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(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$\quad$ andremained stuck. By what factor would this affect the rotational frequency of the Earth $(1$ rev/day $)$ ?

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Physics 101 Mechanics

Chapter 11

Angular Momentum; General Rotation

Moment, Impulse, and Collisions

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

Simon Fraser University

University of Sheffield

Lectures

02:21

In physics, rotational dynamics is the study of the kinematics and kinetics of rotational motion, the motion of rigid bodies, and the about axes of the body. It can be divided into the study of torque and the study of angular velocity.

04:12

In physics, potential energy is the energy possessed by a body by virtue of its position relative to others, stresses within itself, electric charge, and other factors. The unit for energy in the International System of Units is the joule (J). One joule can be defined as the work required to produce one newton of force, or one newton times one metre. Potential energy is the energy of an object. It is the energy by virtue of an object's position relative to other objects. Potential energy is associated with restoring forces such as a spring or the force of gravity. The action of stretching the spring or lifting the mass is performed by a force which works against the force field of the potential. The potential energy of an object is the energy it possesses due to its position relative to other objects. It is said to be stored in the field. For example, a book lying on a table has a large amount of potential energy (it is said to be at a high potential energy) relative to the ground, which has a much lower potential energy. The book will gain potential energy if it is lifted off the table and held above the ground. The same book has less potential energy when on the ground than it did while on the table. If the book is dropped from a height, it gains kinetic energy, but loses a larger amount of potential energy, as it is now at a lower potential energy than before it was dropped.

00:59

11. Suppose a 5.8 x 10l kg…

02:56

Suppose 1*1010 kg meteorit…

03:07

The mass of earth is 5.97 …

01:31

Calculate the angular mome…

04:56

(a) What is the rotational…

05:33

Find the rate at which the…

02:43

Earth is approximately a s…

04:18

Suppose a uniform spherica…

03:14

Assume the Earth is a unif…

02:23

The earth has a radius of …

03:45

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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