A thin, uniform rod has length L and mass M. A small uniform sphe...

Question

A thin, uniform rod has length L and mass M. A small uniform sphere of mass m is placed a distance x from one end of the rod, along the axis of the rod (Fig. E13.34)

. (b) Use Fx = -dU>dx to find the magnitude and direction of the gravitational force exerted on the sphere by the rod (see Section 7.4). Show that your answer reduces to the expected result when x is much larger than L.

Accepted Answer

Everyone in this problem. A regular round bar has a mass M and length L A point mass is placed at a point parallel to the bars length on its axis, an arbitrary distance D from the nearest end of the bar. Using the relationship between force and energy is equal to negative D. You over D R, we are asked to determine the magnitude and direction of the gravitational force due to the bar on the click and then for D much greater than L. Whereas to prove that the answer agrees with the expected result. So we're given a diagram here and we have our bar mass with mass B M and it is a distance D from this point mass M along its axis. And the length of this mass is L. Now we know what to do if these are both point masses. And we know the relationship for the gravitational force. If we were dealing with two point masses, we're dealing with a bar here. So let's imagine we can break this bar up into little pieces. Let's break it up into this tiny piece. And we're gonna call this tiny piece D okay. It's just some small piece of our big mass M and then we're gonna call the length or the distance from the end of our bar mask may the end closest to our point sm to this little piece that we've picked out to be little L. Alright. So now if we're thinking about the gravitational force just from this little piece, we have that the radius R or the distance R from the point mass to this little piece that we have to worry about. It's gonna be L plus D okay because we have the distance D from the point mass to the end of the bar, the distance little L from the end of the bar to the piece of the bar that we're interested in. And note that this piece D M this in blue as well. This piece D M it's gonna have mass, it's gonna be M over L. We've taken the entire mass M divided by these lengths, this entire length L and then we multiply it by D L and D L is just the length of this tiny piece. All right. So now if we think in general about our energy, okay, we're asked to do the gravitational force but were told to think about this relationship between force and energy. So let's start with the energy first. The energy potential energy you I recall in general is given by negative G big. Well, um over our okay and recall that we're saying that our is equal to L plus D. So we have negative G big M little M over R R. Sorry L busty. Alright. And we didn't switch this big M, we need to switch this big M. Okay? And we're gonna call this D um yeah we're gonna we're gonna leave it as a big game for now. That's okay. Alright, so let's move on to the next step. Okay. When we're talking about force, we want to relate this to a derivative, we want to figure out you in more detail. Alright, so let's think about the derivative. We were to take the derivative. Oh There's potential with respect to L, we can write this as a derivative of the potential with respect to our mass DM times the derivative of our mass D M with respect to this length deal. No, no, the derivative of U with respect to big. Um Let's go ahead and do that. That's gonna be negative G little M over L plus D. And then we're gonna multiply by the derivative of big M with respect to L. Now we have this D M is equal to M over L D O. And from there, we can say that D M over de L is then equal to big M over big L. Yes, we're gonna multiply here by big M over a big L. Now we can rearrange some more. So we have that D U is equal to negative G little M big divided by big ill times L plus D, all times deal. And now we can integrate. So if we integrate both sides, we have you on the left hand side, on the right hand side, we're gonna integrate across the entire length of our bar. Okay. So we're gonna integrate from zero from this little D L value of zero. And let me write that. So it's clear little L from zero all the way up to big Al we're taking that tiny little piece starting from the far right end all the way up until the end of our link of negative G little M big M over big L. Steve D O, the negative G A little M big M over L are not affected in this integral. Okay. They're just constant when we're talking about um with respect to little so we can pull them outside of our integral. We have negative G little M big M over big L Times The integral from 0 to L of one over little L plus D D little L. Now we're going to actually integrate, okay. We have the integral of one over a little L plus D. Okay. So you can think of us this is the integral of one over X plus a constant. OK. It's that exact same form. And so this is going to be equal to negative G little um big M over capital L times lawn of little L plus D and this is going to be evaluated from zero to the integral of one over little L plus T and lawn of little L plus T. If we substitute in our end points, we have negative G little M big M over big L times long. The big L plus D minus lawn 50. All right. We can use our log rules to simplify this guy. No, let's continue giving ourselves some more space. We use our log rolls. We have long of big L plus D minus lawn of D. Okay. So we have some long term minus another long term. We can turn this into one single log rhythm where we have division of the arguments. We have negative G M little big M over L times lawn of L plus D over de and we can simplify again inside of those brackets. Negative G little big over big L, I'm blonde, big L over D plus one. All right now haven't in this format. And let's recall that we can approximate Launch one Plus X as X minus X squared over two plus X cubed over three. Okay. This expansion Series expansion -X 4/4 plus dot dot dot So let's go to four terms and expand our quantity. Can we have lawn of big L over D plus one? The X term in our case is gonna be L over D. He's a lawn of one plus L over D. It's going to be equal to, hello birdie minus L over D squared over two plus L over d cubed over three -L over D to the experiment for over four. And we can continue that series expansion. Now, we're asked to approximate what happens for D much bigger than L. So if D is much larger than L okay, these fractions are going to be really, really, really small L over D okay. So L over T to the exponent end when this value L over D is already really small, it's just gonna go to zero. Okay. So as long as N is greater than one, This goes to zero. So we're gonna approximate, we're gonna approximate lawn of one plus over DE as approximately big L over D. If we do that, then the potential energy. Let me write that this was, you are potential energy, you, it's going to be equal to negative G little M big M over L times L over D. Okay. We can cancel these two L terms. We get negative G little M big M over de which is what we expected. Okay. So we have G, the constant, the multiplication or the product of the two mass is the mass of the point mass, the mass of the bar divided by the distance between the point mass in the end of that bar. Okay. So we didn't need to do anything special here. No, let me highlight that result. Yeah, that's what we found for gravitational potential energy. U. Now we're gonna go back up and to work on the force, I'm just gonna erase some of this work we did. So we have some more room to work. So let's start here. All right. No, the problem reminds us that the force F is related to the potential energy through the following negative D you by D R. Now, in this case, the distance R between the point mass and the bar is D and so negative D U D R is equal to negative d'you by D D. So this is gonna be negative. What are we taking the derivative of? Well, our potential function you that we found and we're not gonna take it once we assume d larger than L. Okay, we're gonna take it from before that point before we did our expansion. So we have negative G little um big M over oh Big al Times lawn, big al over D plus. Okay. We're going to take the derivative of this with respect to D and remember that this long term, okay, we had taken to lawns and combined them into one single term in order to do the previous step in this step. Remember that we could separate this into lawn of L plus C minus lawn of D. That's what we had previously. Okay. That's just gonna make it a little bit easier to take the derivative. Alright, so when we take the derivative, we have G little thing over out times one over oh musty -1 over D. Alright. And this negative we have this negative outside and then this negative G M over L. So those negatives cancel. Okay. Let's go ahead and simplify. Let's find a common denominator for these terms in the brackets. So we have big G little M big M over L times. In order to have a common denominator, let's have D times L plus D. So in the first term, the denominator is L plus D, we need to multiply by D in order to get the common denominator. So we need to do that in the numerator as well. So we end up with D over L plus D times D minus. Same thing here. We're multiplying by L plus D numerator and denominator to get that common denominator. This is going to give us G um big over L times negative L over de L plus D. How do we get negative when we combine these terms with the same denominator? Now, we have D minus L minus D, it just gives us minus okay. The L terms will divide up and we're left with negative G little M big M over L oops not over L over D times L plus D, we can expand negative G M over de O plus D squared. Now, this is a negative force. F is less than zero, which tells us that it's an attraction okay, an attractive force. Now, we want to make the same assumption that D is much larger than L that we did for the potential function or the potential energy. So if D is much larger than L, then the quantity D L plus D square that's in the denominator is just going to approach D squared. Okay. And so our function F is just going to be equal to negative G little M big M over D squared just like we would expect. Alright, so let's go back up to answer traces. We figured out the potential energy, we figured out that force F now. Alright. So the force F we found was negative G little M times big M over D squared plus L D. And when we assume that D is much larger than L, we get an approximation where the force F is equal to negative G little M big M over D squared that corresponds with answer choice B. Thanks everyone for watching. I hope this video helped see you in the next one.

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