Math Problem Statement
Find the mass M and center of mass x overbar of the linear wire covering the given interval and having the given density delta left parenthesis x right parenthesis. delta left parenthesis x right parenthesis equals left brace Start 2 By 2 Matrix 1st Row 1st Column 7 minus x comma 2nd Column 0 less than or equals x less than or equals 1 2nd Row 1st Column x comma 2nd Column 1 less than x less than or equals 7 EndMatrix
Solution
We are given a linear wire covering the interval with a piecewise density function:
\begin{cases} 7 - x, & 0 \leq x \leq 1 \\ x, & 1 < x \leq 7 \end{cases}$$ We need to determine: 1. The total mass $$M$$. 2. The center of mass $$\bar{x}$$. ### **Step 1: Compute the Total Mass $$M$$** The mass of the wire is given by the integral: $$M = \int_0^7 \delta(x) \, dx.$$ Since $$\delta(x)$$ is piecewise, we split the integral into two parts: $$M = \int_0^1 (7 - x) \, dx + \int_1^7 x \, dx.$$ **First Integral:** $$\int_0^1 (7 - x) \, dx = \left[ 7x - \frac{x^2}{2} \right]_0^1$$ $$= \left( 7(1) - \frac{1^2}{2} \right) - \left( 7(0) - \frac{0^2}{2} \right)$$ $$= (7 - 0.5) - 0 = 6.5$$ **Second Integral:** $$\int_1^7 x \, dx = \left[ \frac{x^2}{2} \right]_1^7$$ $$= \left( \frac{7^2}{2} - \frac{1^2}{2} \right)$$ $$= \left( \frac{49}{2} - \frac{1}{2} \right) = \frac{48}{2} = 24$$ Thus, the total mass is: $$M = 6.5 + 24 = 30.5$$ --- ### **Step 2: Compute the Center of Mass $$\bar{x}$$** The center of mass is given by: $$\bar{x} = \frac{1}{M} \int_0^7 x \delta(x) \, dx.$$ Expanding the integral: $$\bar{x} = \frac{1}{30.5} \left[ \int_0^1 x(7 - x) \, dx + \int_1^7 x(x) \, dx \right]$$ $$= \frac{1}{30.5} \left[ \int_0^1 (7x - x^2) \, dx + \int_1^7 x^2 \, dx \right].$$ **First Integral:** $$\int_0^1 (7x - x^2) \, dx = \left[ \frac{7x^2}{2} - \frac{x^3}{3} \right]_0^1$$ $$= \left( \frac{7(1^2)}{2} - \frac{1^3}{3} \right) - \left( \frac{7(0^2)}{2} - \frac{0^3}{3} \right)$$ $$= \left( \frac{7}{2} - \frac{1}{3} \right) - 0 = \frac{21}{6} - \frac{2}{6} = \frac{19}{6}.$$ **Second Integral:** $$\int_1^7 x^2 \, dx = \left[ \frac{x^3}{3} \right]_1^7$$ $$= \left( \frac{7^3}{3} - \frac{1^3}{3} \right)$$ $$= \left( \frac{343}{3} - \frac{1}{3} \right) = \frac{342}{3} = 114.$$ Thus: $$\bar{x} = \frac{1}{30.5} \left( \frac{19}{6} + 114 \right)$$ $$= \frac{1}{30.5} \left( \frac{19 + 684}{6} \right) = \frac{703}{6 \times 30.5}$$ $$= \frac{703}{183} \approx 3.84.$$ --- ### **Final Answer:** - **Mass:** $$M = 30.5$$ - **Center of Mass:** $$\bar{x} \approx 3.84$$ Would you like a more detailed explanation of any step? #### **Related Questions:** 1. How do you find the center of mass for a 2D lamina? 2. How would the solution change if the density function were constant? 3. What happens to $$\bar{x}$$ if we change the density to $$\delta(x) = 2x$$ for all $$x$$? 4. Can we solve for $$M$$ and $$\bar{x}$$ using numerical integration? 5. How do mass and center of mass relate to physics concepts like torque? #### **Tip:** Breaking piecewise functions into separate integrals simplifies computations and reduces mistakes. Always check continuity at breakpoints.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Calculus
Piecewise Functions
Center of Mass
Integration
Formulas
M = ∫_0^7 δ(x) dx
x̄ = (1/M) ∫_0^7 x δ(x) dx
Theorems
Properties of integrals for piecewise functions
Suitable Grade Level
Grades 11-12
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