In this example the friction force between a vehicle’s tires and the road provid

Question

In this example the friction force between a vehicle’s tires and the road provides the radial acceleration. A car rounds a flat, unbanked curve with radius R. (a) If the coefficient of static friction between tires and road is s, derive an expression for the maximum speed vmax at which the driver can take the curve without sliding. (b) What is the maximum speed if R=250m and s=0.90?

SOLUTION

SET UP (Figure 1) shows the situation, plus a free-body diagram for the car. We point the x axis toward the center. Note that the friction between a rolling wheel and the pavement is static friction because the portion of the wheel in contact with the pavement doesn’t slide relative to the pavement. (If the wheel slides or skids, however, the friction force is kinetic.) This friction force fs is responsible for the car’s centripetal acceleration arad=v2/R; it points toward the center of the circle, in the x direction. In the y direction, the car’s weight and the normal force on the car sum to zero.

SOLVE Part (a): The component equations from Newton’s second law are

fs=mv2R,n+(mg)=0

In the first equation, fs is the force needed to keep the car moving in its circular path at a given speed v; this force increases with v. But the maximum friction force available is fs=sn=smg; this limitation determines the car’s maximum speed vmax. When we equate the maximum friction force available to the force mv2/R required for the circular motion, we find

smg=mvmax2R and vmax=sgR

Part (b): If s=0.90 and R=250m, then

vmax=(0.90)(9.8m/s2)(250m)=47m/s

or about 105 mi/h.

REFLECT If the car travels around the curve at less than vmax, fs is less than sn, but it is still true that fs=mv2/R. If the car is moving faster than vmax, not enough friction is available to provide the needed centripetal acceleration. Then the car skids. Because the kinetic friction acting on skidding tires is less than the static friction acting before the car skids, a car always skids toward the outside of the curve.

It’s possible to bank a curve so that at one certain speed no friction at all is needed. Then a car moving at just the right speed can round the curve even on wet ice with Teflon® tires. Airplanes always bank at this angle when they make turns, and bobsled racing depends on the same idea.

QUESTION:

Part A - Practice Problem:

What is the friction force on a 1100 kg car if it rounds a curve with radius 100 m at a speed of 15 m/s ?

Part B - Practice Problem:

What minimum coefficient of friction is needed?

Explanation / Answer

Centripetal acceleration will be provided by friction only,So

Fs = mv^2 / R = 1100× 15×15/100= 2475N

Hence friction force is 2475 N

for minimum coefficient of friction,

us mg =Fs

us × 1100 × 9.8 = 2475

us = 2475/( 1100×9.8)

= 0.23

minimum coefficient of friction is 0.23

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