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Chapter 3: Problem 21

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

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

Step by step solution

01

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.
02

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).
03

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.
04

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.
05

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 \(\pi\).
06

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:\[ y_{CM} = \frac{\sigma R^3}{3M} \int_{0}^{\pi} \sin(\phi) \, d\phi \]
07

Evaluate the Angular Integral

The angular integral becomes:\[ \int_{0}^{\pi} \sin(\phi) \, d\phi = 2 \]Therefore:\[ y_{CM} = \frac{\sigma R^3}{3M} \cdot 2 \]
08

Substitute Back the Mass and Surface Density

Now substitute \(\sigma = \frac{M}{\frac{1}{2} \pi R^2}\), and note \(M\) cancels out:\[ y_{CM} = \frac{R^3}{3 \cdot \frac{1}{2} \pi R^2} \cdot 2 = \frac{2 R}{3 \pi} \]
09

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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鈥檚 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.

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Most popular questions from this chapter

Show that the moment of inertia of a uniform solid sphere rotating about a diameter is \(\frac{2}{5} M R^{2}\). The sum (3.31) must be replaced by an integral, which is easiest in spherical polar coordinates, with the axis of rotation taken to be the \(z\) axis. The element of volume is \(d V=r^{2} d r \sin \theta d \theta d \phi\). (Spherical polar coordinates are defined in Section 4.8. If you are not already familiar with these coordinates, you should probably not try this problem yet.)

[Computer] A grenade is thrown with initial velocity \(\mathbf{v}_{\mathrm{o}}\) from the origin at the top of a high cliff, subject to negligible air resistance. (a) Using a suitable plotting program, plot the orbit, with the following parameters: \(\mathbf{v}_{\mathrm{o}}=(4,4), g=1,\) and \(0 \leq t \leq 4\) (and with \(x\) measured horizontally and \(y\) vertically up). Add to your plot suitable marks (dots or crosses, for example) to show the positions of the grenade at \(t=1,2,3,4 .\) (b) At \(t=4,\) when the grenade's velocity is \(\mathbf{v},\) it explodes into two equal pieces, one of which moves off with velocity \(\mathbf{v}+\Delta \mathbf{v} .\) What is the velocity of the other piece? (c) Assuming that \(\Delta \mathbf{v}=(1,3),\) add to your original plot the paths of the two pieces for \(4 \leq t \leq 9 .\) Insert marks to show their positions at \(t=5,6,7,8,9\). Find some way to show clearly that the CM of the two pieces continues to follow the original parabolic path.

Find the position of the center of mass of three particles lying in the \(x y\) plane at \(\mathbf{r}_{1}=(1,1,0)\) \(\mathbf{r}_{2}=(1,-1,0),\) and \(\mathbf{r}_{3}=(0,0,0),\) if \(m_{1}=m_{2}\) and \(m_{3}=10 m_{1} .\) Illustrate your answer with a sketch and comment.

A particle moves under the influence of a central force directed toward a fixed origin \(O\). (a) Explain why the particle's angular momentum about \(O\) is constant. (b) Give in detail the argument that the particle's orbit must lie in a single plane containing \(O\).

A rocket (initial mass \(m_{\mathrm{o}}\) ) needs to use its engines to hover stationary, just above the ground. (a) If it can afford to burn no more than a mass \(\lambda m_{\mathrm{o}}\) of its fuel, for how long can it hover? [Hint: Write down the condition that the thrust just balance the force of gravity. You can integrate the resulting equation by separating the variables \(t\) and \(m .\) Take \(v_{\text {ex }}\) to be constant.] (b) If \(v_{\mathrm{ex}} \approx 3000 \mathrm{m} / \mathrm{s}\) and \(\lambda \approx 10 \%,\) for how long could the rocket hover just above the earth's surface?
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