Centres of Mass

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#1
Hi, I am going to post below. I am not sure how to do part b
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#2
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#3
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1 month ago
#4
An object will topple if it's center of mass doesn't lie directly over its base of support.
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#5
(Original post by DFranklin)
An object will topple if it's center of mass doesn't lie direc
Will the length along the bottom still be equal to 3r ? If so the, (3-2^0.5)r is beneath the rectangle part of the shape, so I need to only determine if the hypotenuse length of a triangle with sides r and 27/20 r is less than (3-2^0.5) ?
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1 month ago
#6
(Original post by CaesarAugustus1)
Will the length along the bottom still be equal to 3r ? If so the, (3-2^0.5)r is beneath the rectangle part of the shape, so I need to only determine if the hypotenuse length of a triangle with sides r and 27/20 r is less than (3-2^0.5) ?
I don't understand what you mean - surely the "bottom" in the diagram you've drawn has length r sqrt(2)?

Your basic plan seems reasonable however, you just need to be careful with the geometry. (And frankly I'm not going to bother checking it without you posting a decent diagram. To be clear - I'd totally understand you not wanting to bother, but if you don't, I'd have to do one if I want to check carefully, and that's more effort than I want to go to myself).
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#7
(Original post by DFranklin)
I don't understand what you mean - surely the "bottom" in the diagram you've drawn has length r sqrt(2)?

Your basic plan seems reasonable however, you just need to be careful with the geometry. (And frankly I'm not going to bother checking it without you posting a decent diagram. To be clear - I'd totally understand you not wanting to bother, but if you don't, I'd have to do one if I want to check carefully, and that's more effort than I want to go to myself).
By bottom, I was referring to the entire length of the shape horizontally.
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#8
I hope this is more appropriate for you. I am wondering how to calculate a ?
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#9
(Original post by DFranklin)
I don't understand what you mean - surely the "bottom" in the diagram you've drawn has length r sqrt(2)?

Your basic plan seems reasonable however, you just need to be careful with the geometry. (And frankly I'm not going to bother checking it without you posting a decent diagram. To be clear - I'd totally understand you not wanting to bother, but if you don't, I'd have to do one if I want to check carefully, and that's more effort than I want to go to myself).
Is this correct ?
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#10
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1 month ago
#11
I've just scan read this, what did you get for the location of the .com of the body?

Also what angle is the cylinder at compared to the horizontal or vertical?
Last edited by mqb2766; 1 month ago
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#12
x = 27r/20 , y=r

(Original post by mqb2766)
I've just scan read this, what did you get for the location of the .com of the body?

Also what angle is the cylinder at compared to the horizontal or vertical?
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1 month ago
#13
(Original post by CaesarAugustus1)
x = 27r/20 , y=r
X is from base or top or ... What is the angle the cylinder makes?
Last edited by mqb2766; 1 month ago
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#14
(Original post by mqb2766)
X is from base or top or ... What is the angle the cylinder makes?
As the shape is initially pictured, not my diagram. The x .com is 27r/20 along the horizontal
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1 month ago
#15
(Original post by CaesarAugustus1)
As the shape is initially pictured, not my diagram. The x .com is 27r/20 along the horizontal
Your "a" is the horizontal length (height) of the cylinder, have you worked this out in terms of r? If you know the angle (not hard), it's easy.
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#16
A= 2rcos45
(Original post by mqb2766)
Your "a" is the horizontal length (height) of the cylinder, have you worked this out in terms of r? If you know the angle (not hard), it's easy.
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#17
(Original post by mqb2766)
Your "a" is the horizontal length (height) of the cylinder, have you worked this out in terms of r? If you know the angle (not hard), it's easy.
Sqrt(1+(27/20)^2) > sqrt2 , therefore the shape does not topple ?
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1 month ago
#18
(Original post by CaesarAugustus1)
A= 2rcos45
Sure or sqrt(2)*r.
So does the (horizontal projected) com lie to the left or right of the toppling point?
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#19
(Original post by mqb2766)
Sure or sqrt(2)*r.
So does the (horizontal projected) com lie to the left or right of the toppling point?
It lies to the right I think
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1 month ago
#20
(Original post by CaesarAugustus1)
It lies to the right I think
I've not checked your com calculation, but if your happy with that ...
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