Finding the Centre of Mass and Toppling Point of a Spinning Top

[vic]

A uniform solid spinning top has the shape of an inverted right circular cone of radius 3r and height 4r surmounted by a right circular of base radius 3r and height 6r. Find the position of the centre of mass of the spinning top and hence show that if it is placed with the curved surface of the cone on a horizontal plane the top will topple.


This is my answer
y = 0 by symmetry

54[Pi]r^3 (3r) + 3[Pi]r^2 (7r) = [54 [Pi] r^3 + 3[Pi] r^2] x

54r^2 + 7r = (18r + 1) x

x = (54 r^2 + 7r) /(18r +1)

But the textbook says the answer is (75/33)r from joint face


Please Help me.

 

[uart]

I got the same answer as the book. To make it easy I used normalized length units with one of my units equalling "4r" of their units. Here's a really quick rough outline of my calculations.

x_c = integral(x f(x) dx) / integral(f(x) dx), where f(x) is the density function, normalized to any level you want.

integral(x f(x) dx) = integral(x^3, x=0..1) + integral(x,x=1..2.5) = 23/8

integral( f(x) dx) = integral(x^2, x=0..1) + integral(1, x=1..2,5) = 11/6

So x_c = 23*6/(11*8) in my normalized units.

In the original units that x_c = (69/11)r from the cones apex which is (25/11)r from the join.

 

[vic]

How do you know that f(x) is x^2 +1 ??

 

[uart]

How do you know that f(x) is x^2 +1 ??

It's not, but all you need is something that is propotional to the true density since the center of mass will be the same whether it's made of lead or plastic, so long as it's the same uniform material throughout. Actually it's only the linear density I'm considering here, that is mass per unit length, that's all you need consider due to the symmetry in the other two dimensions.

So the linear density, "f(x)", is just a function proportional to the cross-sectional area at the position "x". That's all you need to do and it makes the problem fairly easy.

I made it even easier by choosing length unit of 1 unit = 4r of the original length units. This makes the width of the cylindrical portion be exactly 1 unit and the width of the conic portion changing linearly from 0 units to one unit width over a length of exactly one unit.

So in terms of my length units the cross-section of the conic part is proportional to x^2 and the cross-section of the cylindrical section is proportional to unity (with the same constant of proportionality in each case).

So actually the function I used for f(x) was :

f(x) = x^2 : for 0 <= x <= 1
f(x) = 1 : for 1 < x <= 2.5
f(x) = 0 : otherwise

 

[vic]

I got it now. Thanks a lot. :)