7 CIRCULAR MOTION 7.7 Motion on curved surfacesR car^
of curvature
mgr(^).
r
g
2
/
2
centre
banked curve
Answer: Consider a car of mass m going around the curve. The car’s weight, m g,
acts vertically downwards. The road surface exerts an upward normal reaction R
on the car. The vertical component of the reaction must balance the downward
weight of the car, so
R cos θ = m g.
The horizontal component of the reaction, R sin θ, acts towards the centre of
curvature of the road. This component provides the force m v^2 /r towards the
centre of the curvature which the car experiences as it rounds the curve. In other
words,
R sin θ = m
v
,
r
which yields
tan θ =
v
,
r g
or
Hence,
tan−^1
v^2
− 1 (40^ ×^ 1000/3600)^2
◦^
θ = tan (^60) × 9.81 = 11.8.
Note that if the car attempts to round the curve at the wrong speed then m v^2 /r =
m g tan θ, and the difference has to be made up by a sideways friction force
exerted between the car’s tires and the road surface. Unfortunately, this does not
always work—especially if the road surface is wet!
θ =