2014_SP08_Sol10

2014_SP08_Sol10 - 13.5. Model: Spinning skater, whose arms...

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Unformatted text preview: 13.5. Model: Spinning skater, whose arms are outstretched, is a rigid rotating body. Visualize: Solve: The speed v = r , where r = 140 cm/2 = 0.70 m. Also, 180 rpm = (180)2 /60 rad/s = 6 rad/s. Thus, v = (0.70 m)(6 rad/s) = 13.2 m/s. Assess: A speed of 13.2 m/s 26 mph for the hands is a little high, but reasonable. 13.8. Model: The turbine is a rigid rotating body. Solve: The known values are i = 3600 rpm = (3600)(2 )/60 = 120 rad/s, t i = 0 s, t f = 10 min = 600 s, f = 0 rad/s, and i = 0 rad. Using the rotational kinematic equation f = i + ( t f t i ), we get 0 rad = (120 rad/s) + (600 s 0 s). Thus, =- 0.628 rad/s 2 . Now, 2 f i i f i f i 2 2 1 ( ) ( ) 2 1 0 rad (120 rad/s)(600 s 0 s) ( 0.628 rad/s )(600 s 0 s) 2 113,100 rad 18,000 rev t t t t = +- +- = +- +-- = = Assess: 18,000 revolutions during 10 minutes when the starting angular velocity is 3600 rpm is reasonable. 13.10. Visualize: Please refer to Figure Ex13.10. The coordinates of the three masses m A , m B , and m C are (0 cm, 0 cm), (10 cm, 10 cm), and (10 cm, 0 cm), respectively. Solve: The coordinates of the center of mass are A A B B C C cm A B C A A B B C C cm A B C (100 g)(0 cm) (200 g)(10 cm) (300 g)(10 cm) 8.33 cm (100 g 200 g 300 g) (100 g)(0 cm)...
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This homework help was uploaded on 04/17/2008 for the course PHYS 2014 taught by Professor Nandi during the Spring '08 term at Oklahoma State.

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2014_SP08_Sol10 - 13.5. Model: Spinning skater, whose arms...

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