311_Physics ProblemsTechnical Physics

311_Physics ProblemsTechnical Physics - Chapter 10 P10.74...

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Chapter 10 313 P10.74 Consider the total weight of each hand to act at the center of gravity (mid-point) of that hand. Then the total torque (taking CCW as positive) of these hands about the center of the clock is given by τθ θ =− F H G I K J F H G I K J + mg L L g mL m L h h hm m mh h h m m m 22 2 sin sin sin sin bg If we take t = 0 at 12 o’clock, then the angular positions of the hands at time t are θω hh t = , where ω π h = 6 rad h and mm t = , where ωπ m = 2 r adh Therefore, τ F H G I K J + L N M O Q P 490 600 270 6 100 2 .. . s i n s i n m s kg m kg 4.50 m 2 af t t or F H G I K J + L N M O Q P 794 6 278 2 N m sin . sin t t , where t is in hours. (a) (i) At 3:00, t = 300 . h , so F H G I K J + L N M O Q P 794 2 6 794 Nm sin . sin (ii) At 5:15, t =+ = 5 15 60 525 h h h . , and substitution gives: 2510 N m (iii) At 6:00, =⋅ 0 N m (iv) At 8:20,
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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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