10_InstSolManual_PDF_Part5 - established by the force and...

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Solve: (a) and (b) Calculate the acceleration of the elevator: gives for the elevator gives and for the pulley gives With this becomes for the counterweight gives and and Reflect: The tension in the cable must be different on either side of the pulley in order to produce the net torque on the pulley required to give it an angular acceleration. The tension in the cable attached to the elevator is greater than the weight of the elevator and the elevator accelerates upward. The tension in the cable attached to the counterweight is less than the weight of the counterweight and the counterweight accelerates downward. 10.14. Set Up: The free-body diagram for the disk is shown in Figure 10.14. and For a uniform disk Use the coordinates shown and take clockwise rotation to be positive. Figure 10.14 Solve: (a) Apply to the motion of the center of mass: and Apply to the rotation about the center of mass: and Note that is not equal to but is twice that. (b) and as before. In this case but that relation is
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Unformatted text preview: established by the force and torque equations and not by a restriction on the motion. a 5 a cm / R , a cm 5 25.0 m / s 2 a 5 F MR 5 167 rad / s 2 . I cm 5 MR 2 a cm / R a a 5 2 1 100.0 N 2 1 4.00 kg 21 0.150 m 2 5 333 rad / s 2 FR 5 A 1 2 MR 2 B a FR 5 I cm a g t 5 I a a cm 5 F M 5 100.0 N 4.00 kg 5 25.0 m / s 2 . F 5 Ma cm g F x 5 ma x a x F R a cm y I cm 5 1 2 MR 2 . F 5 100.0 N. R 5 0.150 m w 5 3.16 3 10 4 N. m 2 5 T 2 g 2 a 5 2.67 3 10 4 N 9.80 m / s 2 2 1.50 m / s 2 5 3.22 3 10 3 kg m 2 g 2 T 2 5 m 2 a g F y 5 ma y T 2 5 T 1 1 1 2 Ma 5 2.60 3 10 4 N 1 1 2 1 875 kg 21 1.50 m / s 2 2 5 2.67 3 10 4 N T 2 2 T 1 5 1 2 Ma . a 5 a / R 1 T 2 2 T 1 2 5 A 1 2 MR 2 B a . g t 5 I a T 1 5 m 1 1 a 1 g 2 5 1 2300 kg 21 1.50 m / s 2 1 9.80 m / s 2 2 5 2.60 3 10 4 N T 1 2 m 1 g 5 m 1 a g F y 5 ma y a 5 2 1 y 2 y 2 t 2 5 2 1 6.75 m 2 1 3.00 s 2 2 5 1.50 m / s 2 y 2 y 5 v y t 1 1 2 a y t 2 Dynamics of Rotational Motion 10-5...
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