Question

A right circular cone with a base diameter of 3 cm and height of 6 cm is cut from a solid cylinder of diameter 5 cm and height of 12cm. Find the position of the center of gravity of the rest of the body.

Answer 1

Hint: In order to solve this problem we need to know about the center of gravity of the cylinder. For a cylinder center of gravity lies at the midpoint of the axis of the cylinder. That is nothing but the total surface area of the remaining solid.

Formula used: To find the slant height of the conical part the formula is,
\[l = \sqrt {{r^2} + {h^2}} cm\]
Where,
r is radius
h is height

Complete step by step solution: Height of the cylindrical part = height of the conical part \[ = 12cm\]
The diameter of the cylindrical part \[ = 5cm\]
Therefore, the radius of the cylindrical part \[ = \dfrac{5}{2}\]\[ = 2.5cm\]
The slant height of the conical part \[ = \sqrt {{r^2} + {h^2}} cm\]
\[ = \sqrt {{{\left( {2.5} \right)}^2} + {{\left( {12} \right)}^2}} \]
\[ = \sqrt {({0.25}) + ({144})} \]
\[ = \sqrt {150.25} \]
\[ = 12.257cm\]
We need to find the position of center of gravity for the rest of the body. For a cylinder center of gravity lies at the midpoint of the axis of the cylinder. That is nothing but the total surface area of the remaining solid.
Then the total surface area of the remaining solid = CSA of cylindrical part + CSA of conical part + base area of the circular part.

\[ = 2\pi rh + \pi rl + \pi {r^2}\]
Here, \[r = 2.5cm\], \[h = 12cm\], \[l = 12.25cm\]
Substitute the value in above equation we get,
\[ = \left( {2\pi \times 2.5 \times 12} \right) + \left( {\pi \times 2.5 \times 12.25} \right) + \pi {\left( {2.5} \right)^2}\]
\[ = \left( {2 \times 3.142 \times 2.5 \times 12} \right) + \left( {3.142 \times 2.5 \times 12.25} \right) + 3.142 \times {\left( {2.5} \right)^2}\]
\[ = 188.52 + 96.223 + 19.637\]
\[ = 304.308c{m^2}\]
Therefore, the total surface area of the remaining solid is \[304.308c{m^2}\]

Note: Here in the given problem it is important to remember that the equation for the center of gravity of a cylinder. That is nothing but the total surface area of the remaining solid. So, we need to find the total surface area of remaining solid.