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**Physics for Scientists and Engineers, 6e**

Chapter 11 - Angular Momentum

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**Which of the following is equivalent to the following scalar product: (A × B) · (B × A)?**

(A × A) · (B × B) (A × B) · (A × B) -(A × B) · (A × B) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

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**This result can be obtained by replacing B × A with -(A × B), according to Equation 11.4.**

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**|A × B| is smaller than AB **

Which of the following statements is true about the relationship between the magnitude of the cross product of two vectors and the product of the magnitudes of the vectors? |A × B| is larger than AB |A × B| is smaller than AB |A × B| could be larger or smaller than AB, depending on the angle between the vectors |A × B| could be equal to AB 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

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Because of the sin θ function, |A × B| is either equal to or smaller than AB, depending on the angle θ.

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**impossible to determine**

Recall the skater described at the beginning of section 11.2 (see figure). Let her mass be m. What would be her angular momentum relative to the pole at the instant she is a distance d from the pole if she were skating directly toward it at speed v? zero mvd impossible to determine 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

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**If p and r are parallel or antiparallel, the angular momentum is zero**

If p and r are parallel or antiparallel, the angular momentum is zero. For a nonzero angular momentum, the linear momentum vector must be offset from the rotation axis.

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**impossible to determine**

Consider again the skater in question 3. What would be her angular momentum relative to the pole at the instant she is a distance d from the pole if she were skating at speed v along a straight line that would pass within a distance a from the pole? zero mvd mva impossible to determine 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

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The angular momentum is the product of the linear momentum and the perpendicular distance from the rotation axis to the line along which the linear momentum vector lies.

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**They both have the same angular momentum. impossible to determine**

A solid sphere and a hollow sphere have the same mass and radius. They are rotating with the same angular speed. The one with the higher angular momentum is the solid sphere the hollow sphere They both have the same angular momentum. impossible to determine 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

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**The hollow sphere has a larger moment of inertia than the solid sphere.**

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**is impossible to determine**

A competitive diver leaves the diving board and falls toward the water with her body straight and rotating slowly. She pulls her arms and legs into a tight tuck position. Her angular speed increases decreases stays the same is impossible to determine 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

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**The diver is an isolated system, so the product Iω remains constant**

The diver is an isolated system, so the product Iω remains constant. Because her moment of inertia decreases, her angular speed increases.

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**is impossible to determine**

Consider the competitive diver in question 6 again. When she goes into the tuck position, the rotational kinetic energy of her body increases decreases stays the same is impossible to determine 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

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As the moment of inertia of the diver decreases, the angular speed increases by the same factor. For example, if I goes down by a factor of 2, ω goes up by a factor of 2. The rotational kinetic energy varies as the square of ω. If I is halved, ω2 increases by a factor of 4 and the energy increases by a factor of 2

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