- #1
highc
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I'm a little unsure with my some of my answers here, mainly because it seems too simple. The Questions ask:
A race car driver is driving her car at the record breaking speed of 225 km/h. The first turn on the course is banked at 15 degrees, and the car's mass is 1450 kg.
a) Calculate the radius of curvature for this turn.
v = 225 km/h = 62.5 m/s
m = 1450 kg
theta = 15 degrees
Since, v^2 = grtan theta
Therefore, r = v^2/g tan theta
= (62.5)^2/(9.8)(tan15)
= 1487.6 m
b) Calculate the centripetal acceleration of the car.
a(c) = v^2/r
= (62.5)^2/1487.6
= 2.6 m/s^2
*Thus far, I feel pretty good about my solutions but here's where my confidence is quickly stripped from me.
c) If the car maintains a circular track around the curve (does not move up or down the bank), what is the magnitude of the force of static friction?
For y:
F(y) = F(n) - F(gy) = 0
F(n) - F(g)cos15 = 0
F(n) = mgcos15
F(n) = 13725.8 N = 1.4*10^4 N
For x:
F(x) = F(f) - F(gx) = 0
F(f) - F(g)sin15 = 0
F(f) = mgsin15
F(f) = 3677.8 N = 3.7*10^3 N
Therefore, the magnitude of the force of static friction is 3.7*10^3.
d) What is the coefficient of static friction necessary to ensure the safety of this turn?
u(s) = F(s)/F(n)
= 3.7*10^3/1.4*10^4
= 0.26
Sorry about the size of the question, hoping someone can catch any errors and help to guide me to the correct solution before I send this in. I'm asking alot, any input is much appreciated.
A race car driver is driving her car at the record breaking speed of 225 km/h. The first turn on the course is banked at 15 degrees, and the car's mass is 1450 kg.
a) Calculate the radius of curvature for this turn.
v = 225 km/h = 62.5 m/s
m = 1450 kg
theta = 15 degrees
Since, v^2 = grtan theta
Therefore, r = v^2/g tan theta
= (62.5)^2/(9.8)(tan15)
= 1487.6 m
b) Calculate the centripetal acceleration of the car.
a(c) = v^2/r
= (62.5)^2/1487.6
= 2.6 m/s^2
*Thus far, I feel pretty good about my solutions but here's where my confidence is quickly stripped from me.
c) If the car maintains a circular track around the curve (does not move up or down the bank), what is the magnitude of the force of static friction?
For y:
F(y) = F(n) - F(gy) = 0
F(n) - F(g)cos15 = 0
F(n) = mgcos15
F(n) = 13725.8 N = 1.4*10^4 N
For x:
F(x) = F(f) - F(gx) = 0
F(f) - F(g)sin15 = 0
F(f) = mgsin15
F(f) = 3677.8 N = 3.7*10^3 N
Therefore, the magnitude of the force of static friction is 3.7*10^3.
d) What is the coefficient of static friction necessary to ensure the safety of this turn?
u(s) = F(s)/F(n)
= 3.7*10^3/1.4*10^4
= 0.26
Sorry about the size of the question, hoping someone can catch any errors and help to guide me to the correct solution before I send this in. I'm asking alot, any input is much appreciated.
