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 τ θ θ θ θ = − F H G I K J F H G I K J = − + m g L m g L g m L m L h h h m m m h h h m m m 2 2 2 sin sin sin sin b g If we take t = 0 at 12 o’clock, then the angular positions of the hands at time t are θ ω h h t = , where ω π h = 6 rad h and θ ω m m t = , where ω π m = 2 rad h Therefore, τ π π = − F H G I K J + L N M O Q P 4 90 60 0 2 70 6 100 2 . . . sin sin m s kg m kg 4.50 m 2 a f a f t t or τ π π = − F H G I K J + L N M O Q P 794 6 2 78 2 N m sin . sin t t , where t is in hours. (a) (i) At 3:00, t = 3 00 . h, so τ π π = − F H G I K J + L N M O Q P = 794 2 2 78 6 794 N m N m sin . sin (ii) At 5:15, t = + = 5 15 60 5 25 h h h . , and substitution gives:
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