Toppling of a road cone that has an axis at an angle $\alpha$ to the horizontal.

Question

A road cone consists of a $45cm$ x $45cm$ square base of height $10cm$, and a conical shell of radius $15cm$ and height $75cm$. The base has a circular hole through it , of radius $15cm$, to aid stacking. The base is made of plastic of density $1Kg$ per $1000cm^3$ and the conical shell has mass $0.3Kg$. The cone is held with an edge of the base in contact with a horizontal surface, and with the axis of the shell making an angle $\alpha$ with the horizontal. The cone is now released. Find the angle $\beta$ such that the cone assumes the upright position if $\alpha>\beta$ and the cone topples if $\alpha<\beta$.

Answer 1

Taking the point $O$ to be the bottom right 'corner' of the base the centre of mass of the whole cone with the cone standing upright is:

Volume of the base = $20250 cm^3$

Volume of the circular section that is cut out of the base = $7068.58 cm^3$

Volume of the base - circular section = $13181.42 cm^3$

Mass of base = $13.181 Kg$

Mass of conical section (hollow cone) = $0.3 Kg$

Therefore the mass of whole cone = $13.481 Kg$

The $y$ co-ordinate of centre of mass of whole cone is:

$$\frac{(5*13.181) + (35*0.3)}{ (13.481)} = 5.67 cm$$

Now taking moments about $O$ when the whole cone is tilted and the weight is split into separate horizontal (parallel to the axis of the cone) and vertical (parallel to the base) components gives:

Clockwise moments = anticlockwise moments

$$5.67*mgcos(\beta) = (45/2)*mgsin(\beta)$$

so $$tan(\beta) = \frac{(5.67)}{45/2} = 0.252$$

so $$\beta = 14.14^o$$

where, if $\alpha$ exceeds the value of $\beta$ i.e. ($\alpha > \beta$), the cone returns to the upright position and if $\alpha < 14.14^o$, the cone would topple.