Consider the ringshaped body of Fig. E13.35<IMAGE>. A particle with mass m is placed a distance x from the center of the ring, along the line through the center of the ring and perpendicular to its plane. (a) Calculate the gravitational potential energy U of this system. Take the potential energy to be zero when the two objects are far apart. (b) Show that your answer to part (a) reduces to the expected result when x is much larger than the radius a of the ring. (c) Use Fx = -dU>dx to find the magnitude and direction of the force on the particle (see Section 7.4). (d) Show that your answer to part (c) reduces to the expected result when x is much larger than a. (e) What are the values of U and Fx when x = 0? Explain why these results make sense.

Video transcript

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use. In order to solve this problem on a horizontal surface. A steel ring is lying flat. The steel ring has a mass of capital M and a radius of lower case a, a small lead ball with a massive lowercase M is placed at a distance of Z above the center of the ring along a vertical line that passes through the center of the ring and is perpendicular to its plane, taking the potential energy to be zero when the two objects are infinitely separated, derive an expression for the gravitational potential energy U of the system. So that's our angle. Our angles are trying to derive an expression for the gravitational potential energy you of this particular system provided the conditions set to us or then provided the conditions that are set in this particular problem. So with that in mind, let's look at our diagram that's provided to us by the problem itself. So at the very top, we have our silver lead ball that has a massive lowercase M. And then we have a dotted black line that is the distance Z above the center of the ring where we have our ring represented in red and A and that has a massive capital M. So from where this dotted line Z is going from the center of our steel ball going down to the center of our ring, the center of the ring to the outer edge of the ring is our radius A of the ring. And now that we have a visualization of what's going on in this particular problem, let's read off our multiple choice answers to see what our final answer might be. A is negative capital G multiplied by lowercase M multiplied by capital M all divided by A plus Z B is negative capital G multiplied by lower case M multiplied by capital M all divided by A squared plus Z squared. C is negative capital G multiplied by lower case M multiplied by capital M all divided by the square root of Z squared plus A squared. And finally D is negative capital G multiplied by lower case M multiplied by capital M all divided by Z. OK. So with that in mind, we first off, we need to recall and note that the potential energy due to a small segment of DM of the ring. So considering that we need to note that the gravitational potential energy between it infinitely or infant mass element DM of the ring and the particle of mass lowercase M or I should be specific. So let's rewind here for a second. So the gravitational potential energy between an infinitesimal mass element DM where M in this case is capital M, that's important of the ring and the particle of mass lowercase M is. So this expression, we could write D capital U. So DU is equal to negative capital G multiplied by lowercase M multiplied by D multiplied by capital M all divided by lower case R. So this is the gravitational potential energy between an infinitesimal mass element D capital M. So DM of the ring and the particle of mass lower case M where capital G is the gravitational constant and R is the distance between the mass element D capital M so DM and the particle. So with that in mind, we need to note that and we'll get rid of this multiplication sign means we're talking about D MD capital M. So before we go any further, let's draw a more detailed diagram of, you know, let's add more detail to the diagram that's provided to us. So we can see what R looks like in respect to our original figure that's provided to us. So in this case, in our figure, our ring is represented in orange and we also have our steel ball that's also represented in orange. So R in this case where we know that R is the distance between the mass element DM and the particle. So it's from our steel ball going down to the outer edge of our ring and looking at a ring, a small little piece of our ring is representing DM Awesome. So with that in mind, we need to recall a note that for any mass element D capital M so DM on the ring, the distance R to the particle is, and we can, as we should recall, we can write this expression as R is equal to the square root of Z squared plus a square. And this distance as we should recall is the same for all mass elements of the ring since the ring is symmetrical. So therefore, as we should recall and note the total gravitational potential energy which is denoted by capital U can be obtained by integrating DU over the entire mass of the ring like soap. So capital U is equal to the interval of DU which is equal to the integral of negative capital G multiplied by M lowercase M to be precise multiplied by D capital M divided by R. So substituting in the value for R, we will find that U is equal to negative capital G multiplied by lower case M divided by the square root of Z square plus A squared multiplied by the interval of D capital M which is all equal to. So that will mean that U is equal to negative capital G multiplied by lowercase M multiplied by capital M all divided by the square root of Z square plus a sw. And that's it we've solved for this problem. Hooray, we did it. So with that in mind, let's look at our multiple choice sensors to see which one matches the answer that we just found together. And it appears the correct answer has to be the letter C which states that it's negative capital G multiplied by lower case M multiplied by capital M all divided by the square root of Z squared plus A squared. Thank you so much for watching. Hopefully that helped and I can't wait to see in the next video. Bye.