Chapter 497P4.35r=250. m,a=15 0. ms2(a)aac==°=cos..cos.30 015 03013 0o22±msejaf(b)avrc=2so vrac2130325=...m msms222v32 55 70msFIG. P4.35(c)aaatr222=+so a=−=−=2150750.2ejP4.36(a)See figure to the right.(b)The components of the 20.2 and the 22 52along the rope togetherconstitute the centripetal acceleration:ac=°−°+°=22 590 036 920 236 929 7.cos...cos..2(c)avrc=2so varc=29 71 506 67...m smm s tangent to circle2v667.m s at 36.9 above the horizontalFIG. P4.36*P4.37Let ibe the starting point and fbe one revolution later. The curvilinear motionwith 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 .