2.) Recently during a weekend part at the fraternity (frat name 1), a keg of swe
ID: 2135730 • Letter: 2
Question
2.) Recently during a weekend part at the fraternity (frat name 1), a
keg of sweetened iced tea was donated by some friends from
(other frat name). 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?
(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
function of y, the height of the remaining iced tea.
(e) Find y when the com reaches its lowest point.
Explanation / Answer
# Keg filled with iced tea can be treated as a system. the system contains the keg whose COM is constant with respect to time and the second thing is iced tea whose COM depends upon the amount of iced tea remaining and will be equal to half of height remaining.
# Consider a system of masses m1 and m2 with COM y1 and y2 respectively.
The equivalent COM for the system is given by (m1*y1 + m2*y2)/(m1+m2)
a) Before the holes were drilled, the keg was full of iced tea.
mass of keg = m1 = 7 kg
COM of keg = y1 = 0.3 m
mass of tea kg - 7 kg = 10.7 kg
COM of tea = 0.6m/2 = 0.3 m
COM of the system = (m1*y1 + m2*y2)/(m1+m2)
= (7*0.3 + 10.7*0.3)/(7+10.7)
= 0.3 m
b) After all tea has drained out:
mass of keg = m1 = 7 kg
COM of keg = y1 = 0.3 m
mass of tea kg
COM of tea = 0m/2 = 0 m
COM of the system = (m1*y1 + m2*y2)/(m1+m2)
= (7*0.3 + 0*0)/(7 + 0)
= 0.3 m
c)Initial COM is at 0.3 m. As tea starts draining out, the COM goes below and after a while starts rising and reaches 0.3 m again once whole tea has drained out.
d) Consider a system of masses m1 and m2 with COM y1 and y2 respectively.
The equivalent COM for the system is given by (m1*y1 + m2*y2)/(m1+m2)
COM of keg remains constant i.e 0.3 m all the time.
weight of keg = 7kg
COM of tea = half the height of remaining tea.
Suppose at some random instant of time, y height of tea is remaining.
weight of tea remaining = 10.7 X(y/0.6) = 17.833y
COM for tea = 0.5y
COM of the system as a function of y = (7*0.3 + 17.833y*0.5y)/(7 + 17.833y)
e) When COM reaches the lowest point, differential of COM with respect to y will be equal to zero.
First differentiating the COM function w.r.t y and then equating to Zero, we get an equation of the form :
107y^2 + 84y - 25.2 = 0
The roots are 0.2316m and -0.107m
y cannot be negative. Thus neglecting the negative values, we get y = 0.2316 m when the COM reaches its lowest point.
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