Get Your Receipt for Work Done for R&D Effortlessly

Receipt for work done for R&D

this is kind of just a fun but useful problem uh I'm going to show that the work done by the gravitational force does not depend on path by giving three different examples so let's talk about work why do we care about work well work is important in the work energy principle which says that the work is a change in energy for some system why is it important to look at this for Gravity well if you can show that of the work done by force does not dependent on a path you can actually move that work done by that to the other side and call it a potential energy with the negative sign and so that's what makes it possible that you can have the change in gravitational potential energy as mg change in y for for gravity and so we can show this but the first thing is to show that it doesn't depend on the path now just showing three different paths doesn't necessarily mean it's path independent but it's it's a good it's a good hint so let's pick Let's imagine the following path I'm going to have I'm going to go from a to c so for this first path I'm going to go straight down to point B and these are all this is uh one and this is one just to make things easier so I'm going to go straight down to point B and then straight over to point C and I'm going to calculate that work and then I'm going to go from here to there along a straight line and calculate the work and then I'm going to go along this curve path that's going to be fun I guess you could go this way too that'd be pretty fun hmm I might be able to do that one okay so let's get to it we Define work for some suppose I have some displacement Delta r and I have some force acting on it f well how do I calculate the work along that path it's just F Delta dot Delta R where this is the dot product so if you have a as the vector a x a y a z those are three components of a and b is b x that's an x b y b z then a dot b is the scalar ax times B X plus a y times b y plus a z times b z now another way to find the dot product is take the magnitude of f the magnitude of the displacement Vector times the cosine of theta where Theta is the angle between them okay so we can do it both ways we'll just do whatever is useful I like the dot product way just because I like dot products but let's do this first path a path one I'll call Path one so here's I'll write this path one A to B to C and uh so let's just do this so work equals work A to B Plus work B to C so I have two paths so I need two works so uh for the force I'm just gonna have mg or G is the vector 0 negative 9.80 Newtons per kilogram so now I need Delta R A to B so what's the vector going from A to B now there's a couple ways you could do this I don't think it's too much of a stretch to write it as the vector 0 negative one zero is that okay can we just do that I mean if I'm going from there to there that's a vector so now uh I can go ahead and do work A to B it's going to be M times 0 negative nine point eight zero dot zero negative one zero so if I do the dot product let's say that I have a mass of one kilogram so it's going to be 0 times 1 times 0 so I get 1 times 0 times 0 minus 1 times 9.8 times -1 right because there's a minus in the gravitational force and there's a minus in the direction and then plus 1 times 0 times 0. 0 times 0. so that's 0 that's 0 and I get positive 9.8 plus 9.8 joules so in this case the work done by gravity is positive because the force and the displacement are in the same direction that's what you would expect okay now let's do from B to C so work B to C I need to know Delta R B to C that's going to be the vector 1 0 0 right so if I'm going from here to there I'm moving one in the X Direction now if I do the dot product with this I can get 1 times 0 times 9.8 and 0 times 0 so that's going to give me zero so the work done on BC is 0 joules and the total work is going to be 9.8 joules now we could do this the other way I could say work A to B is mg times the displacement of one meter times cosine of zero right so in this case this is the magnet to the gravitational force that's a scalar value that's positive 9.8 that's the magnitude of the displacement and this is the aim between them so since they're both going down both G and Delta are down then Theta would be zero the cosine zero is one let's just gets one times nine point eight times one and then w b c is going to be equal to mg times 1 times cosine of 90 or pi over 2. because in this case Delta R is that way f is that way they're perpendicular to each other the angle between them is 90 degrees and the cosine of 90 is zero so either way I get the work is equal to 9.8 joules okay now let's do path two get a new piece of paper it's stuck okay path two here's my diagram going from a to see so this is uh let's just say this is the vector uh one zero zero and this is the vector 0 1 0. for that one right there and I want to go like that so we could find this angle here it's actually pretty easy that if this is one and that's one then this angle is 45 degrees and I'll do it both ways but I can also get Delta R Delta R is going to be my final position which is 1 0 0 minus my initial position of 0 1 0. it Delta R is change in position final position initial position so if I do this product to get one minus 0 is 1 0 minus one is negative one zero minus zero is zero and so that makes sense right I'm going down one over one so or over one and down one and that that Vector does agree with what I'd expect right there so that's my displacement now I can do work it's going to be equal to uh I'm going to multiply the one inside so I have the force is zero negative 9.8 Newtons 0 dot 1 negative one zero so now if I do the dot product I get zero times one is zero times one negative nine point eight times one is going to be let's write that as plus negative 9.8 times negative one plus zero times zero so zero times one is zero negative nine point eight times negative one is nine point eight zero same thing okay so now let's do it the other way if that's my gravitational force and that's Delta R then the co the angle between them is Theta okay but now I have to deal with this total distance so if I do work is f Delta r cosine Theta well f is mg Delta R is going to be the hypotenuse of this so it's actually going to be equal to the square root of 1 squared plus 1 squared plus 0 squared right it's going 1 in The X Direction 1 the y direction is zero in the Z Direction so this is the square root of 2. so work is going to be mg which is 1 times 9.8 and these are positive numbers right because I'm using the magnitude Delta R is going to be the square root of 2. cosine of 45. so the cosine of 45 is yes it's 1 over the square root of 2. if you want to put that in your calculator you don't trust me that's fine but you can so I get 9.8 joules okay so work path one work path two the same thing work path three is tricky now there is a way to do this with calculus but we're not going to do that so let's start right here here's point a here's Point C and this is a circle oh missed it okay uh so we have a problem here because yes the gravitational force at this point is down and the displacement is right that way but up here the displacement's that way so the displacement keeps changing directions uh the whole time it's moving so you can't do I can't do work equals F dot Delta r that is only true if F and Delta R are constant vectors but here even though the force is a constant value the displacement Vector Delta R changes in Direction so we have we have a problem now you could set up an integral to do this but we're going to do it in Python instead so we're going to do the following I'm going to do this so here's a I'm going to draw it like this and there's C so what if I just went broke this into steps like this is this is a very few steps okay so if I break this going into this this piece then I have Delta R is that and I can approximate Delta R's constant along that path and I can calculate just a little bit of work along that path I'll call it DW so in that case DW IS F dot Dr now I can move my position right here and have a new Dr going this way now if I assume Delta R is a constant over this path I can calculate another work and then I can add it to the total and then I can calculate the work along this path and this path so I can break it into a whole bunch of short paths that are approximately constant and so of course we could do this on paper but it'd be easier to do it in Python so here's what we're going to do I'm going to start with this Vector r as the location of my object with some angle Theta uh and I can then say d Theta is 0.1 radians and just keep increasing Theta until I get down to here so a lot of times we measure Theta from the x-axis I'm going to start up here where Theta is zero just because I like it that way so I'm going to go from Theta uh equals 0 to theta equals pi over 2. and then all I need to do is to calculate the vector Direction d r for each one of those paths so if I have step size of Delta Theta at the step angle I can calculate the length of this right here I can say uh s is r Delta Theta right because that's the Arc Length of that piece but I actually need Delta r as a vector so here's a tiny little trick so I can say Delta R you could do it manually if you wanted to it's going to be s the length times R hat cross Z hat I know that's a lot but so the cross product gives you uh the direction of a resultant Vector Z hat is a magnitude of one our hat has a magnitude of one because I'm making it that way um and then I multiply it by my unit Vector DS and in fact I could just say Dr it's going to be equal to I can bring that RN I can say Delta Theta R cross Z even though R is one so there's no Z hat so the the right hand rule says that if I have uh R right here uh R Cross C has to be perpendicular to R and Z so Z is coming out of the paper like this so R cross Z Would using the right hand rule uh come up this way and be in the right direction so now we have a vector for that and as we change Theta we change R we change we can find new Delta r it might make more sense when you put it in Python and so I can we can visualize it but hopefully at the end we should get a work done by adding all the works at something close to 9.8 it won't be exact because we're making an approximation okay so let's jump over here to python okay so I'm in close group V Python and I will give you this code uh let's just start off with some stuff uh g equals vector 0 negative 9.8 0 and mass equals one now the first thing I just want to do is to make a ball I'm going to make a visual ball and have it move along that angle so I'm going to say theta equals 0 D theta equals 0.1 we can change these things later if we don't like it and I'm going to calculate the vector R right I'm going to say R is going to be equal to 1. let's just say this R equals one that's the the length the radius of the circle R is the vector R times Vector the nice one of the nice things I like about glow script V python is that it has vectors built in boom okay so it's going to be the vector uh sine Theta cosine Theta 0 and like I said sine is in the X Direction because I measuring the angle from the Y Direction so if you look at the picture that shouldn't be too difficult now let's put a ball there ball is an object of type sphere it's position is R its radius is let's say zero point zero five and let's do this make Trail equals true okay so all I want to do is move that ball along that path I'm not going to calculate work I'm just moving the ball so let's say while Theta is less than pi over 2. uh I'm going to give it a rate of 10. this just says don't do more than 10 calculations per second so we can see it moving you don't have to have that in there but I like it okay so the first thing we're going to do is to um put the ball at the position of theta so I'm going to say even though I've already done this I'm going to say ball dot POS equals uh let's calculate r I'm just going to recalculate R up here and you may get into these order programming order problems but the order of calculations can always be fixed by having smaller step sizes so it's not a big deal okay so then the ball position is going to be r and then I'm going to increase Theta theta equals theta plus d Theta and then that's it it should it should move in a quarter of a circle I think there you go nice very nice okay uh so let's go back over here let's save it uh work done by gravity Circle I'll give you the code don't worry about that you're going to get the code okay so it's moving along the path what I want to do now is to draw an arrow representing Dr or call it DL just so that we can visualize that so let's say DL I'm going to actually make it an arrow equals an error an arrow its position is going to be the ball and its axis it's going to so when we make an arrow in V python we have two important properties one is the position the location of where that Arrow starts and the other is the the actual length of the arrow so this is going to be equal to D Theta just like I said times the cross product of r and the vector Z which is going to be Vector 0 0 1. I think that will work okay and I'll should I make it yellow let's make it yellow color equals color dot yellow I think that should work okay so now down here after I move the ball I need to move the arrow so I'm going to say d l dot POS equals ball dot pus and I need to change the direction so DL dot axis equals I'm just going to do the same thing I had a 4D Theta times cross not course no cross R Vector 0 0 1. and so our changes so so that cross product changes so it should change let's see what happens okay it is yellow because it's a little small um and that's fine let's just make it bigger just for effect okay it's the actual size of the step that's why it's so small um so let's just say uh SC for the scale I'm just going to make it twice as big so down here it's going to put in SC I'm going to put in SC okay now let's see what it looks like nice it's going in the right direction everything seems to be fine so now I can calculate the work so up here I'm going to say work equals zero because remember I'm going to calculate the work along each little piece I'll call it DW and then I need to add it to this so I need to have I need to start it as something I'm going to start it at zero so let's just do there's my R let's go ahead and calculate the work DW equals uh let's just say m times G that's the force oh no I'm sorry dot m times G d l d l dot axis so DL is an object of an arrow the actual D oh okay I'm going to take off this so it's going to be wrong because I have that SC is 2. so that would make it twice as much work because the I'm doing the actual length so I'm switching the arrow back to its real length it's the real length okay um so that's the work done along that path the dot products are built-in function into closed group V python just like the cross product is which is really kind of nice um so now I just say work equals work plus DW and that should be it I do need to print that print work equals w ools 8.6 okay so it's not it's not that it's not the right answer but if I take my D Theta and make it uh 10 times smaller and run this 10 times as fast I have a lot smaller steps you can't even see the arrow it's so small and I get 9.7 so it's still a little small but again it's approximation so we can fix that let's just go go even smaller I'll just do half a small and run this 10 times as fast okay so I think it's pretty good I'm pretty happy now I did say one thing right what if I what if I want to go from here and go around the other way could I do that well I think I can so if I have Theta as that um everything's the same but what if I take what if I do this Theta equals Theta minus D Theta so then it's going to go negative until it will eventually get and as long as Theta is greater than uh which that'd be pi is it three-fourths no two pi uh so while Theta is less than no minus moves the other way I don't think I want to do that I want to do um let's do this as plus let's just let's just imagine Theta is going the other way then my X component is going to be negative sign so I think I just come up here and say uh R is going to just have negative sign and this is negative sign and I want to go until 3 4 of 2 pi so it's 3 4 of 2 pi so three halves pi so let's listen three has five I think this may work I don't know and I think I have the Dr wrong but let's just run it okay so it did it and all I need to do is change this right now my Dr is going the opposite direction so I can just say negative negative 1 up here and that'll be fine negative one 9.8 boom I'm pretty happy with that so that was actually four pass even though I said I was only gonna do three three pass work is the same so now you can use well plausibly at least you can use the work done by gravity and move it over as a potential energy which is pretty nice you don't worry about the path then you can just use the change in potential okay that's that