| Posted on Monday, 12 May, 2003 - 11:05 am: |
Please help with this question;
A vase is made of sheet metal of uniform thickness. It is composed of two hollow cylinders. The lower cylinder has diameter 5cm and length 30cm. The upper cylinder has diameter 4cm and length 6cm. The upper cylinder rests on top of the lower cylinder. The vase is open at the top (ie the top of the upper cylinder).
Find the height of the centre of mass above the base.
| Posted on Monday, 12 May, 2003 - 11:25 am: |
Imagine your vase is lying on its side, and let the base of the whole vase be at the point x=0. Do you know the formulae:
"Total moment = Sum of the individual moments,"
and,
"Moment = distance from the pivot times the perpendicular force" ?
Let the centre of mass be a distance h away from the base. Now all you need to do is work out the mass of the two portions of the vase, and their sum, then apply the two formulae, which will then give you h. Does that give you enough of a hint? If not, post back, and I will add a bit more!
| Posted on Monday, 12 May, 2003 - 03:12 pm: |
Yes, but working out the mass is a little tricky! So far I have let k = mass of 1 sq cm of metal Then mass of lower cylinder = ((10px 30) + (25p) + (25-4)p)k. This equates to (300+25+21)pk or 346pk. The mass of the upper cylinder would just be (4p*6)k = 24pk. Thus the total mass is 370 pk. Then using x¢ = [(346pk x 15) + (24p k x 33)] / 370pk gives x¢ = (5190pk + 792pk) / 370pk gives x¢ = 5982 pk / 370 pk Cancelling p and k gives x¢ = 5982/370 = 16.16756757 Is this correct?
| Posted on Monday, 12 May, 2003 - 03:13 pm: |
The correct answer appears to be 16.0 to 3sf
| Posted on Monday, 12 May, 2003 - 05:32 pm: |
Greg, you have gone slightly wrong, but don't worry! Let me explain:
First of all, you say that the DIAMETER is 5cm, not the radius. When you work out the surface area of the lower portion of vase, you have a 25p in there. I assume (correct me if I am wrong) that that comes from the area of the base of the vase, ie 52 p , in which case you have mistaken diameter for radius!
Let me work out the surface area of the lower vase for you, and then hopefully you can do the upper part yourself:
Very bottom base: p *2.5*2.5 = 6.25p
Side of bottom cylinder: 2*2.5*p *30 = 150p
Top of bottom cylinder = area of large circle - area of small circle (you also went slightly wrong here before!) = 2.5*2.5*p - 2*2*p = 2.25p .
So total surface area = 6.25p + 150p + 2.25p = 158.5p .
Can you finish it off from here? Let me know if not, and I will help you out some more.
P.S. If you do post again, could you be a bit more specific as to which bottoms are there and not there? Thanks.
| Posted on Monday, 12 May, 2003 - 06:49 pm: |
Sorry my mistake! Diameter is indeed 10cm.
Bottom base: 25p
Side of bottom cylinder: 300p
Top of bottom cylinder: (25-4)p
Total surface area = 25p + 300p + 21p = 346p
OK so far?
| Posted on Monday, 12 May, 2003 - 07:32 pm: |
< font size="2" face= "Verdana,Arial,Helvetica» Nearly, but not quite: Bottom base: fine Side of bottom cylinder: fine Top of bottom cylinder: this is slightly out. If I am reading correctly what you wrote in your last post, you have a large circle of radius 5, and you want to take out a circle in the middle of radius 4. Thus, the area required is
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| Posted on Monday, 12 May, 2003 - 08:59 pm: |
actually diameter is 4, so radius 2! (see first post).
| Posted on Monday, 12 May, 2003 - 09:02 pm: |
that said, is my answer of 16.17 correct? to 2 dp
| Posted on Monday, 12 May, 2003 - 10:20 pm: |
Ah! Ok, I misinterpreted what you said! I thought you had misquoted both the radii as diameters.
Ok, so they have radius 5 and 2, thus, you were quite correct before - the area of the top of the bottom part is (25-4)p =21p .
And yes, if we assume that the smaller top portion of the vase does not have any bottoms, and thus it has surface area 2*2p *6 = 24p , then yes, I get 16.16756757.......=16.17 to 2dp.
Hmmmm..... your book gave 16.0 didn't it? I must admit, I agree with your answer of 16.17 - perhaps they give the smaller tube a bottom, so in fact you just have a small vase with a big sealed bottom part acting as a big base unit!!!! Try it and see if that extra surface area makes the difference!
Well done for sticking the problem out though - I know it can be annoying when we don't give much away at first, but I am sure you can see, you will have gained far more benefit by working through it yourself with gentle hints rather than me just telling you how it is done straight away.
Please let me know if there is something that you have misinterpreted, which would give you the answer given by your book. It would be interesting to know if the book's answer was wrong or not
!
| Posted on Tuesday, 13 May, 2003 - 01:15 pm: |
I see the computation as follows:
Assume radii are 5 and 2, and k is as Greg defined it.
Total mass, M = (mass of bottom cylinder base)+(mass of bottom cylinder)+(mass of middle cylindrical disc)+(mass of upper cylinder)
M=(p 52 k)+(2p 5.30k)+(p [52 -22 ]k)+ (2p 2.6k)
M=p k(25 + 300 + 21 + 24)
M=370p k
This, Greg, is as you had in your second post. Where we disagree is in your computation of x?. You have taken the bottom cylinder, and the two discs (the solid one at the bottom and the one with a hole at the top) and regarded them as being at a distance 15 from the base. This is not true, since the discs are not the same ? one has a hole. The best thing to do is to treat them separately:
x' = [(mass of bottom cylinder base).distance+(mass of bottom cylinder).distance+(mass of middle cylindrical disc).distance+(mass of upper cylinder).distance]/M
x' = [(25p k).0+(300p k).15+(21p k).30+(24p k).33]/[370p k]
x' = [0 + 4500 + 630 + 792]/[370] = 16.005
Andre
| Posted on Tuesday, 13 May, 2003 - 01:40 pm: |
D'oh! Thanks Andre - I have been a complete muppet and also forgotten that the middle disc is not the same as the base disc. So what Andre has said is the right answer. Sorry about that Greg!
| Posted on Tuesday, 13 May, 2003 - 03:43 pm: |
Thats ok, thanks for all your help guys!
I'm one of those people who don't like leaving a problem until it is completely finished, so i'm glad that you have got 16.0.
| Posted on Tuesday, 13 May, 2003 - 03:48 pm: |
Its so easy, because there are two cylinders, to only consider these, whereas, as you say, it needs to be split up further into the individual parts.
| Posted on Tuesday, 13 May, 2003 - 03:49 pm: |
Yes, I agree it is nice to completely finish a problem - as you can now see, the safest way to do these is to always split the object you are considering into all the small parts that make up the whole thing (unlike me, who made the same mistake as you and considered the bottom part as a symmetric shape mass-wise.) If you get one of these in the exam, do split the shape up into all its little bits, as this will hopefully give you less chance of missing something out (which is only too easy as I remember from my A-level days
)