Chapter 12 - 12.1. Model: A spinning skater, whose arms are...

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12.1. Model: A spinning skater, whose arms are outstretched, is a rigid rotating body. Visualize: Solve: The speed , vr ! " where 140 cm/2 0.70 m. r "" 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.
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12.2. Model: Assume constant angular acceleration. Solve: (a) The final angular velocity is !" f 2 rad min 2000 rpm 209.4 rad/s. rev 60 s # $ % &% & ’’ ( )( ) * +* + The definition of angular acceleration gives us fi 209.4 rad/s 0 rad/s 419 rad/s 0.50 s tt , - . . -- The angular acceleration of the drill is 2 4.2 10 rad/s. / (b) ! "! " 22 11 0 rad 0 rad 419 rad/s 0.50 s 52.4 rad i 0 1 - 1 - 1 1 The drill makes rev (52.4 rad) 8.3 revolutions. 2 rad %& () *+
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12.3. Model: Assume constant angular acceleration. Visualize: Solve: (a) Since , t ar ! " find first. With 90 rpm 9.43 rad/s " and 60 rpm 6.28 rad/s, " 2 9.43 rad 6.28 rad/s 0.314 rad/s 10 s t # $% " " " $ The angular acceleration of the sprocket and pedal are the same. So &’ & ’ 22 0.18 m 0.314 rad/s 0.057 m/s t " " " (b) The length of chain that passes over the sprocket during this time is . Lr ( " $ Find : $ & ’& ’ 2 f i i 2 2 fi 1 2 1 6.28 rad/s 10 s 0.314 rad/s 10 s 78.5 rad 2 tt " ) $ ) $ % " $ " ) " The length of chain which has passed over the top of the sprocket is (0.10 m)(78.5 rad) 7.9 m L " "
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12.4. Model: Assume constant angular acceleration. Visualize: Solve: The initial angular velocity is !" i 2 rad min 60 rpm 2 rad/s. rev 60 s # $# % &% & ’’ ( )( ) * +* + The angular acceleration is 2 fi 0 rad/s 2 rad/s 0.251 rad/s 25 s t $ , -- ’ - . The angular velocity of the fan blade after 10 s is ! " 2 f i 0 2 rad/s+ 0.251 rad/s 10 s 0 s 3.77 rad/s tt / - - - The tangential speed of the tip of the fan blade is ! "! " 0.40 m 3.77 rad/s 1.51 m/s t vr (b) ! " ! "! " 2 2 2 f i i 11 0 rad 25 s 0.251 rad/s 25 s 78.6 rad 22 0 / . / . / / - The fan turns 78.6 radians 12.5 revolutions while coming to a stop.
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12.5. Model: The earth and moon are particles. Visualize: Choosing 0 m E x ! sets the coordinate origin at the center of the earth so that the center of mass location is the distance from the center of the earth. Solve: " #" # " #" # 24 22 8 E E M M cm 24 22 EM 6 5.98 10 kg 0 m 7.36 10 kg 3.84 10 m 4.67 10 m m x m x x mm $ % $ $ % !! % $ % $ !$ Assess: The center of mass of the earth-moon system is called the barycenter, and is located beneath the surface of the earth. Even though E 0 m x ! the earth influences the center of mass location because E m is in the denominator of the expression for cm . x
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12.6. Visualize: Please refer to Figure EX12.6. The coordinates of the three masses AB , , mm and C m are (0 cm, 0 cm), (0 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)(0 cm) (300 g)(10 cm) 5.0 cm (100 g 200 g 300 g) (100 g)(0 cm) (200 g)(10 cm) (300 g)(0 cm) 3.3 cm (100 g 200 g 300 g) m x m x m x x m m m m y m y m y y m m m ! ! ! ! " " " ! ! ! ! ! ! ! ! " " " ! ! ! !
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12.7. Visualize: Please refer to Figure EX12.7. The coordinates of the three masses AB , , mm and C m are (0 cm, 10 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 (200 g)(0 cm) (300 g)(10 cm) (100 g)(10 cm) 6.7 cm (200 g 300 g 100 g) (200 g)(10 cm) (300 g)(10 cm) (100 g)(0 cm) 8.3 cm (200 g 300 g 100 g) m x m x m x x m m m m y m y m y y m m m ! ! ! !
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This note was uploaded on 02/19/2010 for the course MTHSC 104 taught by Professor Simms during the Spring '10 term at Clemson.

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Chapter 12 - 12.1. Model: A spinning skater, whose arms are...

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