Chapter 09, Problem 008 A metal soda can of uniform composition has a mass of 0.

Question

Chapter 09, Problem 008 A metal soda can of uniform composition has a mass of 0.143 kg and is 14.8 cm tall (see the figure). The can is filled with 1.09 kg of soda. Then small holes are drilled in the top and bottom (with negligible loss of metal) to drain the soda. What is the height h of the center of mass of the can and contents (a) initially and (b) after the can loses all the soda? (c) If x is the height of the remaining soda at any given instant, find x when the center of mass reaches its lowest point (a) Number (b) Number (e) Number Units Units Units

Explanation / Answer

CoM of Soda can and soda it contians is

h = m*H/2 + M* H/2 /(m+M)

here m = 1.09 kg

M= 0.143 kg

H = 14.8 cm

so

h = H/2 = 14.8 /2 = 7.4 cm

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part B : now Consider can alone,

So that COM is H/2 = 7.4 cm above the base

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part C:

here also h = H/2 = 7.4 cm

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for x in lowest position

use the formula

x = (MH/m) (-1 + sqrt(1 + m/M)

x = (0.143* 0.148/1.09) (-1 * sqrt(1 + 1.09/0.143)

x = 3.75 cm

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