ch11-p008 - 8. Using the floor as the reference position...

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which leads to 2.7 0.7 2.7 . hRrR =− With R = 14.0 cm , we have h = (2.7)(14.0 cm) = 37.8 cm. (b) The energy considerations shown above (now with h = 6 R ) can be applied to point Q (which, however, is only at a height of R ) yielding the condition gR v g R 6 7 10 bg =+ com 2 which gives us vg R com 2 = 50 7. Recalling previous remarks about the radial acceleration, Newton’s second law applied to the horizontal axis at Q leads to () 2 com 50 7 v gR Nm m R rR r == −− which (for R r >> ) gives 42 2 50 50(2.80 10 kg)(9.80 m/s ) 1.96 10 N. 77 mg N × ≈= = × (b) The direction is toward the center of the loop. 8. Using the floor as the reference position for computing potential energy, mechanical energy conservation leads to 22 release top top com 11 2. UK U m g h m vI m g R ω ¡ + Substituting Im r = 2 5 2 (Table 10-2(f)) and = vr com (Eq. 11-2), we obtain 2 2 com com
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