Answer:
(a) The distance is \(\dfrac{11}{4}r\).
(b) The object can rest in equilibrium on the curved surface of the cone.
(a) Measure distances from the end of the cylinder that is not attached to the cone.
The cylinder has height \(4r\), so its centre of mass is at distance
\(2r\)
from this end.
The cone is attached to the other end of the cylinder. Its height is \(4r\), and the centre of mass of a solid cone is \(\dfrac14\) of the height from the base. Therefore the cone's centre of mass is
\(4r+\frac14(4r)=5r\)
from the free end of the cylinder.
The cylinder and cone have the same radius \(3r\). Their volumes are proportional to their heights multiplied by the same base area, with the cone having the factor \(\dfrac13\). Thus
\(V_{\text{cylinder}}:V_{\text{cone}}=4r:\frac13(4r)=3:1.\)
Let \(\bar{x}\) be the distance of the centre of mass of the whole object from the free end of the cylinder. Taking moments,
\(\bar{x}=\frac{3(2r)+1(5r)}{3+1}.\)
Therefore
\(\bar{x}=\frac{6r+5r}{4}=\frac{11r}{4}.\)
(b) Now measure the distance of the centre of mass from the vertex of the cone. The total distance from the free end of the cylinder to the vertex of the cone is \(8r\), so
\(OG=8r-\frac{11r}{4}=\frac{21r}{4},\)
where \(O\) is the vertex of the cone and \(G\) is the centre of mass.
The cone has height \(4r\), radius \(3r\), and slant height
\(\sqrt{(4r)^2+(3r)^2}=5r.\)
Let \(OA\) be the slant height of the cone. The angle between the axis and the slant side has cosine
\(\frac{4r}{5r}=\frac45.\)
For the object to rest on the curved surface of the cone, the vertical through the centre of mass must be able to pass through the contact region. The required condition is
\(OG\cos\phi\lt OA,\)
where \(\phi\) is the semi-vertical angle of the cone. Here this becomes
\(\frac{21r}{4}\cdot\frac45\lt 5r.\)
The left hand side is
\(\frac{21r}{5}.\)
Since
\(\frac{21r}{5}\lt 5r,\)
the condition is satisfied. Therefore the object can rest in equilibrium with the curved surface of the cone in contact with a horizontal surface.
