96_Physics ProblemsTechnical Physics

# 96_Physics ProblemsTechnical Physics - Chapter 4 P4.35 r =...

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Chapter 4 97 P4.35 r = 250 . m , a = 15 0 . ms 2 (a) aa c == ° = cos . . cos . 30 0 15 0 30 13 0 o2 2 ±ms ej af (b) a v r c = 2 so vr a c 2 130 325 = .. . m m s m s 22 2 v 32 5 5 70 ms FIG. P4.35 (c) aaa tr 222 =+ so a =− = = 2 150 750 . 2 e j P4.36 (a) See figure to the right. (b) The components of the 20.2 and the 22 5 2 along the rope together constitute the centripetal acceleration: a c ° + ° = 22 5 90 0 36 9 20 2 36 9 29 7 . cos . . . cos . . 2 (c) a v r c = 2 so va r c = 29 7 1 50 6 67 ... m s m m s tangent to circle 2 v 667 . m s at 36.9 above the horizontal FIG. P4.36 *P4.37 Let i be the starting point and f be one revolution later. The curvilinear motion with constant tangential acceleration is described by
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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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