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
Step by step solution
Understand the Geometry and Symmetry
Define Polar Coordinates
Identify the Area Element
Write the Expression for Surface Mass Density
Setup the Integral for CM Position
Perform the Integration
Evaluate the Angular Integral
Substitute Back the Mass and Surface Density
Conclusion: Final Result
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
semicircular sheet
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
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
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
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 \(\pi\).