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(III) $\mathrm{A} 230$ -kg beam 2.7 $\mathrm{m}$ …

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Problem 49 Hard Difficulty

(II) Suppose a $5.8 \times 10^{10} \mathrm{kg}$ meteorite struck the Earth at the equator with a speed
$v=2.2 \times 10^{4} \mathrm{m} / \mathrm{s},$ shown in Fig. 37$\quad$ and
remained stuck. By what factor would this affect the rotational frequency of the Earth $(1$ rev/day $)$ ?

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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

can apply the angular Momenta. You comply the conservation of angular momentum to the earth meteorite system, Um, before and after the collision. So we can say that the initial angular momentum equals the final angular momentum. And this would be well to the moment of inertia for the earth multiplied by the initial angular velocity minus the mass of the media or times. Uh, they're the mass of the earth. My apologies, Times the radius of the earth times the sign 45 degrees. And then this would be equaling the moment of inertia for the Earth, plus the moment of inertia for the meteorite multiplied by the final angular velocity. And so we can say that the final angular velocity would be equaling the moment of inertia for the Earth times the initial angular velocity minus the mass of the Earth times the radius of the earth, the sign of a 45 degrees. This is gonna be divided by the moment of inertia for the Earth, plus the moment of inertia for the meteorite, this would be equaling. We can say we can model the earth of this fear, and this would be 2/5 times the mass of the earth radius of the earth squared times the initial angular velocity minus, um, our sub e be signed 45 degrees, and so we can. Then this would be divided by 2/5 times. I'm sub e r c e squared plus m r sub d squared. I said that lower case M was the mass of the meat of the mass of the earth. My apologies, Lower case M is actually going to be the mass of the meteorite. My apologies, because initially we have to account for the mass of the meteorite because it is also going at a linear speed towards earth. So my apologies, lower case them is indeed the master of the meteorite. And so we can say then that omega final over a mega initial would be equaling. This would be 2/5 times the mass of the earth radius of the earth squared minus the mass of the meteorite times the radius of the earth multiplied by the initial. Rather the linear velocity of the meteorite, divided by the additional angular velocity multiplied by one over radical, too. And then this would be divided by the radius of the earth. squared, multiplied by 2/5 times the mass of the Earth plus the mass of the meteorite. And so we can say that the final angular velocity divided by the initial angular velocity would be equaling the radius of the earth squared, multiplied by 2/5 times the mass of the earth minus the mass of the meteorite times a linear speed over radical to omega initial our sub e. And then this entire term would be divided by the radius of the earth squared, multiplied by and plus 2/5 times the mass of the Earth. Of course, these are going to the radius of the earth squared terms gonna cancel out for both the numerator and the denominator. So we can then say that here the change in the angular velocity divided by the initial angular velocity. This would be equaling the final angular velocity minus the initial angular velocity over the initial angular velocity. And this would equal the final angular velocity departed by the initial angular velocity minus one. And so we can say that the change in the angular velocity divided by the initial angular velocity would be equally 2/5 times the mass of the earth minus, um V over radical, two times the initial angular velocity times the radius of the earth. And then this term would be divided by the mass of the meteorite, plus 2/5 times the mass of the earth minus one. And so this would be equal to negative. This would be negative, the over radical, two times the initial angular velocity times a radius of the earth plus one and then divided by one plus 2/5 times the mass of the earth divided by the mass of the media re. And at this point, we can solve. So Delta Omega over Omega initial, this is gonna equal. This is gonna be quite large. So for the numerator, we have 2.2 times 10 to the fourth meters per second. This would be divided by radical two multiplied by two pi over 86,400 seconds or one day radiance for a second. This would be multiplied by the radius of the earth, 6.38 times, 10 to the sixth meters. And then this would be plus one and then this would be divided by one plus two fits multiplied by the Mass of the earth, 5.97 times, 10 to the 24th kilograms divided by the mass of the meteorite. 5.8 times, 10 to the 10th kilograms. And so we find that the change of the change in the angular velocity with respect to the initial or rather over the initial velocity would be equaling negative 8.39 times 10 to the negative 13th. And this would be our ratio. That is the end of the solution. Thank you for

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Physics for Scientists and Engineers with Modern Physics

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Problem

(III) $\mathrm{A} 230$ -kg beam 2.7 $\mathrm{m}$ …

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Physics for Scientists and Engineers with Modern Physics

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Watch More Solved Questions in Chapter 11

Video Transcript

Douglas C. Giancoli

Physics for Scientists and Engineers with Modern Physics

Christina Krawiec

Andy Chen

Farnaz Mohseni

Aspen Fenzl

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

Physics for Scientists and Engineers with Modern Physics

Christina Krawiec

Andy Chen

Farnaz Mohseni

Aspen Fenzl

Douglas C. GiancoliPhysics for Scientists and Engineers with Modern Physics

Christina Krawiec

Andy Chen

Farnaz Mohseni

Aspen Fenzl

Douglas C. Giancoli

Physics for Scientists and Engineers with Modern Physics