Question 1205006: The volume of a frustum of a cone of height h is given by the formula
V = ⅓πh(R² + Rr + r²)
Where R and r are the radii of its circular ends.
Hollow metal hemispheres of internal and external diameters 12.5cm and 17.5cm respectively, are to be cast from molten contents of a ladle in the shape of a frustum of a cone of depth 1.2m and radii of ends 0.6m and 0.635m. Calculate the number of hemispheres that can be cast.
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
This can be done on a regular calculator, but I'll turn to GeoGebra instead.
Here is the particular GeoGebra workbook that I created for this problem.
https://www.geogebra.org/calculator/gwhxa25b
Normally I use GeoGebra to make some kind of graph.
However, the graph part is blank. Focus on the column of items on the left side.
I'll explain the inputs below. Each input is highlighted in blue
R = 0.635 and r = 0.6 represent the radius values of the conical frustum.
h = 1.2 is the height or depth of the ladle.
So far everything is in meters, so I'll convert the diameters "12.5cm and 17.5cm" to meters
12.5 cm = 0.125 m
17.5 cm = 0.175 m
Divide by 100 to convert from cm to meters.
The formula Vf = (1/3)*pi*h*(R^2+R*r+r^2) represents the volume of the conical frustum.
It is the formula your teacher gave you.
GeoGebra will replace the letters h, R and r with their proper values.
The result of the calculation is roughly Vf = 1.43788 when rounding to 5 decimal places.
So far we've just taken care of the conical frustum only.
Recall that the volume of a full sphere is (4/3)*pi*(radius)^3
The larger sphere has volume (4/3)*pi*(external radius)^3
The smaller sphere has volume (4/3)*pi*(internal radius)^3
The volume of a spherical outer shell, ignoring the hollow empty space, is:
shell volume = (larger sphere volume) - (smaller sphere volume)
shell volume = (4/3)*pi*(external radius)^3 - (4/3)*pi*(internal radius)^3
shell volume = (4/3)*pi*( (external radius)^3 - (internal radius)^3 )
shell volume = (4/3)*pi*( (eR)^3 - (iR)^3 )
where,
eR = external radius
iR = internal radius
In this case:
eR = 0.175/2 = 0.0875
iR = 0.125/2 = 0.0625
both are in meters.
Keep in mind the 0.175 m and 0.125 m represent diameters.
Cut those in half to get their corresponding radius values.
The formula (4/3)*pi*( (eR)^3 - (iR)^3 ) represents a full spherical shell. But we want a hemispherical shell instead. We'll cut that in half to get (2/3)*pi*( (eR)^3 - (iR)^3 )
So that's where the formula Vh = (2/3)*pi*( (eR)^3 - (iR)^3 ) comes from.
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Here's a quick recap of the inputs to type in (type them in the order mentioned)
R = 0.635
r = 0.6
h = 1.2
Vf = (1/3)*pi*h*(R^2+R*r+r^2)
That takes care of the frustum
Then for the hemispheres we have these inputs
eR = 0.0875
iR = 0.0625
Vh = (2/3)*pi*( (eR)^3 - (iR)^3 )
GeoGebra should display these approximate values
Vf = 1.43788
Vh = 0.00089
which represent the frustum volume and hemisphere volume respectively.
The last calculation to make is to type in ratio = Vf/Vh and the result should be roughly 1612.41248
Round this down to the nearest whole number and it shows we can make 1612 hemispheres.
This assumes that there is no waste.
Again this could have been done with any calculator, but I used GeoGebra so we could name the inputs and use them at a later time. Likely other calculators have this capability as well.
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Answer: 1612
Answer by ikleyn(52852)  (Show Source):
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