Remoulding

What is Remoulding?

Remoulding refers to the process of reshaping or reforming a solid object into a new shape without altering its material. This concept is commonly used in geometry problems where a solid is melted or reshaped, and the total volume remains constant.

Key Principles of Remoulding

Volume Conservation:

The volume of the original object equals the volume of the remoulded object.

V_{\text{original}} = V_{\text{remoulded}}

Shapes May Change, but the Material Stays the Same:

The remoulded object may have a different surface area or dimensions, but the volume remains constant.

Common Scenarios:

Melting a solid to form another shape (e.g., melting a sphere into a cylinder).
Cutting a shape and rearranging it into another form.

Worked Examples

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Example 1: Cylinder to Sphere

Problem: A cylindrical block of wax with a radius of 5 cm and height 12 cm is melted to form a sphere.

Find the radius of the sphere.

Solution:

Step 1: Calculate the volume of the cylinder:

V_{\text{cylinder}} = \pi r^2 h = \pi (5)^2 (12) = :highlight[300\pi \, \text{cm}^3]

Step 2: Equate to the volume of the sphere:

V_{\text{sphere}} = \frac{4}{3} \pi r^3

300\pi = \frac{4}{3} \pi r^3

Step 3: Solve for $r$ :

r^3 = \frac{300 \times 3}{4} = 225 \quad \Rightarrow \quad r = \sqrt[3]{225} \approx :success[6.13 \, \text{cm}]

Answer: The radius of the sphere is approximately 6.13 cm

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Example 2: Cube to Cylinder

Problem: A cube with a side length of 10 cm is remoulded into a cylinder with a height of 15 cm.

Find the radius of the cylinder.

Solution:

Step 1: Calculate the volume of the cube:

V_{\text{cube}} = a^3 = 10^3 = :highlight[1000 \, \text{cm}^3]

Step 2: Equate to the volume of the cylinder:

V_{\text{cylinder}} = \pi r^2 h

1000 = \pi r^2 (15)

Step 3: Solve for $r$ :

r^2 = \frac{1000}{15\pi} = \frac{200}{3\pi} \quad \Rightarrow \quad r = \sqrt{\frac{200}{3\pi}} \approx :success[4.61 \, \text{cm}]

Answer: The radius of the cylinder is approximately 4.61 cm

Summary

Remoulding Principle: The volume of the original and remoulded shapes is always equal.
Volume Conservation Formula: $V_{\text{original}} = V_{\text{remoulded}}$
Common applications involve transitioning between spheres, cylinders, cubes, and other shapes.
Use the specific volume formula for each shape to solve remoulding problems.
Practice with various geometrical shapes to enhance problem-solving skills.

Remoulding Simplified Revision Notes for Leaving Cert Mathematics