Question #100563
To test the performance of its tires, a car
travels along a perfectly flat (no banking) circular track of radius 368 m. The car increases
its speed at uniform rate of
at ≡d |v|dt = 3.3 m/s2
until the tires start to skid.
If the tires start to skid when the car reaches
a speed of 31.5 m/s, what is the coefficient of
static friction between the tires and the road?
The acceleration of gravity is 9.8 m/s2
1
Expert's answer
2019-12-23T11:41:19-0500

The Newton's second law states


ma=Fnetma=F_{\rm net}

In this case


ma=Ffric=μN=μmgma=F_{\rm fric}=\mu N=\mu mg

The total acceleration


a=an2+aτ2=(v2R)2+aτ2a=\sqrt{a_n^2+a_{\tau}^2}=\sqrt{\left(\frac{v^2}{R}\right)^2+a_{\tau}^2}

Hence


a=μga=\mu g

Finally, the coefficient of static friction


μ=(v2R)2+aτ2g\mu=\frac{\sqrt{\left(\frac{v^2}{R}\right)^2+a_{\tau}^2}}{g}

=(31.52368)2+3.329.8=0.435=\frac{\sqrt{\left(\frac{31.5^2}{368}\right)^2+3.3^2}}{9.8}=0.435


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Comments

Alex
17.01.23, 04:51

THANK YOU

nate
29.03.21, 04:07

thank u!

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