Radius(Curve), r = 25 m
Acceleration, g = 9.8 m/s2
Speed, v = 36 km/h =
= 10 m/sec
To find the Banking angle use:
Answer: 22o
How we get the banking angle formula:

Fnet = Fcentripital
A frictionless road turn is circular of radius 25 metres is designed to have a maximum...
A frictionless highway turn is circular with radius 36 metres is banked at an angle of 18°. What will be the maximum posted speed for the turn to the nearest km/h? You may assume the gravitational acceleration is equal to 9.8 m/s. Maximum Posted Speed v= km/h
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A banked circular highway curve is designed for traffic moving at 62 km/h. The radius of the curve is 214 m. Traffic is moving along the highway at 41 km/h on a rainy day. What is the minimum coefficient of friction between tires and road that will allow cars to take the turn without sliding off the road? (Assume the cars do not have negative lift.)
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A car at speed v takes a turn of radius R on a banked road of
angle
. What is the angle that the road must be banked ti bit require
the driver to turn the steering wheel? For circular motion, the
centripetal acceleration is
Now the road has a coefficient of friction of
with
. What is the maximum velocity that the driver can take the
turn?
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What is the radius of the turn of the angle is 20 degrees. (
assuming the car continues in uniform circular motion around the
turn?
t Banked Frictionless Curve, and Hlat Curve A car of mass M- 800 kg traveling at 65.0 km/hour enters a banked turn covered with ice. The road is banked at an angle θ, and there is no friction between the road and the car's tires as shown in (Figure 1). Use g 9.80 m/s throughout...
A car travels on a circular road with a radius R, and banking angle theta. At what speed v_max the car begins to skid if the static friction coefficient is mu_s?