Grinding Wheel
The Wizard of Oz and Ross, both respectable residents of Puzzlaria, have each started small businesses in the basement of their homes. While walking one day, and enjoying a smoke of one of the strange herbs which grows wildly in the fields surrounding Puzzlaria, they see a yard sale sign.
One item for sale is a large grinding wheel, 22 inches in diameter, with a 3 1/7" bore. Neither has a particular use for it, but the price is right, so they split the cost and make the purchase. The flip of a coin decides that Wizard will be the first owner of the wheel. It is decided that once he uses half of the wheel, he will give it to Ross.
What will be the diameter of the wheel when Ross receives it?
(Amendment: wheel bore has been added)
One item for sale is a large grinding wheel, 22 inches in diameter, with a 3 1/7" bore. Neither has a particular use for it, but the price is right, so they split the cost and make the purchase. The flip of a coin decides that Wizard will be the first owner of the wheel. It is decided that once he uses half of the wheel, he will give it to Ross.
What will be the diameter of the wheel when Ross receives it?
(Amendment: wheel bore has been added)
posted by Zaux at
6:10 AM
21 Comments:
about 15.5 inches in diameter
May be I'm being silly but I interpreted this as Wizard will use the wheel till half of it (in terms of area) is eroded due to grinding. The answer in that case is:
The diameter is ~15.56 inches when Ross gets it.
Total area: 22/7 * (22)^2 = 379.94
Half area: ~190 sq In
New radius = Under root (190*7/22)
= 7.78
Therefore diameter = 15.56
everyone is pretty close, but not exact
how exact do you want?
i can say 15.55634919 inches in diameter haha
Most grinding wheels have an unusable core ... at the very least, the diameter of the hole where the wheel is attached to the motor that spins it will not be usable. So take off a half inch diameter there?
well without knowing the diameter of the unusuable core we cannot find an appropriate answer.
I worked through this problem without looking at the published answer ... I got 15.7142 inches. I checked the published solution ... it is 15 5/7" or 15.7142.
I understand the possibility of the published solution being inaccurate ... but, if it is wrong, it's rather weird that I made the same error to four decimal places.
I will show my method later today.
Let the diameters of the outer and inner useful part of the wheel be Do and Di. Let Dh correspond to the half way point.
Then want π(Di² - Dh²)/4 = π(Dh² - Di²)/4
=> Dh = Sqrt((Do² + Di²)/2)
If Di = 0 and Do = 22, then Dh = 22/√2 ≈ 15.55635.
15 5/7" implies an useful inner diameter of 3 1/7" = 22/7" (exactly). This is one seventh (exactly) of the outer diameter.
This post has been removed by the author.
I can't see how you got 15.7142 inches Zaux.
I'm with everybody else: I get 15.55634919 inches if there is no inner portion of the wheel that is to be ignored.
I'm intrigued to see your method later.
Wow... looks like everyone's in agreement ... if all think the sky is blue, I certainly will not declare it red ...
I will show my method after lunch ... 11:54AM here.
As I've stated before, my math skills are not at the level of most of the regulars here at TOM.
After looking at the manner in which you guys approached the problem, it makes my approach look quite tedious.
Please understand that I am not contradicting what you guys are saying ... I simply would like you to analyze my solution to help me learn. I will over-explain some points for clarity.
So here goes:
Given:
dia. of grinding wheel - 22"
center bore - 3 1/7" (22/7")
Question:
What is the diameter of the wheel when half used?
Process:
(Area of circle defined by the wheel's circumference) - (are of the bore) = usable area [technically, if you could use it all].
1/2 the usable area + area of the bore would define a circle with the diameter we are seeking.
Math:
Area of circle defined by wheel=
π(11")² = 380.1336 sq. in.
Area of bore = π (11/7)²
= 7.7578 sq. in.
Usable area = 380.1336 - 7.7578 =
372.3758 sq. in.
Using half the wheel,
372.3758/2 = 186.1879 sq. in.
The half used wheel + the bore identify a circle with the desired diameter:
186.1879 (wheel material) + 7.7578 (bore area) = 193.9457 sq. in. The diameter of the circle with area 193.9457 sq. in. is the diamter we seek.
193.9457 = π r²
r² = 193.9457/3.1416 = 61.7347
r = 7.8571
diamter = 7.8571 x 2 =
15.7142 sq.in.
Answer" = 15.7142 sq. in.
This diameter agrees with the published solution. We all know that doesn't necessarily mean it is correct ... but I think it does justify another look. Thanks ... and I hope to learn from this. I'm begining to understand Chris's rant that the method is far more interesting than the answer.
Zaux. You didn't post the centre bore (3 1/7") in the problem. That completely explains the discrepancy.
Hi Zaux. LOL, I just saw the "Chris's rant". Of course the result is totally uninteresting. The method is everything. There are a few exceptions e.g when the result are wildly different to your first guess.
Jeez! ... Chris, you are so right ... I an't believe I missed putting the bore in the statement ... I will correct it.
so basically everyone was right ... you guys with the info you had and me with the info I had... so sorry ... guess that means I'm human (don't tell my wife :))
That's fine for you, Zaux, but I'm the one getting stuck with less than half the usable part of the grinding wheel :-)
Ross ...aw man... sorry ... I'm part owner of the Puzzlaria Tavern...stop by for a free pitcher of your favorite ... unless it's single malt scotch :)
Simple is best. Forget π and r²
Diameter-Wall(Wall)=area
(22-66/7)(66/7) = 5808/49
Half the wheel = 2904/49
(110/7-66/7)(66/7)
(44/7)(66/7)
2904/49
And thus, 110/7 is the diameter of half the wheel.
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