An airplane traveling at 510 km/h needs to reverse its course....

Question

An airplane traveling at 510 km/h needs to reverse its course. The pilot decides to accomplish this by banking the wings at an angle of 38° while moving in a level circular path. Find the time needed to reverse course.

Accepted Answer

Welcome back, everyone in this problem. How long does it take for a glider traveling at 511 kilometers per hour and banking at an angle of 39 degrees to navigate a sharp turn on a level circular path. For our answer choices. A says it's 50 seconds. B 56 seconds C 60 D 65 seconds. Now, if we're going to figure out how long it takes for our glider to navigate this sharp turn on our level circular path. First, let's ask ourselves, what do we know about time and circular motion? Well, for one, we know that the time it's going to take, ok? Is going the time that is the time it's going to take to reverse course is going to be a half period of the circular motion. So in other words, we are solving for a time. OK. That's going to be a half of a half period of the circular motion or the period divided by two. Now, what do we know about the period that can help us? Well, we also know, ok, we can recall that our period is equal to two pi R divided by V now, we know the speed is 500 kilometers per hour, but we don't know the radius for our circular motion. So if we can find that, then we should be able to solve for our half period. Now, how can we do that? Well, here we have a diagram that we can label our forces and try to understand what forces are acting here, turning it into a free body diagram and thus resolved forces and eventually R will come along when we are resolving. And, and, and let's start to resolve and then see where it takes us from there. OK. So, so far, we know that we have our gravitational force acting downwards which is MG, we have the force of the lift here acting as F, OK? And we have our angle, theta, we know theta is 39 degrees. If we think about it here, we can extend our vertical upwards and thus F makes an angle of theta degrees with the vertical. Now, if we are going to resolve our forces, let's think about what's happening in the vertical and horizontal directions. So in other words, let's look at the sum of the forces in the Y direction and the sum of the forces in the X direction. Now, starting with the forces in the Y direction, I should probably have put that, put that in red here, OK? We know that the sum of those forces are going to be equal to zero OK. Now, if we think about it acting upwards, if we take the wards direction as positive, we have the vertical component of our force. OK? Which is going to be F CAS theta because our anger theta is adjacent to the vertical, its vertical component is adjacent and acting downwards. We have the weight MG. So F CAS theta minus MG equals zero, which tells us then that F CAS theta equals MG. Next. when we think about our forces in the X direction, a centripetal acceleration will result from the horizontal component of that lift force. So recall that Newton's second law tells us that the sum of forces or the net force is equal to the mass multiplied by the acceleration. In this case, this is going to be our centripetal acceleration. OK. So we can simplify that to say the sum of the forces in the X direction, which in this case would just be the X component of the lift force. F sine theta is going to be equal to the mass multiplied by the centripetal acceleration, which is V squared divided by R know that we have two expressions and we have one in terms of R let's divide both of these equations for the sum of the forces in the vertical and horizontal directions. Cause now when we do that, OK, then we're going to have FS theater. Let's divide the X, the equation for the X forces by the equation for the Y forces. So let me explicitly write that here. So now when we do that, we're going to have F sine theta. OK. F sine theta being equal to MV squared divided by R. And that equation is divided by F cos theta equals MG. So we're dividing MV squared divided by R by MG, which we could simply just write as MV squared divided by RMG. Now notice that F cancer in both terms, M also cancer and the sine of theta divided by the cosine of theta is defined as the tangent. Therefore, the tangent of theta equals V squared divided by RG, which then tells us that R, the variable we want is equal to V squared divided by G 10. The no, this is pretty good because no, we have an expression for R. We know V we can take the acceleration due to gravity. G as 9.8 m per second squared. And we know our anger theta is 39 degrees. So we can substitute it into our formula to solve for our period and thus our half period. So now doing that, OK, then this tells us that our period which was two pi divided by V is now going to be two pi divided by V multiplied by this expression for RV squared divided by G tan theta. Now V goes into V squared V times. So we can rewrite that as two pi V divided by G tan theta. OK. And now if we're finding our half period, remember T is our half period, then we can multiply both sides by a half. OK? And two multiplied by a half is one. OK. So that means our half period T divided by two equals pi V divided by G 10 theta. Let's substitute our values first. We know our speed was 511 kilometers per hour. So we can write it and then convert it to SI units. So we're going to multiply by 1000 m divided by 3600 seconds. OK. And that's multiplied by pi dividing it by G, take G to be 9.8 m per second squared, multiplied by the tangent of our angle 39 degrees. Now we can go ahead and solve for our half period. And when we do that, when we substitute this into our, uh when we use our calculator, then we should get it to the nearest second to be approximately equal to 56 seconds. Therefore, that tells us it takes the glider 56 seconds traveling at that speed to navigate a sharp turn on ac on a level circular path. If we go back to our answer choices, that means B is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

An airplane traveling at 510 km/h needs to reverse its course. The pilot decides to accomplish this by banking the wings at an angle of 38° while moving in a level circular path. Find the time needed to reverse course.

Key Concepts

Banking of Wings

Centripetal Acceleration

Time to Reverse Course

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