car on a banked currve Consider a car going around a 25 degree banked curve at t
ID: 2256283 • Letter: C
Question
car on a banked currve
Consider a car going around a 25 degree banked curve at the highest possible speed of 54 m/s without sliding where the static coefficient of friction during dry conditions of the tire on the road is mu =0.75 and during icy conditions the static coefficient of friction is mu =0.00, all in a configuration shown below: What is the radius of curvature of the turn at the given angle of theta =25 degree during dry conditions when the car is at the maximum speed? If the angle goes down to zero, theta =0 degree , then during dry conditions the frictional force keeps the car on the turn, what is the radius of the turn for this case of maximum speed? When the road is at theta =25 degree and is covered with ice the coefficient of friction will go to zero, mu =0, what is the radius of the turn in this case for the maximum speed? If mass of the car is doubled by what factor does the radius of the turn change? Calculate the acceleration experienced by the driver for the conditions given by part (a) above if the driver has a mass of 65 kg and is held rigidly in place.Explanation / Answer
v = 54 m/s
a)
the maximum speed through which the car can take the curve is
V = sqrt[ g*R*(sin(theta) + mue_s*cos(theta))/( cos(theta) - mue_s*sin(theta)) ]
v^2 = [9.8*R*(sin(25) + 0.75*cos(25))/(cos(25) - 0.75*sin(25)) ]
54^2 = 18.33*R
==> R = 54^2/18.33
= 159 m
b) when theta = 0
v = sqrt(mue_s*g*R)
==> R = v^2/(mue_s*g)
= 54^2/(0.75*9.8)
= 396.74 m
c) when, theta = 25 degrees, mue_s = 0
v = sqrt(g*R*tan(theta))
==> R = v^2/(g*tan(theta))
= 54^2/(9.8*tan(25))
= 638 m
d) radius of the turn does not depend on mass of the car.
e) a_rad = v^2/R = 54^2/(159) = 18.34 m/s^2
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.