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Unformatted text preview: 45. (a) Eq. 435 gives T = 2πR/v = 2π(10 m)/(6.1 m/s) = 10 s. (b) The situation is similar to that of Sample Problem 67 but with the normal force direction reversed. Adapting Eq. 619, 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 67) and adapt Eq. 619 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.
 Spring '08
 Any
 Physics, Force

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