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3 years ago

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A car enters a level, unbanked semi-circular hairpin turn of 100 m radius at a speed of 28 m/s. The coefficient of friction betw

een the tires and the road is μ = 0.800. If the car maintains a constant speed of 28 m/s, it will A. first veer toward the center for the first quarter-circle, then veer toward the outside for the second quarter circle. B. tend to veer toward the outside of the circle.

1 answer:

meriva3 years ago

8 0

Answer:

As 28m/s = 28m/s

Explanation:

r = the radius of the curve

m = the mass of the car

μ = the coefficient of kinetic friction

N = normal reaction

When rounding the curve, the centripetal acceleration is

$a = \frac{v^{2}}{r}$

therefore

$\mu mg = m \frac{v^{2}}{r} \\\\ \mu = \frac{v^{2}}{rg}$

$v = \sqrt{\mu rg}$

$\mu = \sqrt{0.8 \times 100\times9.8} \\\\= 28m/s$

As 28m/s = 28m/s

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Part b)

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$\frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{r})^2 = 12.7$

$\frac{1}{2}(1.815) v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2 = 12.7$

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