Problem 3.125

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Problem *3.125 [Difficulty: 3]
V = 5 m/s 
d 
d 
R= 1 m 
rT 
Given: Pail is swung in a vertical circle. Water moves as a rigid body.
Point of interest is the top of the trajectory.
Find: (a) Tension in the string
(b) Pressure on pail bottom from the water.
Solution: We will apply the hydrostatics equations to this system.
agp GG ρρ =+∇−Governing Equations: (Hydrostatic equation)
rr agr
p ρρ =+∂
∂− (Hydrostatic equation radial component)
Assumptions: (1) Incompressible fluid
(2) Rigid body motion
(3) Center of mass of bucket and water are located at a radius
of 1 m where V = rω = 5 m/s
Summing the forces in the radial direction: T− Mb Mw+( ) g⋅− Mb Mw+( )ar= where ar V2r−=
Thus the tension is: T Mb Mw+( ) V2r g−
⎛⎜⎝
⎞
⎠⋅= where: Mb 15 N⋅
s2
9.81 m⋅×
kg m⋅
N s2⋅
×= Mb 1.529 kg⋅=
and: Mw ρ V⋅= ρ
π
4
⋅ d2⋅ h⋅= Mw 999
kg
m3
⋅ π
4
× 0.4 m⋅( )2× 0.2× m⋅= Mw 25.11 kg⋅=
Now we find T: T 1.529 25.11+( ) kg⋅ 5 m
s
⋅⎛⎜⎝
⎞
⎠
2 1
1 m⋅× 9.81
m
s2
⋅−
⎡⎢⎢⎣
⎤⎥⎥⎦
× N s
2⋅
kg m⋅×= T 405 N⋅=
r
Vg
r
p 2ρρ −=−∂
∂− ⎟⎟⎠
⎞
⎜⎜⎝
⎛ −=∂
∂ g
r
V
r
p 2ρIf we apply this information to the radial hydrostatic equation we get: Thus:
If we assume that the radial pressure gradient is constant throughout the water, then the pressure gradient is equal to:
pr 999
kg
m3
⋅ 5 m
s
⋅⎛⎜⎝
⎞
⎠
2 1
1 m⋅× 9.81
m
s2
⋅−
⎡⎢⎢⎣
⎤⎥⎥⎦
× N s
2⋅
kg m⋅×= pr 15.17
kPa
m
⋅=
and we may calculate the pressure at the bottom of the bucket: ∆p pr ∆r⋅= ∆p 15.17
kPa
m
⋅ 0.2× m⋅= ∆p 3.03 kPa⋅=