Recently during a weekend party at a fraternity, a keg of sweetened ice tea was
ID: 2134515 • Letter: R
Question
Recently during a weekend party at a fraternity, a keg of sweetened ice 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 bottum of the keg in order to drain the keg of it's contents. However, an argument about the center of mass (com) of the keg broke out and the hosts et 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 has a mass of 7.00 kg, and was in shape of a cylinder of height L=0.600m. The total mass of the ice tea before the holes were drilled were 17.7kg
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 the iced tea is drained out
c) Describe quantatively what happens to the com, h, of the keg and iced tea combination as the ice 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
a) at the centre of the cylinder .3 m above base and at its centre
b) com will remain same if the cylinder is completely filled with ice tea ie .3m above base and at its centre
c) when ice tea starts draining the com comes down fron centre of the cylinder and keeps moving down untill it hits a critical point. after the critical point com again moves up till all ice tea is drained . After that com will again be at centre.
d) total hieght= .6m total weight= 17.7
remaining height = y
remaining weigt of ice tea= y/.6 *17.7 M1
weight of keg=7 kg M2
com of M1 ie ice tea= y/2
com of M2 ie keg= .3
total com= (M1 *com1 + M2*com2)/ (M1+M2)
= ( y/.6 *17.7 *y/2 + 2.1) /( y/.6 *17.7+7)
total com = (14.75y^2 +2.1) / (29.5y +7)
e) lowest y = when mass of ice tea = mass of keg
ie when
y/.6 *17.7 =7
y= .237 metres above base
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