A curve of radius 24 m is banked so that a 940 kg car traveling at 43.2 km/h can
ID: 1438298 • Letter: A
Question
A curve of radius 24 m is banked so that a 940 kg car traveling at 43.2 km/h can round it even if the road is so icy that the coefficient of static friction is approximately zero. You are commissioned to tell the local police the range of speeds at which a car can travel around this curve without skidding. Neglect the effects of air drag and rolling friction. If the coefficient of static friction between the road and the tires is 0.300, what is the range of speeds you tell them? I got the V Maximum which is 58.32 Km/h, I need the V Minimum please.
Explanation / Answer
radius of the curve, r=24m
mass of the car, m=940 kg
speed of the car, v=43.2 km/hr
co-efficient of static friction, us=0.3
use,
tan(theta)=v^2/r*g
tan(theta)=(43.2*(1000/3600))^2/(24*9.8)
====>
banking angle, thea=31.47degrees,
if the car is traveling around the curve at minimum speed,
along horizontal line,
m*v_min^2/r=N*sin(theta)-fs*cos(theta)
m*v_min^2/r=N*sin(theta)-us*N*cos(theta)
m*v_min^2/r=N*(sin(theta)-us*cos(theta)) ------(1)
and
along vertical line,
m*g=N*cos(theta)+fs*sin(theta)
m*g=N*cos(theta)+us*N*sin(theta)
m*g=N*(cos(theta)+us*sin(theta)) ---------(2)
from (1) and (2)
v_min^2/(r*g) = (sin(theta)-us*cos(theta))/(cos(theta)+us*sin(theta))
v_min^2/(24*9.8) = (sin(31.47)-0.3*cos(31.47))/(cos(31.47)+0.3*sin(31.47))
=======>
v_min = 7.87 m/sec or 28.35 Km/hr