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(1I) Suppose the man at $\mathrm{B}$ in Fig. 26 t…

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(III) Let $\vec{\mathbf{g}}^{\prime}$ be the effective acceleration of gravity at a point on the rotating Earth, equal to the vector sum of the "true" value $\vec{g}$ plus the effect of the rotating reference frame $\left(m \omega^{2} r\right.$ value $\vec{\mathbf{g}}$ plus the effect of the rotating reference frame $\left(m \omega^{2} r\right.$ term). Sce Fig. $42 .$ Determine the magnitude and direction of $\vec{g}^{\prime}$ relative to a radial line from the center of the Earth $(a)$ at the North Pole, $(b)$ at a latitude of $45.0^{\circ}$ north, and $(c)$ at the equator. Assume that $g$ (if $\omega$ were zero) is a constant 9.80 $\mathrm{m} / \mathrm{s}^{2}$ .

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Physics 101 Mechanics

Physics for Scientists and Engineers with Modern Physics

Chapter 11

Angular Momentum; General Rotation

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Video Transcript

This is the diagram of the system and for part A we consider it that the North Pole uh, the factor, um, omega squared times are the mass times the angular, velocity squared times are is in equal zero. And so there's no effect from the rotating reference frame. And so G prime would be simply pulling g minus omega squared R, which we know to be equaling G minus zero. And so this would simply be equal to G Oh, our 9.80 meters per second squared. The direction would be inward. Or we can say radio Lee radiantly inward and then for part B. To find the direction relative to a radial line, we have to coordinate the coral coordinate system tent along the tangential X and then the radio. Why so why we're going to say was always is always along the radius or essentially always towards the center of the earth. And then we can say that at a specific latitude fi five, describing again the latitude Ah, the true gravity will point purely in the positive y direction. So we can say that G prime would be a sorry the vector G would be equal to the magnitude of G times J hat, and we can then label the effect of the rotating reference frame as G. We can say the, uh, g sub rotate this would r o T. This would be the effect of rotating reference frame. And so, essentially, the effect of the ah, the effect of this vector here can be found by decomposing it along the axes on. And we know that the new our would be equaling the radius of the earth Times co sign of the latitude Fi. And so we can then say that, uh, G some rotate would be equalling our omega squared sine if I all right hat and then minus our Omega squared times co sign of Fei J hat again. We're just looking. We're getting this from the diagram and this would be equaling the radius of the earth Omega squared Times Co sign If I sign If I I hat minus the radius of the Earth Omega Squared co sign squared of fi J hat And so we can then say that the true the true gravity we can say G prime would be equaling G plus geese, a borrowed tea and this would be equaling the radius of the Earth Omega Squared Co sign of Phi Sine. If I I had, this would be the only component in the ex direction and then plus G minus our sub e omega squared co sign squared of Fi J hat. And so here we know that four part B fada the angle of deflection here can be found from the components of G prime. So we can say That's Veda with the equaling arc tanne of G Prime sub X, divided by G prime sub y. We can then say fate that is equaling arc 10 of are some e omega squared co sign If I sign If I divided by G minus the radius of the earth Omega squared co sign squared of five and we can solve So we can say Fada is equaling are 10. This is gonna be quite large. This would be the radius of the earth 6.38 times 10 to the sixth meters, multiplied by two pi radiance divided by one day or 86,400 seconds. Quantity squared, multiplied by co sign of 45 degrees sign of 45 degrees again, 45 degrees would be the latitude in this case, and then this would be divided by 9.80 meters per second squared minus 6.38 times 10 to the sixth meters, most supplied by two Powerade Ian's divided by 86,400 seconds, quantity squared and this would be times co sign squared of 45 degrees close parentheses. This is after solving everything. This is essentially going to 0.988 degrees. This would be the angle of deflection. And so we can find the magnitude of the true for party the the magnitude of of G prime. This would simply teeth from Pythagorean theorem. So this would be G prime sub X squared plus g sub y crime squared. This is equaling. Um, you're going to get these values once you saw for this so solve, uh, solving this first and then solving this will give you the X and Y components. And so this would be 1.687 times 10 to the negative second meters per second squared quantity squared plus 9.78 three meters per second squared, quantity squared and so g prime would equal 9.78 meters per second squared again. This would be the magnitude and then oh, to be point 988 degrees south from and inward radio line. This would be our full answer for Part B and then for part C that we're trying to evaluate this at the equator and so at the equator. For part C, we can say that G prime would be equaling G minus omega squared R and again at the equator. The effect of the rotating reference frame is directly opposite to the true acceleration due to gravity. And so this would be equaling G minus omega squared times the radius of the earth. And so this would be 9.80 meters per second squared minus. And then this would be to pie radiance divided by 86,400 seconds. Quantity squared, multiplied by the radius of the earth, 6.38 times 10 to the sixth meters. And so G prime is equaling 9.77 meters per second squared. This would be in word. Or we can say begin radial e inward or inward along a radio line. That is the end of the solution. Thank you for watching

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Problem

(1I) Suppose the man at $\mathrm{B}$ in Fig. 26 t…

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Physics for Scientists and Engineers with Modern Physics

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Video Transcript

Douglas C. Giancoli

Physics for Scientists and Engineers with Modern Physics

Liev Birman

Marshall Styczinski

Zachary Mitchell

Aspen Fenzl

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Physics for Scientists and Engineers with Modern Physics

Liev Birman

Marshall Styczinski

Zachary Mitchell

Aspen Fenzl

Douglas C. GiancoliPhysics for Scientists and Engineers with Modern Physics

Liev Birman

Marshall Styczinski

Zachary Mitchell

Aspen Fenzl

Douglas C. Giancoli

Physics for Scientists and Engineers with Modern Physics