Problem 48 Hard Difficulty

Integrated Concepts
The axis of Earth makes a $23.5^{\circ}$ angle with a direction perpendicular to the plane of Earth's orbit. As shown in Figure 10.41 , this axis precesses, making one complete rotation in $25,780$ y.
(a) Calculate the change in angular momentum in half this time.
(b) What is the average torque producing this change in angular momentum?
(c) If this torque were created by a single force (it is not) acting at the most effective point on the equator, what would its magnitude be?

Answer

(a) $\Delta L = 5.64 \times 10 ^ { 33 } \mathrm { kg } \cdot \mathrm { m } ^ { 2 } / \mathrm { s }$
(b) $\tau = 1.38 \times 10 ^ { 22 } \mathrm { N }$ in
(c) $F = 2.17 \times 10 ^ { 15 } \mathrm { N }$

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Physics 101 Mechanics

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Chapter 10

Rotational Motion and Angular Momentum

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Problem 48

Video Transcript

So here we can say that the change in angular momentum and half its time is calculated as the angular momentum l equaling the moment of inertia times the angular velocity. Now for part A. We know that from ah, from the formula to find the change in angular momentum, we can say doubt l The change in angular momentum would be two times the angular momentum l multiplied by sine of 23 0.5 degrees. And so we can say that here the change in angular momentum would be equaling two times 7.8 times 10 to the 33rd kilograms, meters squared per second, multiplied by sine of 23.5 degrees. And this is equaling 5.64 times 10 to the 33rd units kilograms, meters squared per second. And so this would be your answer for part A For part B. We're going to find maybe the question asking, what is the average Twerk producing this change and angular momentum will weaken first for part B, convert the time into seconds. So 25,000 780 years and then we're going to This would be equal to Of course, 25,780 years. And then for every there are 3.16 times, 10 to the seventh seconds for every year. And so this is gonna be equal to 8.14 times 10 to the 11th seconds. Therefore, the change in time, Delta T would be equaling 4.7 times 10 to the 11 seconds essentially just half of the half of 25,000, 780 years. And so the torque is gonna be equal to the change in angular momentum. Divide about the change in time. And so this would be equaling toe 5.65 times 10 to the 33rd kilograms, meters squared per second, divided by the time that we just found 4.7 times 10 to the 11th seconds and the torque, the required torque would be 1.39 times 10 to the 22nd Newton meters. So this would be our answer for part B and then four parts. See, we want to find the force and you know the torque. It's simply equal to the force exerted divided by our are in this case would be the radius of the earth and so we could find the force. This would be equal to the torque divided by the radius of the earth. And this would be equaling toe 1.39 times 10 to the 22nd Newton meters, divided by the radius of the earth. 6.38 times, 10 to the sixth meters. And so the force would be equaling 2.17 times 10 to the 15th Newton's. This would be our final answer for part C. That is the end of the solution. Thank you for watching.

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