1. V = 20 m/s magnitude of the velocity vector.
A car drives with speed of 20m/s on a banked track on a circular path, the surface of the road makes 15⁰ angle with horizon. Find the radius of the curve. Ignore the friction between the tires and the road.
Break it down:
V = 20 m/s magnitude of the velocity vector.
A car drives with speed of 20m/s
The surface of the road is not horizontal.
It makes an angle of θ with the horizon.
on a banked track
on a circular path, the surface of the road makes
15⁰ angle with horizon.
θ = 15⁰
r =?
Find the radius of the curve.
Fs = 0 Fs = μS*FN  μS = 0
Although the car is moving, it should not be confused with kinetic friction. Kinetic friction opposes the movement of the car, but static friction keeps the car on the circular path. Static friction is toward the center of the circle.
Ignore the friction between the tires and the road.
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2. Draw Free Body Diagram (FBD) for the car
A car drives with speed of 20m/s on a banked track on a circular path, the surface of the road makes 15⁰ angle with horizon. Find the radius of the curve. Ignore the friction between the tires and the road.
Solution:
Draw a diagram
Draw Free Body Diagram (FBD) for the car
+ Y FN
FN Cos15⁰
15⁰
r =?
FN Sin15⁰
+ X
15⁰
15⁰ mg
Notice that:
FN is perpendicular to the curve surface
“+X” axis is toward the center of the circular path (parallel to r).
Weight (mg) is along “- Y” axis
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3. Calculate Fnet along x and y axis
A car drives with speed of 20m/s on a banked track on a circular path, the surface of the road makes 15⁰ angle with horizon. Find the radius of the curve. Ignore the friction between the tires and the road.
Solution:
Calculate Fnet along x and y axis
FnetX =
FnetY =
This component of FN along x axis makes the centripetal acceleration.
∑Fx = FN sin 15⁰
∑Fy = FN cos 15⁰ – mg
This component of FN along + y axis supports the weight of the car.
Note that the direction of +x axis is toward center of the horizontal path, and acceleration is along +x axis as well.
Apply newton’s second law
Again net force is equal to the centripetal force which is mv 2 / r.
∑FX = mv 2/r  F N sin 15⁰ = m v 2 /r
(1)
The car does not move along y axis, then ∑Fy = 0
∑FY = 0  FN cos 15 – mg =  F N cos 15 = mg
(2)
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4. With the given speed and road slope, the radius of the circular path
A car drives with speed of 20m/s on a banked track on a circular path, the surface of the road makes 15⁰ angle with horizon. Find the radius of the curve. Ignore the friction between the tires and the road.
Solution:
Divide equation (1) by equation (2)
FN Sin15⁰/ FN Cos 15⁰ = [mV2/r]/mg
 tg (15⁰) = V2/ (r * g)
F N sin 15⁰ = m v 2 /r (1)
F N cos 15 = mg (2)
Solve this equation to find the radius (r)
r = V 2 / [g * tg 15⁰]
Now substitute V, tg 15⁰, and g to find the radius (r)
V = 20 m/s θ = 15⁰  tg 15⁰ = g = 9.8 m/s2
With the given speed and road slope, the radius of the circular path
should be 151m that car can drive in order not rely on friction at all.
 r = (20) 2 / [9.8 * (0.27)]
 r = 151 m
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