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Unformatted text preview: 10.23: n = mg cos mg sin  s mg cos = ma g (sin  s cos ) = a (eq.1) n and mg act at the center of the ball and provide no torque. 2 = f = s mg cos R; I = 5 mR 2
2 = I gives s mg cos = 5 mR 2 2 No slipping means = a R , so s g cos = 5 a (eq.2) We have two equations in the two unknowns a and s . Solving gives 5 2 2 a = 7 g sin and s = 7 tan = 7 tan 65.0 = 0.613 2 b) Repeat the calculation of part (a), but now I = 2 mR . 3 a = 3 g sin and s = 2 tan = 2 tan 65.0 = 0.858 5 5 5 The value of s calculated in part (a) is not large enough to prevent slipping for the hollow ball. c) There is no slipping at the point of contact. ...
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 The Hollow, mg cos

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