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Old Oct 9th 2017, 10:13 PM   #1
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Unbanked Circular Turn

A car is safely negotiating an unbanked circular turn at a speed of 26 m/s. The road is dry, and the maximum static frictional force acts on the tires. Suddenly a long wet patch in the road decreases the maximum static frictional force to one-ninth of its dry-road value. If the car is to continue safely around the curve, to what speed must the driver slow the car?

I'm have no idea how to solve this problem.
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Old Oct 10th 2017, 12:34 AM   #2
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Originally Posted by leh3188 View Post
A car is safely negotiating an unbanked circular turn at a speed of 26 m/s. The road is dry, and the maximum static frictional force acts on the tires. Suddenly a long wet patch in the road decreases the maximum static frictional force to one-ninth of its dry-road value. If the car is to continue safely around the curve, to what speed must the driver slow the car?

I'm have no idea how to solve this problem.
The sideways force that friction must resist comes from the centrifugal acceleration due to the turn.

$f_c = m \dfrac{v^2}{R}$

the frictional force is given by

$f_f = \mu m g$ where $\mu$ is the static friction coefficient

if the car is not to skid

$f_c \leq f_f$ so

$m \dfrac{v^2}{R} \leq \mu m g$

$v \leq \sqrt{\mu g R}$

thus $v$ varies as the square root of $\mu$

so if $\mu$ is reduced by 9, $v$ must be reduced by $\sqrt{9}=3$

and thus the speed must be reduced to $\dfrac{26}{3}~m/s$
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Old Oct 10th 2017, 12:41 AM   #3
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Originally Posted by leh3188 View Post
A car is safely negotiating an unbanked circular turn at a speed of 26 m/s. The road is dry, and the maximum static frictional force acts on the tires. Suddenly a long wet patch in the road decreases the maximum static frictional force to one-ninth of its dry-road value. If the car is to continue safely around the curve, to what speed must the driver slow the car?

I'm have no idea how to solve this problem.
when negotiating a curve , the sideways push on the car is ... m v2/r ( that's m times v squared ,divided by r) ....

so this sideways push is proportional to velocity squared ..

If friction is suddenly reduced to one ninth , the velocity must be reduced to one third to stay on the road ...
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