Problem

A radio transmission tower has a mass of 80 $\mat…

08:40

Question

Answered step-by-step

Problem 79 Hard Difficulty

Competitive ice skaters commonly perform single, double, and triple axel jumps in which they rotate $1 \frac{1}{2}, 2 \frac{1}{2},$ and 3$\frac{1}{2}$ revolutions, respectively, about a vertical axis while airborne. For all these jumps, a typical skater remains airborne for For all these jumps, a typical skater remains airborne for about 0.70 s Suppose a skater leaves the ground in an "open" position (e.g. arms outstretched) with moment of inertia $I_{0}$ and rotational frequency $f_{0}=1.2 \mathrm{rev} / \mathrm{s},$ maintaining this position for 0.10 $\mathrm{s}$ . The skater then assumes a
"closed" position (arms brought closer) with moment of inertia $I,$ acquiring a rotational frequency $f,$ which is maintained for 0.50 s. Finally, the skater immediately returns to the "open" position for 0.10 s until landing (see Fig. $49 ) .$ (a) Why is angular momentum conserved during the skater's
jump? Neglect air resistance. (b) Determine the minimum rotational frequency $f$ during the flight's middle section for the skater to successffully complete a single and a triple axel. (c) Show that, according to this model, a skater must he able to reduce his or her moment of inertia in midflight by a factor of about 2 and 5 in order to complete a single and triple axel, respectively.

Video Answer

Solved by verified expert

Textbook Answer

Official textbook answer

Video by Averell Hause

Numerade Educator

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

Related Courses

Physics 101 Mechanics

Physics for Scientists and Engineers with Modern Physics

Chapter 11

Angular Momentum; General Rotation

Discussion

You must be signed in to discuss.

Top Physics 101 Mechanics Educators

Christina Krawiec

Rutgers, The State University of New Jersey

Watch More Solved Questions in Chapter 11

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

Problem 51

Problem 52

Problem 53

Problem 54

Problem 55

Problem 56

Problem 57

Problem 58

Problem 59

Problem 60

Problem 61

Problem 62

Problem 63

Problem 64

Problem 65

Problem 66

Problem 67

Problem 68

Problem 69

Problem 70

Problem 71

Problem 72

Problem 73

Problem 74

Problem 75

Problem 76

Problem 77

Problem 78

Problem 79

Problem 80

Problem 81

Problem 82

Problem 83

Problem 84

Video Transcript

during the jump. The Onley force on the skater is airborne, the only force on the skaters gravity because she is airborne. So when she is airborne, she's essentially a projectile in free fall. And so there's no there's no there's no there's not gonna be any force on the skater except for gravity. And this is acting through the center of Mass. So essentially, we can say that, um, there is no torque about the center of mass, and so we can say that angular mo mentum is conserved during the jump for Part B. We know that for a single axle, the skater must have one and 1/2 total revolutions. We know that the number of evolution's during each phase of the motion as the rotational frequency times the elapsed time. And so we can say that we also note that the rate of rotation is the same for both occurrences of the open position, so creating the relationship we have 1.2 revolutions per second multiplied by a point 10 seconds plus the frequency multiplied by 0.50 plus again, 1.2 revolutions per second multiplied by again point 10 seconds and we know that this must equal 1.5 revolutions. And so we can say that the frequency must equal 1.5 revolutions minus two times 1.2 revolutions, 0.10 seconds and then divided by 0.50 seconds. And so this is equaling approximately 2.5 revolutions per second. This would be, um, the frequency essentially for a single axle. And so the calculation would be the exact same for a triple axel. So continuing part B, we have 1.2 revolutions per second, multiplied by 0.10 seconds plus frequency triple axel times 0.50 seconds plus 1.2 revolutions per second multiplied by again 0.1 point +10 seconds. And this must now equal 3.5 revolutions because this is a triple axel. And so we find that the frequency would be equaling 3.5 evolutions, minus two times 1.2 revolutions for a second multiplied by 0.10 seconds. And then this would be divided. My 0.50 seconds. This is giving us approximately 6.5 revolutions her second. This would be our final, our second final answer for part B and then for part C, we're simply going thio upside the angular momentum, the conservation of angular ruin some to relate the moments of inertia. And so, for part C, we have moment of inertia, of the single open equals moment of inertia of the single closed. And this would give us the moment of inertia of the single opened, multiplied by the anger angular velocity of the single open equals moment of inertia. Single closed, angular velocity single closed. And so we find that the moment of inertia of the single closed, divided by the moment of inertia. For these single open, this would be Omega sub s O divided by Omega's of S C. This is equaling F sub s O divided by F. S. C. And this is equaling 1.2 revolutions for a second, divided by 1.52 revolutions per second will around at the very end. This is equal in point for seven six and we can essentially round to approximately 1/2. So this would be the ratio of the moment of inertia of the single close versus the single open. This is essentially me, meaning that the single axle moment of inertia must be reduced by a factor of two. Um, in order to do the triple axel calculation, it's the exact same thing. So you could say for part C moment of inertia for the triple axel closed position divided by the triple axel open position. This would be equal to F sub s. Sorry, have some tea. Oh, divided by f so tc and this is equaling 1.2 revolutions per second, divided by 6.52 revolutions per second. This is giving us 0.184 or we can say approximately the 1/5. So essentially, the triple axel moment of inertia must be reduced by a factor of about five in order to make the triple axel. So this would be your second final answer for part C. And this would be, ah, your first answer for part C. That is the end of the solution. Thank you for watching

Get More Help with this Textbook

Douglas C. Giancoli

Physics for Scientists and Engineers with Modern Physics

View More Answers From This Book

Find Another Textbook

Study Groups

Study with other students and unlock Numerade solutions for free.

Problem

A radio transmission tower has a mass of 80 $\mat…

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

Related Courses

Physics for Scientists and Engineers with Modern Physics

Related Topics

Discussion

Top Physics 101 Mechanics Educators

Christina Krawiec

Farnaz Mohseni

Zachary Mitchell

Meghan Miholics

Physics 101 Mechanics Courses

Recommended Videos

Watch More Solved Questions in Chapter 11

Video Transcript

Douglas C. Giancoli

Physics for Scientists and Engineers with Modern Physics

Christina Krawiec

Farnaz Mohseni

Zachary Mitchell

Meghan Miholics

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

Physics for Scientists and Engineers with Modern Physics

Christina Krawiec

Farnaz Mohseni

Zachary Mitchell

Meghan Miholics

Douglas C. GiancoliPhysics for Scientists and Engineers with Modern Physics

Christina Krawiec

Farnaz Mohseni

Zachary Mitchell

Meghan Miholics

Douglas C. Giancoli

Physics for Scientists and Engineers with Modern Physics