circular motion and gravitation A driver is turning around a curve of radius 300

Question

circular motion and gravitation A driver is turning around a curve of radius 300 m on flat ground. The 1500 kg car is traveling a constant speed of 30 in/s. What is the magnitude and direction of the force causing the car to move in a circular path? What type of force is this? (Normal, friction, tension, etc...) If the static friction coefficient is 0.7, how fast could the car take the turn without slipping or deviating from the 300 m radius turn? Mid-curve, the friction coefficient suddenly drops to 0.2 due to water on the roadway, and the car slides off the road. What is the maximum speed that would have allowed the car to continue safely making the turn? If the car were driving in a straight line, what is the shortest distance in which it could slow down from 30 m/s to the speed in part (c) if mu_s = 0.7? If mu_s = 0.4?

Explanation / Answer

(A)
Force causing motion = Centripetal Force = mv^2/r

Magnitude = 1500*30^2/300 = 4500 N

Direction towards center of the circle.

(B)

Friction Force = u*N
Friction Force = u*m*g

This force should be equal to the Centripetal Force.

mv^2/r = u*m*g
v = sqrt(u*g*r)
v = sqrt(0.7*9.8*300)
v = 45.36 m/s

(C)
Now,

Friction Force = u*N
Friction Force = u*m*g

This force should be equal to the Centripetal Force.

mv^2/r = u*m*g
v = sqrt(u*g*r)
v = sqrt(0.2*9.8*300)
v = 24.25 m/s

(D)

Final Velcoity = 24.25 m/s
Initial Velocity = 30 m/s

Friction Force = u*m*g
acceleration = F/m
acceleration = u*g

v = u - a*t
24.25 = 30 - u*g*t
t = 5.75/u*g

Time taken when us = 0.7

t = 5.75/0.7*9.8
t = 0.84s

Time taken when us = 0.4

t = 5.75/0.4*9.8
t = 1.46 s

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