ch09-p008

# ch09-p008 - 8. (a) Since the can is uniform, its center of...

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The positive root is used since x must be positive. Next, we substitute the expression found for x into h = ( MH 2 + mx 2 )/2( MH + mx ). After some algebraic manipulation we obtain (12 cm)(0.14 kg) 1.31kg 1 1 1 1 2.8 cm. 1.31 kg 0.14 kg HM m h mM §· § · =+ = + = ¨¸ ¨ ¸ ¨ ¸ ©¹ © ¹ 8. (a) Since the can is uniform, its center of mass is at its geometrical center, a distance H /2 above its base. The center of mass of the soda alone is at its geometrical center, a distance x /2 above the base of the can. When the can is full this is H /2. Thus the center of mass of the can and the soda it contains is a distance h MH mH Mm H = + + = // 22 2 bg above the base, on the cylinder axis. With H = 12 cm, we obtain h = 6.0 cm. (b) We now consider the can alone. The center of mass is H /2 = 6.0 cm above the base, on the cylinder axis. (c) As x decreases the center of mass of the soda in the can at first drops, then rises to
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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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