Posted by
**William** on
.

A car moves with a speed v on a horizontal circular highway turn of radius is R = 100. Assume the height of the car’s center of mass above the ground is h = 1 m, and the separation between its inner and outer wheels (car's width) is w = 2 m. The car does not skid (static friction works between tires and road surface).

With what maximum speed (in mph) can the car negotiate the turn with without losing equilibrium and starting to overturn? Use g = 10 m/s^2.

Hint 1: car is in circular motion with constant speed. Hint 2: car starts to rotate about its center of gravity when net torque about its center of gravity becomes non-zero. Hint 3: draw an extended free body diagram and identify two torques counteracting each other: torque due to friction force on the inner tire versus torque due to normal force on the outer tire (assume that at the moment the car lifts its inner tires, the normal force becomes equal to the car’s weight: N = mg). Hint 4: friction force is in perpendicular direction to the direction of car’s velocity (it is towards the center of the circle). So, Newton’s 2nd law: F_net = mv2/R. Hint 5: friction force is the only force acting on the car in horizontal direction: F_net = F_fr.

Given

Radius , r = 100 m

Height , h = 1 m

Width ,w = 2 m

We will find angle of banking

sinθ = 1 / 2

θ = 300

The speed of the car is

v = gr tan30

= 10 m/s^2 * 100 m *0.5773

v = 24 m/s

53.686471 mph

That answer is wrong, it has to be in MPH btw, please help me.