Question
A cylinder with diameter 2R and height H is completely filled with water. A solid sphere of radius R and density same as that of water is fully immersed in it to the bottom. Find the work done to bring the sphere just out of water surface.
A cylinder with diameter 2R and height H is completely filled with water. A solid sphere of radius R and density same as that of water is fully immersed in it to the bottom. Find the work done to bring the sphere just out of water surface.
Answer
Refer to the figure below.
For the figure with calculations in brief refer to the figure in the link
http://www.flickr.com/photos/52771834@N00/4888459981/
(Click on magnifier for better visibility.)
1)
Method base on average force x displacement
No work is done to bring the spere from the fully
submerged position at the bottom of the tank to the
position where its top is touching the water surface.
Volume of the sphere = (4/3) πR^3
Volume of water surrounding the sphere in the tank
= π * (2R)^2 * 2R - (4/3)πR^2
= (20/3) πR^3
Refer to the figure below.
For the figure with calculations in brief refer to the figure in the link
http://www.flickr.com/photos/52771834@N00/4888459981/
(Click on magnifier for better visibility.)
1)
Method base on average force x displacement
No work is done to bring the spere from the fully
submerged position at the bottom of the tank to the
position where its top is touching the water surface.
Volume of the sphere = (4/3) πR^3
Volume of water surrounding the sphere in the tank
= π * (2R)^2 * 2R - (4/3)πR^2
= (20/3) πR^3
When the sphere is out, this water will be contained in the height h from the level touching the bottom of the sphere given by
π(2R)^2 * h = (20/3) πR^3
=> h = (5/3) R.
=> water level falls by 2R - (5/3)R = R/3
=> displacement of the sphere = 2R - R/3 = 5R/3
The average force on the sphere is calculated as under.
Consider two cross-sections of the sphere at equal distances "x" above and below its center.
Force on the cross-section at a height "x" above its center when it is level with water surface
= Vdg - V'dg, where
V = volume of the sphere,
V' = volume of water displace by the sphere, and
d = density of the sphere
Force on the cross-secion at a height "x" below its center when it is level with water surface
= Vdg - (V - V')dg
=> average force
= (1/2)[Vdg - V'dg + Vdg - (V - V')dg]
= 1/2)Vdg
= mg/2
Considering all complementary cross-sections of the sphere equidistant from its center, the average force during its displacement out of water
= (1/2)mg
=> work done = (1/2)mg * (5R/3) = (5/6) mgR
The average force on the sphere is calculated as under.
Consider two cross-sections of the sphere at equal distances "x" above and below its center.
Force on the cross-section at a height "x" above its center when it is level with water surface
= Vdg - V'dg, where
V = volume of the sphere,
V' = volume of water displace by the sphere, and
d = density of the sphere
Force on the cross-secion at a height "x" below its center when it is level with water surface
= Vdg - (V - V')dg
=> average force
= (1/2)[Vdg - V'dg + Vdg - (V - V')dg]
= 1/2)Vdg
= mg/2
Considering all complementary cross-sections of the sphere equidistant from its center, the average force during its displacement out of water
= (1/2)mg
=> work done = (1/2)mg * (5R/3) = (5/6) mgR
2)
Alternate method based on PE:
Increase in PE of the sphere
= mg * (2R - R/3) = (5/3) mgR
Decrease in PE of water:
Mass of water surrounding the sphere in the cylinder = (20/3) πR^3 * d = 5mg
Original average height of this water from the surface of water at the bottom of the sphere when the top surface of the sphere is touching the water
= R
Final average height of this water
= (1/2) (2R - R/3) = 5R/6
Decrease in PE of water
= mass of water x fall in height
= 5mg * (R - 5R/6) = 5mgR/6
Net gain of PE
= mgR (5/3 - 5/6)
= (5/6)mgR
= work done.
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