Chapter 8237P8.51m=mass of pumpkinR=radius of silo topFma nmgmvRrr∑=⇒−=−cosθ2When the pumpkin first loses contact with the surface, n=0.Thus, at the point where it leaves the surface: vRg2=cos.FIG. P8.51Choose Ug=0 in the =°90 0.plane. Then applying conservation of energy for the pumpkin-Earthsystem between the starting point and the point where the pumpkin leaves the surface givesKU KUmvmgRmgRfgfigi+=++1202cosUsing the result from the force analysis, this becomes12mRgmgRmgRcoscosθθ, which reduces tocos=23, and gives ==°−cos.123482bgas the angle at which the pumpkin will lose contact with the surface.P8.52(a)UmgRA=0 2009 800 3000 588....kgm smJ2ejaf(b)KUKUAABB+=+KKUUmgRBAAB=+−==0588. J(c)vKmBB=2205880200242...
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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 .