2
$\begingroup$

How can a distinction be made between the centre of mass of a right circular cone

  1. As a uniform solid ($\frac{h}{4}$ from base, where is $h$ is height) and
  2. That composed of infinitesimally small thickness of right triangles of the same height and base (same as radius of cone). In which case it apparently appears to be $\frac{h}{3}$ from base.
$\endgroup$

1 Answer 1

0
$\begingroup$

You are missing the fact that point 2 is not a valid separation of a cone into sub-parts. It should instead be separated into triangular-based pyramids with small-angled isosceles triangle at the base. The center of mass of that thing is at h/4.

This is how a small element would look:

conesector

The vertex of the cone is up at the top {x,y,z}=={2,0,0}. The base of this element is a sector of a circle.

$\endgroup$
4
  • $\begingroup$ You are right ! Seems to explain. One more aspect needs to be addressed in this context, how come rectangles be able to make up a right circular cylinder ? Though sometimes wrong approximations may result in right answers, is this then the right way to conclude ? $\endgroup$
    – ArKE
    Commented Oct 16, 2015 at 3:34
  • $\begingroup$ @ArKE Well, a rectangle is a 2D figure, whereas for the purposes of calculating the center of mass you need to take elements of volume. It is true, that a cone is the product of the revolution of a rectangle, but when revolving it, its outer edge covers a larger distance, thus more mass is closer to the base of the cone, which is why the final center of mass ends up closer to the bottom. $\endgroup$
    – LLlAMnYP
    Commented Oct 16, 2015 at 10:30
  • $\begingroup$ @LLIAMnYP .. I mean for cylinder .. it is $h/2$ and for a rectangle also it is $h/2$. So if we consider turning a rectangle (width of radius of base for cylinder) by one rotation we can get a cylinder. $\endgroup$
    – ArKE
    Commented Oct 19, 2015 at 3:33
  • $\begingroup$ @ArKE that's a matter of coincidence, because for a small cylindrical sector the center of mass is also at $h/2$ $\endgroup$
    – LLlAMnYP
    Commented Oct 19, 2015 at 6:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.