brett h pyta:
What is the required banking angle for the curve?
Find the required banking angle for a curve of radius 400.0 m of the curve is to be negotiated at a speed of 60.0 km/hr without the need of a friction force.
What is the required banking angle for the curve?
Find the required banking angle for a curve of radius 400.0 m of the curve is to be negotiated at a speed of 60.0 km/hr without the need of a friction force.
#1 by kuiperbelt2003 on April 25, 2010 - 4:11 pm
Quote
draw a diagram of the bank, showing the directions of the normal force (perpendicular to the banked road), weight,(straight down), and the centripetal force (to the center of the circle)
we will decompose these forces into x and y components, and apply Newton’s second law
call theta the angle of the bank; then, use geometry principles to show that theta is also the angle between a vertical line and the direction of the normal force
this enables us to write the x and y components of the normal force as:
x component = N sin (theta)
y component = N cos (theta)
the y component must balance the weight, so we have
N cos (theta) = mg (eq 1)
and the x component must produce the centripetal force, so we have
N sin(theta) = mv^2/r (eq 2)
divide equation (2) by equation (1) to get:
sin (theta)/cos(theta) = v^2/rg
or tan(theta) = v^2/rg
for the data here, we have
tan(theta) = (16.7m/s)^2/(400m x 9.8m/s/s)
tan(theta) = 0.07 or theta = 4.05 deg
(note that you should convert km/hr to m/s)