321_Physics ProblemsTechnical Physics

# 321_Physics ProblemsTechnical Physics - 2 toward the center...

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Chapter 10 323 P10.90 Fm a xx = reads −+ = fTm a . If we take torques around the center of mass, we can use τα = I , which reads +− = fR TR I 21 α . For rolling without slipping, α = a R 2 fR TR Ia R I Rm Tf fR m TR R m IT If f I mR T I mR R f Im RR Im R T 21 22 2 2 12 2 2 12 12 2 2 −= = −= += + = + + F H G I K J bg ej bg Since the answer is positive, the friction force is confirmed to be to the left. f n T mg FIG. P10.90 ANSWERS TO EVEN PROBLEMS P10.2 (a) 822 rad s 2 ; (b) 4 21 10 3 . × rad P10.28 1 2 2 ML P10.4 (a) 120 10 2 . × rad s; (b) 25.0 s P10.30 168 N m clockwise P10.6 226 rad s 2 P10.32 882 N m P10.8 13 7 . r ads 2 P10.34 (a) 1.03 s; (b) 10.3 rev P10.10 (a) 2.88 s; (b) 12.8 s P10.36 (a) 21 6 . kgm 2 ; (b) 3 60 . Nm ; (c) 52.4 rev P10.12 (a) 0180 . r a d s ; P10.38 0.312 (b) 810 . m s 2 toward the center of the track P10.40 104 10 3 . ×
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Unformatted text preview: 2 toward the center of the track P10.40 1 04 10 3 . J P10.14 (a) 0 605 . m s ; (b) 17 3 . rad s ; (c) 5 82 . m s ; P10.42 149 rad s (d) The crank length is unnecessary P10.44 (a) 6.90 J; (b) 8 73 . rad s; (c) 2 44 . m s; P10.16 (a) 54.3 rev; (b) 12 1 . rev s (d) 1 043 2 . times larger P10.18 0.572 P10.46 2 36 . m s P10.20 (a) 92 0 . kg m 2 ; 184 J; P10.48 276 J (b) 6 00 . m s ; 4 00 . m s ; 8 00 . m s ; 184 J P10.50 (a) 74.3 W; (b) 401 W P10.22 see the solution P10.52 7 10 2 Mv P10.24 1 28 . kg m 2 P10.54 The disk; 4 3 gh versus gh P10.26 ~10 kg m 2...
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## This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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