Question

9. A NASCAR racecar rounds one end of the Martinsville Speedway. This end of the track is a turn with radius approximately 57.0 m. If we approximate the track to be completely flat and the racecar is traveling at a constant 26.5 m / s (about 59 mph) around the turn, what is the racecar\'s centripetal (radial) acceleration?

? m/s^2

What is the force responsible for the centripetal acceleration in this case?

a) weight b) normal c) gravity d) friction

To keep from skidding into the wall on the outside of the turn what is the minimum coefficient of static friction between the racecar\'s tires and track?

?

10. A jet airplane is in level flight. The mass of the airplane is m = 8990 kg. The plane travels at a constant speed around a circular path of radius R = 8.01 miles and makes one revolution every T = 0.106 hours. What is the magnitude of the lift force acting on the plane?

L = ? kN

At what angle is the airplane banked?

θ = ? degree

11. A 2030-kg space station orbits Earth at an altitude of 519000 m. Find the magnitude of the force with which the space station attracts Earth. The mass and mean radius of Earth are 5.98 × 1024 kg and 6370000 m, respectively.

? N

Answer 1

a = v²/R = (26.5m/s)² / 57.0m = 12.3 m/s² ◄

µ = a / g, so to keep from skidding, the minimum
µ = 12.3 / 9.8 = 1.26 ◄

friction force is responsible for centripetal acceleration

10)

The vertical component of the lift is equal to the weight:
F*cosΘ = mg
The horizontal component is equal to the centripetal force:
F*sinΘ = mv²/r

Divide horizontal by vertical:
FsinΘ / FcosΘ = mv²/r / mg
tanΘ = v² / gr
8.01 mi = 12890.85m
v = 2πr/t = 2π*12890.85m / 0.106*3600s = 212.25 m/s

Then
tanΘ = (212.25m/s)² / 9.8m/s²*12890.85m = 0.3566
Θ = 19.62º ◄ bank angle

F*cos19.62º = 8890kg * 9.8m/s²
F = 92492 N ◄ lift force

11)

  We can take a shortcut with this by knowing
that GM/r^2 = 9.8 m/s^2, then adjust for the different r:
9.8 m/s^2 times (6370^2/6889^2) = 8.38 m/s^2
F = ma = 2030 kg times 8.38 m/s^2 = 17.011 kN