A homogeneous liquid whose density is 300 kg/ m^3 fills three buried containers.
ID: 3001606 • Letter: A
Question
A homogeneous liquid whose density is 300 kg/ m^3 fills three buried containers. The containers are each 10 meters tall. The top of each container is at ground level. All three containers have the same volume. The middle container is a cylinder, and the other two are circular cones - the one on the left has the base on ground level and the one on the right has the point at ground level. Which container needs the least amount of work to empty (that is, to pump the liquid to ground level)? Which container needs the most work to empty? Justify your assertions by computing the work necessary in each case. Discuss why your answer is correct.Explanation / Answer
The density of the liquid is defined as: = 300 kg/m³ (we'll ignore gravity since it's a constant and since the density is irrelevant when comparing the work required for each case)
To determine the actual work required in all three cases, we need to calculate the relative geometry of each shape. The equation for the volume of a cylinder, Vs, and thevolume of a cone, Vc, are essentially identical, except that a cone of the same radius, r, and height, h, will have 1/3 the volume of an identically dimensioned cylinder:
Vs = (rs)²h
Vc = (1/3)(rc)²h
Since the volumes are equal, Vs = Vc:
(rs)²h = (1/3)(rc)²h
(rs)² = (1/3)(rc)²
Therefore, the radius of an equivalent cone, rc, would be:
(rc)² = 3(rs)²
rc = 3(rs)²
rc = (rs)3
Now that we know the relative radii, we can calculate the work required to empty the three volumes.
1. Work to empty the cylinder is evaluated over the interval [0,10]:
dV = (rs)² dy
W = (10 - y) dV = (10 - y) (rs)² dy = (rs)² [0,10] 10 - y dy
Integrating:
W = (rs)²[10y - (1/2)y²] evaluated over the interval [0,10]
Evaluating:
W = (rs)²[10(10) - (1/2)(10)²]
W = 50(rs)² kg-m
2. Work to empty the inverted cone is evaluated over the interval [0,10]:
rc = (rs)3
The radius, r, at any height, y, is determined by proportions:
r/y = (rs)3/h ==> r = [(rs/h)3]y = [(rs)(3)/10]y
dV = (r)² dy = {[(rs)(3)/10]y}² dy = (3/100)(rs)²y² dy
W = (10 - y) dV = (10 - y) (3/100)(rs)²y² dy = (3/100)(rs)² [0,10] 10y² - y³ dy
Integrating:
W = (3/100)(rs)²[(10/3)y³ - (1/4)y^4] evaluated over the interval [0,10]
Evaluating:
W = (3/100)(rs)²[(10/3)(10)³ - (1/4)(10)^4] = (3/100)(rs)²[10,000/12]
W = 25(rs)² kg-m
3. Work to empty the un-inverted cone is evaluated over the interval [0,10]:
rc = (rs)3
The radius, r, at any height, y, is again determined by proportions:
r/(h - y) = (rs)3/h ==> r = [(rs)3](h - y)/h = [(rs)3](10 - y)/10
dV = (r)² dy = {[(rs)3](10 - y)/10}² dy = (3/100)(rs)²(10 - y)² dy
W = (10 - y) dV = (10 - y) (3/100)(rs)²(10 - y)² dy = (3/100)(rs)² (10 - y)³ dy
Expanding (10 - y)³ for ease of integration:
W = (3/100)(rs)² [0,10] 1,000 - 300y + 30y² - y³ dy
Integrating:
W = (3/100)(rs)²[1,000y - 150y² + 10y³ - (1/4)y^4] evaluated over the interval [0,10]
Evaluating:
W = (3/100)(rs)²[1,000(10) - 150(10)² + 10(10)³ - (1/4)(10)y^4] = (3/100)(rs)²[2,500]
W = 75(rs)² kg-m
Thus, a cylinder requires twice the work of an inverted cone, and a non-inverted cone requires thrice the work of an inverted cone (or requires 150% of the work of a cylinder) when volumes and heights are identical.
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