Problem 13 - Basic Engineering Mathematics

Question

A solid metal cylinder of radius \(6 \mathrm{~cm}\) and height \(15 \mathrm{~cm}\) is melted down and recast into a shape comprising a hemisphere surmounted by a cone. Assuming that \(8 \%\) of the metal is wasted in the process, determine the height of the conical portion, if its diameter is to be \(12 \mathrm{~cm}\).

Step-by-Step Solution

Answer

The height of the conical portion is \(29.4\mathrm{~cm}\).

Step 1: Calculate the Volume of the Original Cylinder

The volume of a cylinder is given by the formula \( V = \pi r^2 h \), where \( r \) is the radius and \( h \) is the height. For our cylinder, \( r = 6 \mathrm{~cm} \) and \( h = 15 \mathrm{~cm} \). Thus, the volume of the metal cylinder is \( V = \pi (6)^2 (15) = 540\pi \mathrm{~cm}^3 \).

Step 2: Account for Metal Wastage

Since \(8\%\) of the metal is wasted, only \(92\%\) of the original volume is used for the new shape. The volume of the new shape is \(0.92 \times 540\pi = 496.8\pi \mathrm{~cm}^3 \).

Step 3: Set Up Volume Equation for Hemisphere and Cone

The new shape consists of a hemisphere and a cone with a shared base diameter of \(12 \mathrm{~cm}\). The radius of both shapes is \(6 \mathrm{~cm}\) (since the diameter is \(12 \mathrm{~cm}\)). The volume of the hemisphere is \( \frac{2}{3} \pi r^3 \) and the volume of the cone is \( \frac{1}{3} \pi r^2 h \). Let the height of the cone be \( h_c \). Then, the equation for the volume of the new shape is:\[ \frac{2}{3} \pi (6)^3 + \frac{1}{3} \pi (6)^2 h_c = 496.8\pi \]

Step 4: Calculate Volume of the Hemisphere

To solve this, calculate the volume of the hemisphere: \[ \frac{2}{3} \pi (6)^3 = \frac{2}{3} \pi (216) = 144\pi \mathrm{~cm}^3 \].

Step 5: Solve for the Height of the Cone

Substitute the volume of the hemisphere in Step 4 back into the equation from Step 3:\[ 144\pi + \frac{1}{3} \pi (6)^2 h_c = 496.8\pi \].\[ 144 \pi + 12\pi h_c = 496.8\pi \].Remove \(144\pi\) from both sides:\[ 12\pi h_c = 352.8\pi \].Divide by \(12\pi\):\[ h_c = \frac{352.8\pi}{12\pi} = 29.4 \mathrm{~cm} \].

Step 6: Conclude the Solution

The height of the conical portion is \(29.4\mathrm{~cm}\), after considering the wastage in the recasting process.

Key Concepts

Cylinder

A cylinder is a 3D geometric shape with two parallel circular bases connected by a curved surface. You can think of it like a soup can, with a height and a base radius.
The formula to find the volume of a cylinder is \( V = \pi r^2 h \).

• \(V\) is the volume.
• \(r\) is the radius of the circular base.
• \(h\) is the height of the cylinder.

In the original problem, the cylinder has a radius of \(6\) cm and a height of \(15\) cm, resulting in a volume of \(540\pi\) cm³.
Such calculations are essential when determining the amount of material required or its capacity. Understanding this allows you to approach similar real-world objects with confidence.

Hemisphere

A hemisphere is half of a sphere. Imagine cutting a basketball into two equal pieces; each half is a hemisphere.
The formula for the volume of a hemisphere is a derivative of the sphere's volume formula. It is \( V = \frac{2}{3} \pi r^3 \).

• \(V\) is the volume of the hemisphere.
• \(r\) is the radius, which is half the diameter of the full sphere.

In this exercise, the hemisphere has a radius of \(6\) cm. Its volume is calculated as \(144\pi\) cm³.
This is crucial for understanding how much material this part alone uses when reshaped from another object like the cylinder.

Cone

A cone is a 3D shape with a circular base tapering smoothly to a point called the apex. It's similar to an ice cream cone.
The formula for the volume of a cone is \( V = \frac{1}{3} \pi r^2 h_c \).

• \(V\) is the volume of the cone.
• \(r\) is the radius of the circular base.
• \(h_c\) is the height of the cone.

In the given problem, after considering 8% metal wastage, the cone works with the hemisphere to match the remaining volume.
To find the height of the cone, we use the given shared radius of \(6\) cm, and calculate the height \(h_c\) as \(29.4\) cm. Understanding the cone's volume calculation helps in manufacturing contexts where precision is key.