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Find the center of mass of a thin plate of constant density \(\delta\) covering the given region.s The region bounded by the \(x\) -axis and the curve \(y=\cos x\) \(-\pi / 2 \leq x \leq \pi / 2\)

Short Answer

Expert verified
The center of mass is at \((0, \frac{\pi}{4})\).

Step by step solution

01

Understand the Problem

We are asked to find the center of mass of a thin plate with constant density over a region bounded by the x-axis and the curve \(y = \cos x\), where \(-\pi/2 \leq x \leq \pi/2\). The weight distribution is uniform across this region because the density \(\delta\) is constant.
02

Set Up the Bounds

The region of interest lies within the bounds \(-\pi/2 \leq x \leq \pi/2\) and is bounded by the x-axis and the curve \(y = \cos x\). This means we integrate with respect to \(x\) over this interval for calculations.
03

Calculate the Area of the Region

To find the total mass, we first find the area of the region. The area \(A\) under the curve \(y = \cos x\) from \(-\pi/2\) to \(\pi/2\) can be calculated as:\[ A = \int_{-\pi/2}^{\pi/2} \cos x \, dx \]Perform the integration to get:\[ A = [\sin x]_{-\pi/2}^{\pi/2} = \sin(\pi/2) - \sin(-\pi/2) = 1 - (-1) = 2 \]
04

Calculate Total Mass

The total mass \(M\) of the plate, assuming density \(\delta\), is given by the area times density:\[ M = \delta \cdot A = \delta \cdot 2 \]
05

Set Up for the Center of Mass

To find the center of mass, calculate the moments about the x-axis \(M_x\) and y-axis \(M_y\) using:\[ M_x = \delta \int_{-\pi/2}^{\pi/2} \frac{1}{2} (\cos^2 x) \, dx \]\[ M_y = \delta \int_{-\pi/2}^{\pi/2} x \cos x \, dx \]
06

Calculate \(M_x\)

The moment about the x-axis is calculated by:\[ M_x = \delta \int_{-\pi/2}^{\pi/2} \frac{1}{2} (\cos^2 x) \, dx \]Using the identity \(\cos^2 x = \frac{1 + \cos(2x)}{2}\), substitute and integrate:\[ M_x = \frac{\delta}{2} \int_{-\pi/2}^{\pi/2} \left( \frac{1 + \cos(2x)}{2} \right) dx = \frac{\delta}{2} \left[ x + \frac{1}{2} \sin(2x) \right]_{-\pi/2}^{\pi/2} \]Solving, we find:\[ M_x = \frac{\delta}{2} \times \pi \]
07

Calculate \(M_y\)

For the moment about the y-axis, calculate:\[ M_y = \delta \int_{-\pi/2}^{\pi/2} x \cos x \, dx \]Use integration by parts, letting \(u = x\) and \(dv = \cos x \, dx\). This gives us:\[ M_y = \delta \left( x \sin x \bigg|_{-\pi/2}^{\pi/2} - \int_{-\pi/2}^{\pi/2} \sin x \, dx \right) \]Calculating, we find:\[ M_y = \delta (\pi/2 \cdot 0 - (-\pi/2 \cdot 0) - [\cos x]_{-\pi/2}^{\pi/2}) = 0 \]
08

Calculate Center of Mass \((\bar{x}, \bar{y})\)

Finally, calculate the coordinates of the center of mass using the formulas:\[ \bar{x} = \frac{M_y}{M} = \frac{0}{\delta \cdot 2} = 0 \]\[ \bar{y} = \frac{M_x}{M} = \frac{\delta \times \pi/2}{\delta \cdot 2} = \frac{\pi}{4} \]
09

Conclusion

The center of mass of the plate is at the point \((0, \frac{\pi}{4})\).

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Key Concepts

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

Thin Plate
A thin plate refers to a two-dimensional object that is much thinner compared to its other dimensions like length and width. In problems involving thin plates, such as calculating the center of mass, we usually assume that its thickness is negligible. This assumption allows us to focus on the plate's area and the distribution of mass over that area.
The exercise involves a thin plate bounded by the x-axis and the curve given by the equation \(y = \cos x\), within the range \(-\pi/2 \leq x \leq \pi/2\). In this context, the thin plate lies flat on the \(xy\)-plane, and the height from the axis to the curve varies as \(x\) changes within the given interval.
Imagine the plate as a flat sheet of paper. It has no volume, only surface area, which makes calculations less complex. Using a thin plate simplifies the mathematics because the mass is spread evenly across its surface without additional thickness.
Constant Density
The concept of constant density is central to solving problems involving the center of mass. Density \(\delta\) describes how much mass per unit area exists throughout the plate. In this exercise, the density remains uniform across the entire region, meaning each small section of the plate has the same mass.
This uniformity simplifies calculations because it allows us to treat the entire plate as being evenly distributed in terms of mass, without concerning changes in thickness or composition. When we know that density is constant, we can calculate the total mass \(M\) of the plate easily as \(M = \delta \times A\), where \(A\) is the area of the plate.
Constant density ensures that mathematical integrations, like determining moments or centers of mass, are straightforward, avoiding the need for complex calculations that would arise if the density varied from point to point.
Moment of Inertia
Moments are mathematical expressions that help calculate the center of mass by considering how far each piece of mass is from a reference axis. Specifically, the moment of inertia, in this context, involves calculating the moments about the x-axis (\(M_x\)) and y-axis (\(M_y\)).
\(M_x\) measures how mass is distributed vertically, factoring in the distance each vertical strip of the plate is to the x-axis. Typically, this is calculated with an integral over the region bounded by the plate, using an expression that includes \(\cos^2 x\) in this scenario.
\(M_y\) considers the horizontal distribution of mass, acting as the balance point for the plate on the y-axis. It involves integrating \(x \cos x\) over the region using integration by parts. Such moments help us determine how balanced the overall structure of the plate is in relation to these axes, which is crucial for pinpointing the precise center of mass.
Integration by Parts
Integration by parts is a technique used in calculus to simplify complex integrals, particularly when dealing with the product of functions. It is derived from the product rule for differentiation and applies the formula:\[ \int u \, dv = uv - \int v \, du \]In the exercise, integration by parts is prominently used to calculate the moment \(M_y\), where \( u = x \) and \( dv = \cos x \, dx \). Through this method, we break down the integral \(\int x \cos x \, dx\) into more manageable parts.
Here's how it works:
  • Choose \(u\) and \(dv\), then differentiate \(u\) to find \(du\) and integrate \(dv\) to find \(v\).
  • Substitute into the integration by parts formula, simplifying step by step.
  • Solve the resulting integrals for a complete solution.
Integration by parts is particularly useful when one function in the product is easily differentiable while the other is straightforward to integrate, allowing for easier calculations and providing accuracy in computing moments like \(M_y\).

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Most popular questions from this chapter

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