Question

A frictionless test track (Go Physicsland!) is so that cars can move around the track at 37 m/s without slipping up or down the track Knowing that the radius of curvature of the track is 900 m, find the bank angle.
Can you please explain the process of finding the answer

Answer 1

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°