91Ó°ÊÓ

When dealing with problems related to geometric shapes, especially when objects are uniform, it's vital to understand what uniform mass distribution is all about.
**Uniform mass distribution** means the object's mass is spread evenly throughout its shape. This is crucial to imagine because it simplifies many physics problems.
For example, in our circular disc, every region, except for the cut-out section, carries the same mass per unit area. So, when calculating physical properties or behaviors, like center of mass, this assumption helps to consider just the geometric properties and overall mass easily without accounting for varying densities.
In calculus or physics, this concept allows us to assume mass is proportional to the area for twodimensional objects like discs.

The mass of a disc is vital in understanding the dynamics or balance of such an object. Given a **uniform mass distribution**, the mass ( \(m\)) of a disc is calculated using its:**- **Area**: For a disc of radius \(r\), its area \(A\) is given by \(\pi \times r^2\).- **Density**: If the mass density is \(\rho\), the mass of the disc is \(\rho \times A\).
In terms of numbers, let's look at our example: - The full disc has a radius of 6 cm, so its area becomes \(\pi \times 6^2 = 36\pi\).- When multiplied by the uniform density \(\rho\), the total mass \(m_1\) is \(36\pi\rho\).Similarly, the hole, with a 1 cm radius, has a mass calculated as \(\pi \times 1^2 \times \rho = \pi\rho\).
This simple but crucial mass calculation is a building block towards finding more complex properties like the center of mass.

In problems where a part of the object is removed, **mass subtraction** becomes essential to find properties of the object as a whole. By subtracting mass, we focus on what remains rather than what was originally present.
The **remaining mass** is key in our center of mass calculation. By removing the circular hole from the full disc, you have:- Original mass of the disc: \(m_1 = 36\pi\rho\)- Mass of the hole: \(m_2 = \pi\rho\)- Therefore, the mass of the remaining part is \[ m_r = m_1 - m_2 = 36\pi\rho - \pi\rho = 35\pi\rho \]This process of mass subtraction helps to isolate and work only with what's left after changes, ensuring accurate density and mass calculations for any physics-related derivations.

Finding the **center of mass** involves using a specific formula, especially when some portions are removed from a shape.
The general formula for the x-coordinate of the center of mass for the system of masses is:\[ x_{cm} = \frac{m_1 \cdot x_1 + m_2 \cdot x_2 + ... + m_n \cdot x_n}{m_1 + m_2 + ... + m_n} \]In our example, since the mass has been removed (subtraction), the equation is adapted to:\[ x_{cm} = \frac{m_1 \cdot x_1 - m_2 \cdot x_2}{m_r} \]Where:- \( m_1 \) is the full disc's mass (\(36\pi\rho\))- \( x_1 \) is 0, because the center initially is at the origin- \( m_2 \) is the mass of the hole \(\pi\rho\)- \( x_2 \) is the distance of the hole's center from the original disc center (3 cm)- \( m_r \) is the remaining mass of the disc after subtraction.Plugging these values in: \[ x_{cm} = \frac{36\pi\rho \cdot 0 - \pi\rho \cdot 3}{35\pi\rho} = \frac{-3}{35} \text{ cm} \]This gives the x-position of the new center of mass after the hole is removed.