Question

A uniform thin sheet of metal is cut in the shape of a semicircle of radius \(R\) and lies in the \(x y\) plane with its center at the origin and diameter lying along the \(x\) axis. Find the position of the CM using polar coordinates. [In this case the sum (3.9) that defines the CM position becomes a two- dimensional integral of the form \(\int \mathbf{r} \sigma d A\) where \(\sigma\) denotes the surface mass density (mass/area) of the sheet and \(d A \text { is the element of area } d A=r d r d \phi .]\)

Short answer

The center of mass is at \((0, \frac{2R}{3\pi})\).

Step-by-step solution

Understand the Geometry and Symmetry

The problem involves a semicircular sheet with uniform mass distribution. The symmetry of the semicircle suggests that the center of mass (CM) will lie along the y-axis. The x-coordinate of the CM is zero due to symmetry about the y-axis.

Define Polar Coordinates

Use polar coordinates where any point on the sheet is defined as \((r, \phi)\). Here, \(r\) is the radial distance from the origin, ranging from 0 to \(R\), and \(\phi\) is the angular position ranging from 0 to \(\pi\) (since it's a semicircle).

Identify the Area Element

The differential area element in polar coordinates is given by \(dA = r \, dr \, d\phi\). This will be used to set up the integral that calculates the CM position.

Write the Expression for Surface Mass Density

Given the sheet is uniform, the surface mass density \(\sigma\) is constant. You can express it as \(\sigma = \frac{M}{A}\), where \(M\) is the total mass, and \(A = \frac{1}{2} \pi R^2\) is the area of the semicircle.

Setup the Integral for CM Position

The y-coordinate of the CM is given by:\[ y_{CM} = \frac{1}{M} \int y \, \sigma \, dA \]Substitute \(y = r \sin(\phi)\), \(dA = r \, dr \, d\phi\), and \(\sigma\) into the integral, restrict \(r\) from 0 to \(R\) and \(\phi\) from 0 to \(\pi\).

Perform the Integration

Calculate the y-coordinate:\[ y_{CM} = \frac{1}{M} \int_{0}^{\pi} \int_{0}^{R} r \sin(\phi) \sigma r \, dr \, d\phi \]This simplifies to:\[ y_{CM} = \frac{\sigma}{M} \int_{0}^{\pi} \sin(\phi) \int_{0}^{R} r^2 \, dr \, d\phi \]Evaluate the radial integral:\[ \int_{0}^{R} r^2 \, dr = \frac{R^3}{3} \]So the integral becomes:\[ y_{CM} = \frac{\sigma R^3}{3M} \int_{0}^{\pi} \sin(\phi) \, d\phi \]

Evaluate the Angular Integral

The angular integral becomes:\[ \int_{0}^{\pi} \sin(\phi) \, d\phi = 2 \]Therefore:\[ y_{CM} = \frac{\sigma R^3}{3M} \cdot 2 \]

Substitute Back the Mass and Surface Density

Now substitute \(\sigma = \frac{M}{\frac{1}{2} \pi R^2}\), and note \(M\) cancels out:\[ y_{CM} = \frac{R^3}{3 \cdot \frac{1}{2} \pi R^2} \cdot 2 = \frac{2 R}{3 \pi} \]

Conclusion: Final Result

The x-component of the CM is zero due to symmetry. Hence, the center of mass of the semicircular sheet is located at \( (0, \frac{2 R}{3 \pi}) \) along the y-axis.

Key concepts

semicircular sheet

A semicircular sheet is a half-circle shaped flat object, typically used in physics problems to explore concepts like symmetry and center of mass. In the exercise, our semicircular sheet is defined by its radius, denoted as \( R \), and is situated in the \( xy \)-plane, with its center located at the origin of this coordinate system.
The symmetry of the semicircle is crucial in simplifying calculations. Since the sheet is perfectly symmetric about the \( y \)-axis, the center of mass (CM) is expected to lie along this axis. This symmetry also implies that the x-coordinate of the CM is zero, making calculations more straightforward.

polar coordinates

Polar coordinates are a system of defining a point in a plane using a radius and an angle. It's particularly useful in dealing with circular and semicircular shapes. In this problem, each point on the semicircular sheet is represented using polar coordinates \((r, \phi)\).
Here, \(r\) is the radial distance from the origin and ranges from 0 to \(R\). The angle \(\phi\) describes the position along the circular path and ranges from 0 to \( \pi \), covering the semicircle. Using polar coordinates makes it easier to set up the integral needed to find the center of mass by naturally aligning with the shape’s geometry.

surface mass density

Surface mass density \(\sigma\) describes how mass is distributed over a surface area. For a uniform thin sheet, it's constant. Mathematically, it's expressed as \(\sigma = \frac{M}{A}\), where \(M\) is the total mass and \(A\) represents the area of the sheet.
For our semicircular sheet, the area \(A\) is given by \(\frac{1}{2} \pi R^2\). This formula is derived from the area of a full circle. Having uniform distribution simplifies calculations, as the integral over the mass density can treat \(\sigma\) as a constant.

two-dimensional integral

A two-dimensional integral allows us to calculate quantities over a surface area, considering both horizontal and vertical dimensions. In this exercise, the integral is set up to find the y-coordinate of the center of mass.
The integral expression is \( y_{CM} = \frac{1}{M} \int y \, \sigma \, dA \), where \(dA\) is the differential area element given by \( r \, dr \, d\phi \). This formula considers the contribution of small elements of the semicircular sheet to the overall center of mass.

• The integration is done first with respect to \(r\) (radial direction) from 0 to \(R\).
• Then with respect to the angle \(\phi\) from 0 to \(\pi\).

This method effectively calculates how each segment of the sheet contributes to the position of the center of mass. After performing the integral, the resulting y-coordinate of the CM is \( \frac{2R}{3\pi} \), confirming the balance of the semicircle along the y-axis.