Problem 76 1 solution ii 0 fe m ifh the iik jcn

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: over? Solution The conservation of mechanical energy between points Ii 0 and P requires Ko + Uo = K + U, or mgR = ~mv2 + :I mgy (since the bug starts from rest). Thus, v2 2gx (R - y) = 2gd. The bug leaves the surface when the normal force has decreased to zero. From the radial component of Newton's second law, mgcosO - N = . mv2 / R, so this occurs when N = mg cos () - mv2j R 0, or v2 = gRcos(} gy = g(R- d). Combining these results, we find v2 = 2gd = g(R - d), or d ~R. lover, the center of the brick must lie to the $fthe vertical through its 8-cm edge, as shown. citential energy of the brick would have to be dbyUb - Ua mg(Yb - Ya)' If we assume that is conserved between a and b, this is equal to iurmum kinetic energy sought (Ka - Kb = Ub if Kb - 0). The numerical value is just the = = K;:,in. = = = 142 CHAPTER 8 Solution The condition for the rope to be taut going around the rock (at 01 in Fig. 8-13) is 0 ~ T mgcos (h + mvUa (this follows from the radial component of Newton's second iaw), or v~ ~ -gacosOl' Th...
View Full Document

This note was uploaded on 07/27/2009 for the course PHYSICS 101 taught by Professor Wormer during the Spring '08 term at NYU Poly.

Ask a homework question - tutors are online