Draw a diagram showing the banked track (indicate θ), a dot on
the track representing the car, the normal force vector N
perpendiculr to the track (with its tail at the dot) and the
vertically-down force vector of gravity.
The horizontal component of the normal force provides the
centripetal force needed to keep the car moving in a 900m circle.
So
mv²/r = mg sin(θ)
v² / (rg) = sin(θ)
θ = arcsin(v² / (rg) )
For the given numbers,
θ = arcsin(37²/(9.8*900)) ≈ 8.93°
[OR]
The normal force supplies the centripetal acceleration for the car, no reliance on friction at the ideal speed of 37m/s (very fast). The normal force is broken down into nsinϴ in the x-direction and ncosϴ in y. newton’s 2nd law for the net horizontal forces gives this equation:
ΣF(x) = mv²/r = nsinϴ------------------->(1)
Vertically, the forces balance where we have the upward normal
force and the downward weight (mg) of the car. From this we can
find an expression for n:
ΣF(y) = 0 = ncosϴ - mg
n = mg / cosϴ---------------------------->(2)
Plugging (2) into (1) and solving for the angle, we get:
mv²/r = (mg / cosϴ)sinϴ
v²/r = gtanϴ
ϴ= tan⁻¹[v² / gr]
= tan⁻¹[(37m/s)² / (9.81m/s²)(900m)]
= 8.8°
Can you please explain the process of finding the answer A frictionless test track (Go Physicsland!)...
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