HW_12_sol - Solutions for HW 12 P112 S07 1: Apply Newton's...

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Solutions for HW 12 P112 S07 1: Apply Newton’s second law to this new planet. " F = ma : Gm s m v r 2 = m v v 2 r v = 2 # r T Gm s r = 2 r T $ % & ( ) 2 T = 4 2 r 3 Gm s = 4 2 2 3 (5.79 * 10 10 m) [ ] 3 (6.67 * 10 + 11 Nm 2 kg 2 )(1.99 * 10 30 kg) = 4.14 * 10 6 s 1d 86,400 s $ % & ( ) = 47.9 days 2: The period is 0.50 s 440 = 1.14 " 10 # 3 s and the angular frequency is " = 2 T = 5.53 " 10 3 rad s. 3: 4: The foci are F1 (0 , (5) 1/2 ) and F2 (0 , -(5) 1/2 ) Major axis length is 6, minor axis length is 4.
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5: As usual, we have to make some assumptions: let’s say that the arms and hands of the skater weigh 8 kg and span 1.8 m when outstretched. When wrapped, they form a cylinder of radius 25 cm. The moment of inertia of his body is constant and can be approximated by a solid cylinder with I = 0.4 kg m 2 . The skater’s initial moment of inertia is I 1 = (0.400 kg " m 2 ) + 1 12 (8.00 kg)(1.80 m) 2 = 2.56 kg " m 2 , and his final moment of inertia is I 2 = (0.400 kg " m 2 ) + (8.00 kg)(25 # 10 $ 2 m) 2 = 0.9 kg " m 2 . Then " 2 = 1 I 1 I 2 = (0.40 rev s) 2.56 kg # m 2 0.9 kg # m 2 = 1.14 rev s. Note that conversion from to rad s is not necessary.
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This note was uploaded on 03/27/2008 for the course PHYS 1112 taught by Professor Leclair,a during the Spring '07 term at Cornell.

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HW_12_sol - Solutions for HW 12 P112 S07 1: Apply Newton's...

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