
#1
Nov2412, 11:04 AM

P: 147

Here's a derviation from HyperPhysics:
He says: [tex]dV=πy^{2}dz[/tex] However, if we're finding the surface area of the sphere: [tex]dA=2πRdz≠2πydz[/tex] If we cannot use [tex]dA=2πydz[/tex], how come [tex]dV=πy^{2}dz[/tex] is still applicable? 



#2
Nov2412, 11:31 AM

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P: 40,905

What does finding the surface area of the sphere have to do with this? 



#3
Nov2412, 12:11 PM

P: 147

If I want to find the MoI of a hollow sphere. Based on the same assumptions [tex]dA=2πydz[/tex], it should be like that, but it's not . I'm trying to calculate the MoIs of some hollow objects, not sure what kind of area formula I should use. 



#4
Nov2412, 02:49 PM

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P: 40,905

Moment of Inertia Problem 



#5
Nov3012, 09:05 AM

P: 147

As z tends to zero, that particular segment of a sphere should tend toward a segment of a cone. And for a cone, when z tends to zero [tex]A→2πrz[/tex] The difference between the conical area and the area of a cylinder wrapped around it should decrease until they are equal when z tends to zero. 



#6
Nov3012, 09:19 AM

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#7
Nov3012, 09:47 AM

P: 318

Is this for the moment of inertia of a sphere?




#9
Nov3012, 09:56 AM

P: 318

I'm pretty sure it's for a full sphere, that dI there represents the moment of inertia of a single disk at height z, and then he integrates from the bottom to the top.




#10
Nov3012, 10:01 AM

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#12
Dec212, 07:11 AM

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P: 40,905

The volume is the area times thickness: dV = area * dz = πr^{2}dz If we vary the radius by dr, we get: dV = π(r + dr)^{2}dz = π(r^{2} + 2rdr + dr^{2})dz dV = πr^{2}dz + The higher order differentials can be ignored. 



#13
Dec212, 07:47 AM

P: 147

What if the external area is perimeter times thickness: dA = perimeter * dz =2πrdz Varying the radius by dr, we get: dA = 2π(r+dr)dz =(2πr+2πdr)dz = 2πrdz + 



#14
Dec212, 07:56 AM

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#15
Dec212, 09:37 AM

P: 147

but then volume can also be considered area times some funny ratio and not just the vertical height dz 



#16
Dec212, 09:55 AM

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#17
Dec212, 10:29 AM

P: 147

but for external area, you said we must account for the slant, I used (r + dr), doesn't that do the job? I'm afriad my mathematical understanding is really sketchy. 



#18
Dec212, 11:34 AM

Engineering
Sci Advisor
HW Helper
Thanks
P: 6,386

Of course you can use trig and get the length in terms of an angle, if you want. 


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