Question

A thin (one-dimensional) wire of constant density is bent into the shape of a semicircle of radius $a$. Find the location of its center of mass. (Hint: Treat the wire as a thin halfannulus with width $\mathrm{\Delta }a,$ and then let $\mathrm{\Delta }a\to 0$.)

Short answer

Answer: The center of mass of the semicircular wire is located at (0, (2a)/π).

Step-by-step solution

Parametrize the Wire

To describe the points on the wire, let's use polar coordinates $(r,\theta )$. The wire forms a semicircle of radius $a$, so the points on the wire can be described by the equation $r=a$. Since it's a semicircle, the angle $\theta$ will range from 0 to $\pi$.

Define the Differential Mass Element

We will treat the wire as a thin half-annulus with width $\mathrm{\Delta }a$ and then let $\mathrm{\Delta }a$ approach 0. Let's consider the differential mass element $dm$ for a small segment of the wire. The differential mass element can be written as $dm=\rho ds$, where $\rho$ is the constant density of the wire and $ds$ is the differential arc length. In polar coordinates, we know that $ds=ad\theta$. Therefore, $dm=\rho ad\theta$.

Find the y-coordinate of the Center of Mass

Since the wire is symmetric about the y-axis, we only need to calculate the y-coordinate of the center of mass, which we denote as $\overline{y}$. The formula for the y-coordinate of the center of mass is $\overline{y}=\frac{1}{M}{\int }_0^{\pi }ydm$, where $M$ is the total mass of the wire. Using the polar coordinate transformation, we can express y as $y=asin(\theta )$. Now, we can find the total mass of the wire as $M={\int }_0^{\pi }dm=\rho a{\int }_0^{\pi }d\theta =\rho a\pi$.

Calculate the Integral

Now we can calculate the integral for $\overline{y}$: $\overline{y}=\frac{1}{\rho a\pi }{\int }_0^{\pi }a\sin (\theta )\rho ad\theta =\frac{1}{\pi }{\int }_0^{\pi }a\sin (\theta )d\theta$. We can now evaluate the integral: $$\overline{y}=\frac{1}{\pi }{[-a\cos (\theta )]}_0^{\pi }=\frac{1}{\pi }(a\cos (0)-a\cos (\pi ))=\frac{a}{\pi }(1+1)=\frac{2a}{\pi }$$.

Determine the Center of Mass

With our result from Step 4, we conclude that the y-coordinate of the center of mass is $\overline{y}=\frac{2a}{\pi }$. Therefore, the center of mass of the semicircular wire is located at $(0,\frac{2a}{\pi })$.

Key concepts

Polar Coordinates

Understanding polar coordinates is crucial for solving many problems in physics and mathematics, including finding the center of mass in objects with circular symmetry. Unlike the Cartesian coordinates, which use perpendicular axes to define a point in space, polar coordinates describe a point based on its angle and distance from a reference point.

In polar coordinates, any point can be represented as $r,\theta )$, where $r$ is the radius—the distance from the origin—and $\theta$ is the angle measured from the positive x-axis. For the semicircular wire problem, every point on the wire is at the same distance $a$ from the origin (the center of the circle), so $r=a$. This simplifies the parametrization of the wire to just considering $\theta$, which ranges from $0$ to $\pi$ because the wire forms a semicircle. Thus, the careful use of polar coordinates allows us to succinctly describe the position of infinitesimally small segments of the wire in terms of the variable angle and a constant radius.

Differential Mass Element

To find the center of mass, we need to consider the mass distribution of the object, which brings us to the concept of a differential mass element ($dm$). The idea is to divide the object into infinitesimally small segments, each of which contributes to the total mass.

For a uniform wire, $dm$ is proportional to the length of the small segment $ds$ and the density $\rho$ of the material. In our semicircle, $ds$ is an arc length that can be expressed in polar coordinates as $ds=ad\theta$. Thus, $dm=\rho \cdot ds=\rho ad\theta$. Notably, even though $\mathrm{\Delta }a$ is mentioned in the hint, it's not explicitly used in calculations due to the wire being one-dimensional and the density remaining constant. Therefore, we use $dm$ to represent a thin slice of the wire's mass, based on its angle and the constant density. This concept simplifies the calculation of the center of mass into a manageable integral over the angle $\theta$.

Integral Calculus

The center of mass problem is intricately solved using integral calculus, which is a branch of mathematics concerned with the accumulation of quantities. When we seek the y-coordinate of the center of mass ($\overline{y}$), integral calculus allows us to sum up all the infinitesimal contributions of mass-elements along the semicircular wire.

To calculate $\overline{y}$, we set up an integral with the function that represents the y-values over our semicircle ($a\sin (\theta )$) and the differential mass element we previously defined. Integral calculus especially shines in problems like these, where symmetry simplifies the process. Because our wire is a semicircle, we only need to calculate one coordinate—the y-coordinate—due to symmetry about the y-axis. The integral ${\int }_0^{\pi }a\sin (\theta )d\theta$ effectively 'adds up' all the y-contributions from each infinitesimal mass element encapsulated in $dm$, resulting in the expression for $\overline{y}$. Therefore, integral calculus not only finds the sum of continuous distributions but in cases like these, also elegantly uses the properties of symmetry and geometry to give us an accurate and insightful solution.