Question

A uniform, square metal plate with side \(L=5.70 \mathrm{~cm}\) and mass \(0.205 \mathrm{~kg}\) is located with its lower left corner at \((x, y)=(0,0),\) as shown in the figure. \(\mathrm{A}\) square with side \(L / 4\) and its lower left edge located at \((x, y)=(0,0)\) is removed from the plate. What is the distance from the origin of the center of mass of the remaining plate?

Short answer

Question: A square metal plate is 5.70 cm on a side and has mass 0.205 kg. A smaller square section with sides equal to one fourth the side of the large square is removed from one corner. Calculate the distance from the origin to the center of mass of the remaining shape. Solution: Step 1: Find the dimensions of each square The side length of the large square (A+B) is 5.70 cm. The side length of the smaller square (A) is (5.70/4) cm. Step 2: Find the mass of each square The mass of the large square (A+B) is 0.205 kg. The mass of the smaller square (A) is 0.205/(5.70)^2 * (5.70/4)^2 kg. The mass of the remaining plate is the mass of the large square minus the mass of the smaller square. Step 3: Find the center of mass of each square For the large square (A+B): (x1, y1) = (5.70/2, 5.70/2) For the smaller square (A): (x2, y2) = ((5.70/4)/2, (5.70/4)/2) Step 4: Find the center of mass of the remaining shape (x_cm, y_cm) = 1/M_remaining * (M_large(x1, y1) - M_small(x2, y2)) Step 5: Find the distance from the origin D = sqrt((x_cm)^2 + (y_cm)^2) Apply the values obtained in Steps 1-4 into Step 5 to find the distance from the origin to the center of mass of the remaining shape.

Step-by-step solution

Find the dimensions of each square

For the large square (A+B), Side length: \(L_{large} = 5.70\,\text{cm}\) For the smaller square (A), Side length: \(L_{small} = \frac{L_{large}}{4} = \frac{5.70}{4} \,\text{cm}\)

Find the mass of each square

Given the mass of large square (A+B), Mass: \(M_{large} = 0.205\,\text{kg}\) To find the mass of the smaller square (A), use mass density, \(\rho = \frac{M}{L^2} \Rightarrow M = \rho L^2\) For large square (A+B), \(\rho = \frac{M_{large}}{L_{large}^2} = \frac{0.205}{(5.70)^2}\,\text{kg/cm}^2\) For the smaller square (A), \(M_{small} = \rho L_{small}^2 = \frac{0.205}{(5.70)^2}\cdot (\frac{5.70}{4})^2\,\text{kg}\) Then, find the mass of the remaining plate, \(M_{remaining} = M_{large} - M_{small}\)

Find the center of mass of each square

For the large square (A+B), \((x_1, y_1) = (\frac{L_{large}}{2}, \frac{L_{large}}{2})\) For the smaller square (A), \((x_2, y_2) = (\frac{L_{small}}{2}, \frac{L_{small}}{2})\)

Find the center of mass of the remaining shape

Using the equation for the center of mass, \((x_{cm}, y_{cm}) = \frac{1}{M_{remaining}}(M_{large}(x_1, y_1) - M_{small}(x_2, y_2))\) Substitute the values and calculate the final coordinates of the center of mass.

Find the distance from the origin

Finally, to determine the distance from the origin to the center of mass, use the distance formula, \(D = \sqrt{(x_{cm})^2 + (y_{cm})^2}\) Calculate the distance to complete the exercise.

Key concepts

Uniform Density

Understanding the concept of uniform density is critical when calculating the center of mass for various objects. Uniform density means that the mass is evenly distributed throughout the entire volume of an object. In the context of our exercise, we are dealing with a square metal plate assumed to have uniform density, which simplifies the calculation process significantly.

When an object has uniform density, it implies that any piece of the object, no matter how small, has the same density as any other piece of the same size. This characteristic allows us to find the mass of parts of the object, such as the small square removed from the plate, without complicated mass distribution calculations. It's the uniform density that ensures the mass of smaller sections is directly proportional to their size when compared with the whole object.

Mass Distribution

In physics, the term mass distribution refers to how mass is spread out within an object. The mass distribution affects the object's center of mass, which is the point where it can be balanced in all directions. With our square plate, when a smaller square is removed, it alters the mass distribution, shifting the center of mass from its original location at the geometric center of the larger square.

To calculate the center of mass accurately in our given situation, it's essential to consider the mass distribution of both the large square and the small square that was removed. Knowing that the small square is homogeneous and finding its mass allows us to determine how much it influences the mass distribution of the larger figure. This becomes particularly crucial when analyzing complex geometries or when dealing with an irregularly shaped body.

Coordinate System

A coordinate system is an essential tool in physics that allows us to define the position of points in space. For our exercise, we are using a two-dimensional Cartesian coordinate system with x and y axes. This system is incredibly useful for locating the center of mass of objects in a plane and for simplifying calculations.

In the exercise, the lower left corner of the original square (before any section is removed) is aligned with the origin, at (0, 0). This makes it more straightforward to determine the position of the removed square and, subsequently, to calculate the center of mass of the remaining shape. Understanding how to utilize the coordinate system effectively is essential for solving problems related to the center of mass and for visualizing the spatial relationship between an object and its components.

Geometry in Physics

The field of geometry in physics provides us with powerful tools to analyze various physical scenarios, including the calculation of the center of mass. Geometry allows us to describe physical objects, their properties, and relationships through mathematical principles. In this textbook exercise, understanding the geometry of the square metal plate and the smaller square cut from it is integral to finding the position of the center of mass.

For instance, geometric understanding tells us that for square objects, if they have uniform mass distribution, the center of mass lies at the intersection of the diagonals - the geometric center of the square. By applying these principles and recognizing the geometric transformations that occur when part of the object is removed (like the smaller square in our problem), we can accurately compute the center of mass for even more complex systems.