### Video Transcript

Cars travel at 105 kilometres per hour along a circular turn of 1.20-kilometre radius. What is the ideal banking angle for the turn?

Weβre told that the speed of the car is 105 kilometres per hour; weβll call that π£. Weβre also told that the radius of the circular turn the car travels through is 1.20 kilometres; weβll call that π. We want to know the ideal banking angle for the turn; weβll call that angle π. Letβs draw a diagram of the scenario: a cross-sectional picture of the car on the banked turn. This diagram shows a cross-sectional slice of the car as it moves through the banked turn.

Imagine that the car is driving into the page. The curve is banked at an angle π. And weβre told that this angle π is ideal in the sense that at this angle the car does not rely on friction to stay on its path. This means that there are only two forces that act on the car in this cross-sectional plane. First, there is the force of gravity acting down. And second, there is the normal force β the force perpendicular to the surface of the road that acts on the car. We can consider these two forces acting on the car and divide them into horizontal and vertical components. Weβll define forces and motion that are up and to the left as positive.

Letβs start by considering the vertical forces that act on the car. To do this, letβs break the normal force πΉ sub π into its horizontal and vertical components. The triangle that weβve drawn with πΉ sub π at the hypotenuse is a right triangle. And the topmost angle is equal to π because this right triangle is similar to the triangle of the bank. So the vertical component of the normal force is written πΉ sub π times the cosine of π. And the other vertical force acting on the car is the force of gravity, which is acting in a negative direction as weβve defined the positive and negative.

Now, letβs recall Newtonβs second law of motion to show us how these forces interact. Newtonβs second law says that the net force acting on an object is equal to the objectβs mass multiplied by its acceleration. So the vertical forces acting on the car are equal to the carβs mass times its acceleration in the vertical direction, which weβll call π sub π£. But as we realize from this scenario, the car does not accelerate vertically; π sub π£ is zero. This means that the vertical component of the normal force πΉ sub π cosine of π and ππ, the gravity force on the car, balance one another out; they are equal in magnitude.

Now that weβve considered the vertical forces, letβs turn our attention to the horizontal force acting on the car. And indeed there is only one horizontal force that acts on the car; itβs the horizontal component of the normal force πΉ sub π times the sine of π. By the second law, this force is equal to the mass of the car multiplied by its acceleration in the horizontal direction. This is centripetal acceleration, so weβll call it π sub π.

We can recall at this point that the centripetal acceleration π sub π of an object is equal to its speed squared divided by the radius of the circle through which it is travelling. This means we can replace π sub π in our horizontal force equation with π£ squared divided by π.

Now, letβs look back at our vertical forceβs equation. We see that this equality implies that the normal force πΉ sub π is equal to the carβs mass times the acceleration due to gravity π divided by the cosine of π. We can substitute this expression for the normal force into the normal force in the horizontal force equation. So we replace πΉ sub π with ππ over cosine of π. As we look at this equation, we see that we have the sine of π divided by the cosine of π.

As a trigonometric identity, the sine of an angle divided by the cosine of that same angle equals the tangent of that angle. So we can rewrite the left side of our equation as ππ times the tangent of π. As we continue to simplify this equation, we see we can cancel out the mass of the car on either side. And if we then divide both sides of the equation by π, then that term cancels from the left side of our equation. The tangent of π is equal to π£ squared divided by π times π.

Finally, we can take the arc tangent of both sides of our equation. Doing so isolates π on the left-hand side of our equation. So the angle weβre looking for β the ideal banking angle β is equal to the arc tangent of π£ squared divided by π times π. Before we plug in the values for π£, π, and π, we want to convert π£ into units of metres per second from its given units of kilometres per hour and π into units in metres rather than its given units in kilometres. Starting with π£, if we multiply 105 kilometres per hour by one hour per 3600 seconds, the units of hours cancel out.

If we next multiply by 1000 metres per one kilometre, then the units of kilometres cancel out as well. When we multiply these three fractions together, weβre left with a speed of 29.17 metres per second. So weβll replace our value of π£ with this speed. When we next convert π into units of metres, we recall that 1000 metres is equal to one kilometre. And therefore, 1.20 kilometres equals 1.20 times 10 to the third metres.

Now, weβre ready to plug our values π£, π, and π into our equation. π£ is 29.17 metres per second, π is 1.20 times 10 to the third metres, and π we treat as exactly 9.8 metres per second squared. When we compute this fraction and take the arc tangent of this value, we find that to three significant figures π is 4.14 degrees. This is the banking angle at which no other forces are needed in the vertical plane to keep the car in its circular turn.