if you want a proof, here's one by calculus
when the cone is positioned upright, standing on its tip we can find the center of mass by the formulas
x' = ∫(x*da)/∫(da)
y' = ∫(y*da)/∫(da)
---with the limits from
x = 0 to x = b(base)
y = 0 to y = h(height)
by intuition center of mass along x = 0
we can represent the equation of the cone as y = mx, where m is the slope
so finding the center of mass along y we find
∫yda = ∫y*2πxdy, since da=2πxdy from infinitesimal area A = 2πrh
= 2π∫(y^2)dy/m
= 2π(y^3)/3m evaluated from 0 to h
= 2πh^3/3m, we can find m from y = mx ---> h = mb so m=h/b
= 2πb(h^2)/3
then we have to find ∫da
∫da = ∫2πxdy
= ∫2πydy / m
= π(h^2)/m
= πbh
so y' = ∫(y*da)/∫(da) = 2πb(h^2)/3πbh = 2h/3
this means the center of mass along the vertical axis of the cone is 2h/3 when the cone is STANDING ON ITS TIP
y = 2h/3 and x = 0
2006-11-02 22:09:18
·
answer #1
·
answered by Jeremy 2
·
2⤊
0⤋
it will depend on how thick is the wall of your hollow cone. Or did you mean some ideal mathematical cone, with an infinitely thin surface, which somehow would still weigh something?
anyway, without data on the thickness, can't calculate
in principle these things are easy to compute, integrating over the volume or surface the infinitesimal mass elements dm, times the distances to a chosen reference point. In practice however, the algebra can become a bit ugly ;-)
2006-11-02 19:26:23
·
answer #2
·
answered by AntoineBachmann 5
·
0⤊
4⤋
if the rigidity of gravity have been a linear function, you would be attracted to the middle of gravity. in spite of the indisputable fact that it incredibly is a sq. function - it varies with the sq. of the gap, no longer in basic terms with the gap, and so which you would be attracted, as Billruss shows, to the closest interior floor of the hollow sphere, in basic terms placed, the farther surfaces might have disproportionately much less effect than the closer surfaces. Edit! Kirchwey is sturdy. nicely, needless to say i'm no longer at Newton's point! in spite of the indisputable fact that, my misconstruance aside, there could be NO gravitational effect in any respect in the hollow sphere.
2016-10-21 04:41:56
·
answer #3
·
answered by casaliggi 4
·
0⤊
0⤋
there is a formula for it (check an intro physics or engineering book), however, I believe it is the same as for a solid cone, which is along the center axis, 2/3 the way down from the tip to the base.
2006-11-02 19:29:25
·
answer #4
·
answered by rigel828 1
·
2⤊
2⤋
2/3 of the length from the apex.
2006-11-02 19:36:29
·
answer #5
·
answered by ag_iitkgp 7
·
0⤊
0⤋
lies in centre even there is no mass present there
2006-11-02 19:28:15
·
answer #6
·
answered by v_aravind_kulasekhar 1
·
0⤊
1⤋