PHYS_2014_Lecture_20

PHYS_2014_Lecture_20 - Fall 2010 Oklahoma State University...

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Unformatted text preview: Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 20, Slide 1 Torque: dL r F dt τ = × = a a a a For a rigid body rotating about a fixed axis: = I τ α Lecture 20: More on Rotational Motion L r p L I ω = × = a a a Angular Momentum is conserved in the same way as Linear Momentum and Energy = a a i f L L Angular Momentum: Where I is the Moment of Inertia: 2 = ∑ n n I m r Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 20, Slide 2 Two asteroids, one with a mass of 1.6 × 10 14 kg and the other with a mass of 9.2 × 10 13 kg, separated by a distance of 3.8 × 10 5 m, are rotating about their common center of mass with a period of 2.2 h. If the distance between the two asteroids is suddenly decreases to 1.9 × 10 5 m, what will be the period of the asteroids’ rotation? Solution: Before: After: r i = 3.8 × 10 5 m m 1 = 1.6 × 10 14 kg m 2 = 9.2 × 10 13 kg r CMi r f = 1.9 × 10 5 m r CMf Conservation of Angular Momentum: i f i i f f L L I I ω ω = = Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 20, Slide 3 r i = 3.8 × 10 5 m m 1 = 1.6 × 10 14 kg m 2 = 9.2 × 10 13 kg r CMi r f = 1.9 × 10 5 m r CMf i i f f I I ω ω = Conservation of Angular Momentum: We can find ω from the period T: 2 2 i T I π ω π = 2 f i I T π = f f f i I T T T I → = Need to find the Moments of Inertia I i and I f : 2 n n I m r = ∑ where m n is the mass of the n th part of the total system and r n is the distance of the n th part from the center of mass, r CM . That means we need to find r CM . Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 20, Slide 4 r i = 3.8 × 10 5 m m 1 = 1.6 × 10 14 kg m 2 = 9.2 × 10 13 kg r CMi r f = 1.9 × 10 5 m r CMf n n CM n m r r m = ∑ ∑ Center of Mass: Before: from center of smaller asteroid: ( ) ( ) 14 5 13 5 14 13 5 5 5 2 2 2 14 5 13 5 24 2 1.6 10 kg 3.8 10 m 9.2 10 kg 0m 2.4 10 m 1.6 10 kg 9.2 10 kg 3.8 10 m 2.4 10 m 1.4 10 m 1.6 10 kg 1.4 10 m 9.2 10 kg 2.4 10 m 8.4 10 kg m CMi i n n i r I m r I × ⋅ × + × ⋅ = = × × + × × − × = × = = × ⋅ × + × ⋅ × = × ⋅ ∑ from center of larger asteroid: Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 20, Slide 5 After: from center of smaller asteroid ( ) ( ) 14 5 13 5 14 13 5 5 4 2 2 2 14 4 13 5 24 2 1.6 10 kg 1.9 10 m 9.2 10 kg 0m 1.2 10 m 1.6 10 kg 9.2 10 kg 1.9 10 m 1.2 10 m 6.9 10 m 1.6 10 kg 6.9 10 m 9.2 10 kg 1.2 10 m 2.2 10 kg m n n CMf n f n n f m r r m I m r I × ⋅ × + × ⋅ = = = × × + × × − × = × = = × ⋅ × + × ⋅ × = × ⋅ ∑ ∑ ∑ from center of larger asteroid: 24 2 24 2 2.2 10 kg m s 2.2h 3600 902s or 0.25 h h 8.4 10 kg m f f i i f I T T I T = × ⋅ = ⋅ = × ⋅ r f = 1.9 × 10 5 m r CMf Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 20, Slide 6 As with total Angular Momentum and total Torque, the total Kinetic Energy of a rotating object will be the sum of the kinetic energies of all the bits of mass that make it up....
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PHYS_2014_Lecture_20 - Fall 2010 Oklahoma State University...

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