ch06-p045 - 45. (a) Eq. 4-35 gives T = 2πR/v = 2π(10...

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Unformatted text preview: 45. (a) Eq. 4-35 gives T = 2πR/v = 2π(10 m)/(6.1 m/s) = 10 s. (b) The situation is similar to that of Sample Problem 6-7 but with the normal force direction reversed. Adapting Eq. 6-19, we find FN = m(g – v2/R) = 486 N ≈ 4.9 × 102 N. (c) Now we reverse both the normal force direction and the acceleration direction (from what is shown in Sample Problem 6-7) and adapt Eq. 6-19 accordingly. Thus we obtain FN = m(g + v2/R) = 1081 N ≈ 1.1 kN. ...
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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