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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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.

Ask a homework question - tutors are online