An 85,000 kg stunt plane performs a loop-the-loop, flying in a 26...

Question

An 85,000 kg stunt plane performs a loop-the-loop, flying in a 260-m-diameter vertical circle. At the point where the plane is flying straight down, its speed is 55 m/s and it is speeding up at a rate of 12 m/s per second. (b) What angle does the net force make with the horizontal? Let an angle above horizontal be positive and an angle below horizontal be negative.

Accepted Answer

Hi, everyone in this particular practice problem, we are asked to calculate the direction of the net force on the ball with respect to the positive X direction. We will have 125 g ball attached to a light cable and spun counterclockwise in a vertical circle with a radius of 0.5 m. So at the position where the ball is moving straight up, the ball is subject to attention shell force with a magnitude of 3. Newton and the center pedal force of magnitude. For Newton, we want to calculate the direction of the net force on the ball with respect to the positive X direction, either clockwise or counterclockwise with respect to that. So at the position where the ball is moving up, I am going to start us off with drawing a system that we have. So I'm just gonna first draw the access here. This is gonna be just the X and Y access. And then I am going to draw the, I guess the course where the ball is going to be moving. And then, as described in the problem statement, we will have first the tangential force of the 3.2 Newton. So that is going to be this vertical thing here. And that will be, I'm just gonna indicate that with F T and then we will also have the centripetal force with a magnitude of four. So the centripetal force is going to be going into the um center of the circle or the center of the rotational course. So this is going to be F C right here. Yeah, I'm just gonna indicate that with FC. So we know that the attention shell force F T is going to be particularly upward. So F D will equals two plus 3.2 J in the factor notation. And then the centripetal force FC FC is going to be horizontal to the left to the center pointing to the center of the rotational circle. So F C is going to be minus four point oh I in the factor notation. So with F D N F C going like this, the net force is going to be approximately going this way and this will be our F net. So our F net is essentially going to be F T plus F C just like. So, so F net, I'm just gonna include all the factor notation here. So F net is going to be um I'm gonna start with the I notation. So for I plus 3.2 just like. So, so that will be F net. So what we wanna do is to calculate the angle or the direction of the net force on the ball with respect to the positive X direction. So what we are actually being asked is actually this angle right here, which I'm going to indicate that with a five. But the way we want to calculate for five, it will be easier for us to actually calculate the data value here. So we're gonna start with calculating the angle data between F net and FC. And then from there, I will calculate the five. So we want to recall that the data here can be calculated by incorporating Trigana metric formula. So we want to recall that uh 10, Our detention of data is going to equal to FT whatever is in front over FC or whatever is adjacent to the data. So 10 equals opposite over adjacent just like. So, so F D over F C. So now we want to recall that to find the data here, we have to find in first of 10 and in first of 10 is going to be our 10. So in first of Tanner F D over F C and there that is going to be our 10 of FT over FC And I am going to plug in our values. So this is going to be our tend of 3.2 over four. So that will give us a theta value of 38.7° and that will be um the angle from the minus X direction access to the F net. But we are being asked the angle or the direction of the net force on the ball with respect to the positive X direction. So that will be our five, which will equals two B, one A D minus 38.7 degrees, which will be approximately degrees. So 1 41 degrees counterclockwise because it is going to this direction counterclockwise from plus X access. So that will be the answer to our problem, which is 141° counterclockwise with respect to the positive X direction or access, which will correspond to option D. So option D is going to be the answer to this particular practice problem. And that will essentially be all for this video. If you guys still have any sort of confusion on this, please make sure to check out our other lesson videos on similar topics and that will be it. Thank you.

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