Recently during a weekend party a fraternity , a keg of sweetened iced tea was d
ID: 2133872 • Letter: R
Question
Recently during a weekend party a fraternity , a
keg of sweetened iced tea was donated by some friends.
However, a tap for the keg could not be located. Being
resourceful, the partygoers drilled holes on the top and
bottom of the keg in order to drain the keg of its contents.
However, an argument about the center of mass (com) of the
keg broke out and the hosts set about to calculate the com of
the keg as a function of the height of the iced tea in the keg.
The hosts were able to determine that the keg had a mass of
7.00 kg, and was in the shape of a cylinder of height L=0.600 m. The total mass of iced tea
before the holes were drilled was 17.7 kg.
(a) What was the com, h, of the keg and its contents before the holes were drilled? (b) What was the com, h, of the keg after all of the iced tea had drained out?
(e) Find y when the com reaches its lowest point.
(c) Describe qualitatively what happens to the com, h, of the keg and iced tea combination
as the iced tea is drained from the keg.
(d) Starting from the definition of the com, write down an expression for the com as a
(e) Find y when the com reaches its lowest point.
Explanation / Answer
Ycom = (m1 Y1 + m2 Y2) / (m1 + m2) equation for center of mass
The center of mass of both the tea and keg are obviously at
Ycom = .3 m
After the keg is empty the center of mass is again at Y = .3 m
As the tea is being drained
Ycom = (7 * .3 + Mt * Y/2 ) / (7 + Mt)
Note: Mt = 17.7 * Y / .6 the mass of tea remaining when
the height of the tea in the keg is at height Y (the center of mass of
the tea will be at Y /2)
The com will start at .3 m and lower as the tea drains and then will
rise back to .3 m as the last of the drains
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