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(II) A mass $M$ is ring shaped with radius $r .$ A small mass $m$is placed at a distance $x$ along the ring's axis as shown inFig. $27 .$ Show that the gravitational force on the mass $m$due to the ring is directed inward along the axis and hasmagnitude$$F=\frac{G M m x}{\left(x^{2}+r^{2}\right)^{\frac{3}{2}}}$$[Hint: Think of the ring as made up of many small point masses $d M ;$ sum over the forces due to each $d M,$ and use symmetry.]

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$\frac{G M m x}{\left(x^{2}+r^{2}\right)^{3 / 2}}$ From the diagram we see that it points inward towards the center of the ring.

Physics 101 Mechanics

Chapter 6

Gravitation and Newton's Synthesis

Physics Basics

Newton's Laws of Motion

Applying Newton's Laws

Gravitation

Rutgers, The State University of New Jersey

University of Winnipeg

McMaster University

Lectures

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A uniform ring of mass $M$…

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So here we have a diagram of this system here and we can say that a changing force on a changing mass would be equal to the gravitational constant times The mass divided by X squared plus r squared according to the diagram times A change in mass We confined the horizontal component and say that the the M so X would be equal to G m Co sign of theta divided by again X squared plus r squared Time's a Change mass de m. At this point, we can say that a d f g m and then so ex is going to equal the gravitational constant terms the mass dough out of my X squared plus r squared. And then, according to the diagram, we khun sake oh, sign of data is going to be equal to X divided by the square root of X squared plus r squared And so we can say again times tm So we can say that Teo Yeah, to the ex is again equaling. She acts divided by X squared plus r squared to the three halves power times again GM and we can finally ah say that if we want to integrate. We can say that Dfx equals and she, um, ex d m divided by X squared plus r squared to the three house power. And now we can integrate and thank the f of X. And then here we're going to interest the other side and se gi. And then at that point, we can simply solved the integration and say that the force, the horizontal force, is going to be equal to the gravitational, constant times. The large mass terms the second mass time's acts divided by X squared plus R squared quantity to the three halves power. And this the direction would be in words towards center of rain. So this would be our final answer. That is the end of the solution. Thank you for watching.

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