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A 1321.9 kg car rounds a circular turn of radius 21.9 m. If the road is flat and the coefficient of static friction between the tires and the road is 0.56, how fast can the car go without skidding?

Guest May 12, 2014

Best Answer 

 #1
avatar+26397 
+5

The centripetal force required to keep the car from skidding is given by

$$F=m\frac{v^2}{r}$$

where m is the mass of the car, v is it's speed and r is the curve radius.

The frictional force is

$$F=\mu mg$$

where μ is the coefficient of friction and g is gravitational acceleration.

So equating these two forces we have:

$$\frac{v^2}{r}=\mu g\\\\
v=\sqrt{\mu gr}$$

I'll let you put in the numbers.

Alan  May 12, 2014
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1+0 Answers

 #1
avatar+26397 
+5
Best Answer

The centripetal force required to keep the car from skidding is given by

$$F=m\frac{v^2}{r}$$

where m is the mass of the car, v is it's speed and r is the curve radius.

The frictional force is

$$F=\mu mg$$

where μ is the coefficient of friction and g is gravitational acceleration.

So equating these two forces we have:

$$\frac{v^2}{r}=\mu g\\\\
v=\sqrt{\mu gr}$$

I'll let you put in the numbers.

Alan  May 12, 2014

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