An airplane feels a lift force L perpendicular to its wings. In level flight, the lift force points straight up and is equal in magnitude to the gravitational force on the plane. When an airplane turns, it banks by tilting its wings, as seen from behind, by an angle from horizontal. This causes the lift to have a radial component, similar to a car on a banked curve. If the lift had constant magnitude, the vertical component of L would now be smaller than the gravitational force, and the plane would lose altitude while turning. However, you can assume that the pilot uses small adjustments to the plane's control surfaces so that the vertical component of L continues to balance the gravitational force throughout the turn. a. Find an expression for the banking angle θ needed to turn in a circle of radius r while flying at constant speed v.

Video transcript

Welcome back, everybody. We have a spacecraft that is on a mission to explore Mars. Now, before diving into the details here, I'm actually going to draw out a little xy plane here and you'll see why in just a second, I'm going to place the body of our Mars spacecraft at the center of our X Y plane with its wings just like that. We are told that it enters the Martian atmosphere at this angle to the horizontal of alpha. Once it enters the Martian atmosphere, it is just going to travel in a circle over and over again in the xy plane. Now, keep in mind here that lift force for aircraft, let me actually go ahead and make this a solid arrow here. We have that lift force for an aircraft is always perpendicular to the wings of whatever aircraft we are looking at. And we are tasked with finding what our angle alpha is in terms of its velocity, the acceleration due to Martian gravity and the radius of our circle that it is traveling in before we can start playing around with these values. I want to draw in a couple more things here. We know that the force due to gravity always acts downward despite whatever planet we're on. And I actually want to decompose this lift force into its X and Y components. The reason being because we are going to be using Newton's Second law here and that's going to require us to decompose this factor. So knowing that this angle right here is alpha, we can also say that this angle is alpha as well. Meaning that the why component of our lift force is L times the cosine of alpha and our X component is L times the sine of alpha. And now that we've established that we are ready to go ahead and use Newton's second law in the Y direction and the radial direction. But in this case, it's the same thing as the X direction. Let's start with the Y direction here. So we have all the forces in Y are equal to the mass times acceleration. Well, we have L co sign times alpha minus the force due to gravity is equal to, if you'll remember um our aircraft is traveling in, in, in an xy plane. So really what we're looking at here, actually, my mistake not traveling in xy plane is traveling in just the explain, actually, just the explain. So if it's traveling in the explain here, it is not rising or falling in the Y direction, meaning the right side of this equation is going to be zero, I'm just going to add the force due to gravity on both sides. And we have L co sign of alpha is equal to MG. I'm going to name this equation equation one great. So now let's do the sum of all forces in the radial direction. Remember as this, this spacecraft is traveling in a circle, this direction points towards the center of that circle. So we have that the sum of all forces in the radial direction is equal to the mass times the radial acceleration. This just gives us L times sine alpha is equal to mass times or radio acceleration. But I'm going to sub in for radial acceleration, the velocity of our spacecraft divided by the radius of the circle. Wonderful. So then I'm going to name this equation to and we're gonna perform a little bit of algebraic manipulation here. Let me scroll down here. I am going to divide equation two by equation one. You'll see that this gives us on the left hand side, L sine of alpha divided by L co sign of alpha else cancel out. And this is just equal to tangent of alpha which let me move over just a little bit here. On the right hand side of the equation, we are going to of our mass times our velocity squared divided by our radius, all divided by four multiplied by one over M G masses are going to cancel out. And all we get is our velocity squared divided by our radius times our acceleration gravity. Last thing to do is just take the inverse tangent of both sides. And you'll see that we get that our alpha is equal to the inverse tangent of the velocity of our spacecraft squared divided by the radius of its orbit times the acceleration due to Martian gravity, which corresponds to our final answer. Choice of C Thank you all so much for watching. Hope this video helped. We'll see you all in the next one.