# Condition for a rigid body in equilibrium on a horizontal surface

The following question is from a past paper from Further Mathematics and it has been bothering me immensely I have spent hours cracking at a way to make sense of it, the question is in two parts

Part 1:

An object consists of a uniform solid circular cone, of vertical height 4r and radius 3r, and a uniform solid cylinder, of height 4r and radius 3r. The circular base of the cone and one of the circular faces of the cylinder are joined together so that they coincide. The cone and the cylinder are made of the same material.

Find the distance of the centre of mass of the object from the end of the cylinder that is not attached to the cone.

I first roughly sketched the object as such

Then finding the $$\bar{x}$$ of each part seperately:

For the cone I used the standard result of the centre of mass being $$\frac{1}{4}r$$ away from the base:

$$\bar{x}_{Cone}=\frac{1}{4}.4r + 4r=5r$$

For the cylinder I derived $$\bar{x}$$ via integration:

$$y=3r$$

$$\bar{x}_{cylinder}=\frac{\int_0^{4r} xdV}{\int_0^{4r}dV}$$

$$\because dV=y^2 \pi dx$$

$$\implies \bar{x}_{cylinder}=\frac{\int_0^{4r} 9r^2 \pi x dx}{\int_0^{4r}9r^2 \pi dx}$$

$$\implies \bar{x}_{cylinder}=\frac{9r^2 \pi \int_0^{4r} x dx}{9r^2 \pi\int_0^{4r} 1 dx}$$

$$\implies \bar{x}_{cylinder}=\frac{\frac{1}{2}\left[x^2 \right]^{4r}_0}{\left[x \right]^{4r}_0}$$

$$\therefore \bar{x}_{cylinder}=2r$$

Then by taking the weighted average of both objects I arrived at the correct value for the distance of the centre of gravity from the base of the cylinder which was

$$\bar{x}=\frac{11}{4}r$$

However the next part has had me at a complete loss for the better part of the day, it states that:

Show that the object can rest in equilibrium with the curved surface of the cone in contact with a horizontal surface.

I tried coming up with a rough sketch for this also

However I do not understand how to tackle this question at all, all I know is that for a body to be at equilibrium on such a surface, the centre of mass must pass through the point of suspension, however clearly this does not give an insight on how to answer this.

The condition for this question given in the marking scheme is that

Can someone explain what this means and why this is the case?

As you correctly found out, the center of gravity $$G$$ is on the axis of the cone and cylinder ($$BF$$ in the diagram) and $$\frac{11r}{4}$$ from $$F$$. So from the vertex of the cone $$B$$, it is at $$\frac{21r}{4}$$ distance $$(\gt BI = 4r)$$.

We know $$G$$ is between $$F$$ and $$I$$. For the solid to be in equilibrium with the curved surface of the cone touching a horizontal surface (ray $$AB$$), we need to show that $$G$$ is directly above a point on segment $$AB$$, in other words we need to show that $$G$$ is between $$H$$ and $$I$$.

$$\small AB = \sqrt{AI^2 + BI^2} = 5r$$

If $$\small \angle ABH = \theta$$,

$$\displaystyle \small \cos \theta = \frac{BI}{AB} = \frac{4}{5}$$.

Now in $$\triangle ABH, \displaystyle \small \cos \theta = \frac{4}{5} = \frac{AB}{BH} = \frac{5r}{BH}$$

$$\implies \displaystyle \small BH = \frac{25r}{4} \gt \frac{21r}{4} = BG$$

This proves that $$G$$ is between $$H$$ and $$I$$.