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Unformatted text preview: e'01' small circle, the tension must be greater than (or eqm.i:Io) zero; t 2 2 ,I s: T = mvtop Ia - mg ~ 0, or Vtop ~ gao T htfrelore, the mechanical energy at the top is greater tH'an (or equal to) a corresponding value: E = K +U =:~mvlop + mg(2a) ~ !m(ga) + 2mga = ~mga, wherrthe zero of potential energy is the lowest point (as:lin Example 8-6). Since energy is conserved"I!E= Uo = mgi(l - cos 00), where 00 is the release aJgle (see Example 8-6 again), so a ~2E15mg = ~~(1 -cosOo). The greatest upper limit for the radius a ~orresponds to the maximum release angle, as stated ih the ,question above, since cos 90 = O.! .
:1) the subscripts 0 and 1 refer to the initial position (both masses at rest~ and the final position (after each has moved 50cm), respectively. Then Uo - Ul Kl (4 kg)(9.8 ni/i)(O.5 m) = ~(2 kg +4 kg)v2, since the .. potential energy changes only for the falling mass, and both masses have the same speed. Solving for v, we find v = 2.56 m/s. (The kinetic energy of the pulley was assumed to be negligible.) = = Problem
36. The masses shown. in Fig. 8...
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