turn on a horizontal road. Which formula gives the maximum speed, $v_\text{max}$ , in terms of the maximum static coefficient of friction, $\mu_{s}$ ,and the acceleration of gravity , $g$ , so the car is able to take the turn?

Assume uniform circular motion.

v_\text{max} = \sqrt{ \frac{ \mu_{s} r}{ g}}

v_\text{max} = \sqrt{\mu_{s} r g}

v_\text{max} = \sqrt{ \frac{ \mu_{s} g}{ r}}

v_\text{max} = \sqrt{ \frac{1}{ \mu_{s} g r}}

1 solution

Tom Van Lier
Jul 21, 2016

Consider the following free body diagram of the car during the turn (courtesy of the physics classroom ) :

It means that the centripetal force, necessary to take the turn, is the static force of friction. Therefor, applying Newton's second law in this direction, we can find that $F_{f} = m a = \dfrac{mv^2}{r}$ . (1)

Now since $F_{f} = \mu_{s} F_{n}$ , and clearly in the vertical direction $F_{n} = F_{g} = m g$ , we find $F_{f} = \mu_{s} m g$ . (2)

Combining (1) and (2) :

$\dfrac{v^2}{r} = \mu_{s} g$ , or $v^2 = \mu_{s} r g$ and $v = \sqrt{\mu_{s} r g}$ .

For the maximal speed, $\mu_{s}$ has to become maximal, which leads to the correct result.

Very well! But a simple check of the units shows that only choice 2 can be regarded.

Andreas Wendler - 4 years, 10 months ago

True, but how many students have you see do that ?

Tom Van Lier - 4 years, 10 months ago

Don't observe students! But I think this procedure is obvious for an older one. ;-).

Andreas Wendler - 4 years, 10 months ago

i first did by dimensional analyses and then derived it also....

Sargam yadav - 4 years, 10 months ago

Maximal speed for taking a turn

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