181_Physics ProblemsTechnical Physics

181_Physics ProblemsTechnical Physics - Chapter 6 P6.63(a...

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Chapter 6 183 P6.63 (a) The mass at the end of the chain is in vertical equilibrium. Thus Tm g cos θ = . Horizontally a mv r r sin == 2 r r =+ + = 250 400 280 400 517 .s i n . i n. . . af m m m Then a v r = 2 . m . By division tan . a g v g r 2 vg v 2 517 980 280 519 ° = .t a n . a n . . a f ms 22 (b) g cos = T mg ° = cos .. cos . 50 0 9 80 28 0 555 kg m s N 2 bg ej T R = 4.00 m l = 2.50 m r mg FIG. P6.63 P6.64 (a) The putty, when dislodged, rises and returns to the original level in time t . To find t , we use vv a t fi : i.e., −=+− g t or t v g = 2 where v is the speed of a point on the rim of the wheel. If R is the radius of the wheel,
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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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