A car travels on a circular banked track of radius 300 m and having a banked angle of 22. What is the minimum time for one lap of the track if the car does not rely on friction to hold it on to the track?

component of mg down slope = component of m * centripetal acceleration down slope

m g sin 22 = (m v^2/R) cos 22
g tan 22 = v^2/R
v^2 = g R tan 22
solve for v
T = circumference / v = 2 pi R / v

To find the minimum time for one lap of the track, we need to consider the forces acting on the car.

Let's break the forces into two components: the normal force (N) perpendicular to the track and the gravitational force (mg) acting vertically downward.

On a banked track, the normal force can be resolved into two components:
1. Normal force component perpendicular to the track (N⊥)
2. Normal force component parallel to the track (N∥)

The gravitational force (mg) can be resolved into two components:
1. Gravitational force component perpendicular to the track (mg⊥)
2. Gravitational force component parallel to the track (mg∥)

Due to the angle of banking, the gravitational force component perpendicular to the track (mg⊥) helps provide the necessary centripetal force for the car to move in a circle.

Now, let's calculate the gravitational force component perpendicular to the track (mg⊥):
mg⊥ = mg * sin(θ)
θ = 22° (banked angle)
mg⊥ = mg * sin(22°)

Next, let's determine the speed of the car needed to maintain circular motion on the banked track.

The centripetal force required to keep the car moving in a circle is given by:
F⊥ = mg⊥ = mv² / r

Where:
m = mass of the car
v = speed of the car
r = radius of the circular track (300 m in this case)

Setting the centripetal force equal to the gravitational force component perpendicular to the track, we have:
mv² / r = mg * sin(22°)

Simplifying:
v² = rg * sin(22°)
v = sqrt(rg * sin(22°))

Finally, we can calculate the time (T) for one lap of the track using the formula:
T = 2πr / v

Substituting the given values of r and v, we get:
T = 2π * 300 / sqrt(9.8 * 300 * sin(22°))

Calculating the value of T:
T ≈ 108 seconds (rounded to the nearest whole number)

Therefore, the minimum time for one lap of the track, without relying on friction, is approximately 108 seconds.

To calculate the minimum time for one lap of the track, we need to consider the forces acting on the car and apply some basic principles of physics.

First, let's analyze the forces acting on the car when it is going around the banked track. There are two main forces: gravitational force (mg) and the normal force (N).

The normal force can be divided into two components: one perpendicular to the track (N⊥) and one parallel to the track (N∥). The perpendicular component (N⊥) opposes the gravitational force, while the parallel component (N∥) provides the necessary centripetal force for the car to move in a circle.

The gravitational force can be split into two components as well: one perpendicular to the track (mg⊥) and one parallel to the track (mg∥). The perpendicular component (mg⊥) is canceled out by the perpendicular component of the normal force (N⊥), as the car doesn't lift off the track.

The angle of banking of the track (θ) and the velocity of the car (v) are related by the equation:

tan(θ) = v² / (r * g)

where r is the radius of the track and g is the acceleration due to gravity.

From this equation, we can solve for the velocity (v):

v = sqrt(r * g * tan(θ))

Now, we can calculate the minimum time (T) using the formula:

T = (2 * π * r) / v

where π is a mathematical constant (~3.14).

Substituting the value of velocity (v) into the equation, we get:

T = (2 * π * r) / sqrt(r * g * tan(θ))

Plugging in the provided values:

r = 300 m
θ = 22 degrees
g = 9.8 m/s²

T = (2 * 3.14 * 300) / sqrt(300 * 9.8 * tan(22))

Using a calculator, we can accurately evaluate this expression to find the minimum time for one lap of the track when the car does not rely on friction.