Question #339414

Find the Center of mass of a thin plate of constant density š›æ covering the region bounded by the parabola š‘¦ = š‘„2 and the line š‘¦ = 4.


Expert's answer

Solution

Points of intersection of given curves are solution of equation š‘„2 = 4 => x1 = a = -2, x2 = b = 2.

For density š›æ mass of the plate is

M=Γ∫ab(4āˆ’x2)dx=Ī“(4xāˆ’13x3)∣2āˆ’2=Ī“(16āˆ’163)=Ī“323M=\delta\int_{a}^{b}{\left(4-x^2\right)dx=}\delta\left(4x-\frac{1}{3}x^3\right)\left|\begin{matrix}2\\-2\\\end{matrix}\right.=\delta\left(16-\frac{16}{3}\right)=\delta\frac{32}{3}

Equations of Moments

Mx=Ī“āˆ«āˆ’2212[42āˆ’(x2)2]dx=Ī“[8xāˆ’110x5]∣2āˆ’2=Ī“[32āˆ’6410]=25.6Ī“M_x=\delta\int_{-2}^{2}{\frac{1}{2}\left[4^2-\left(x^2\right)^2\right]dx}=\delta\left[8x-\frac{1}{10}x^5\right]\left|\begin{matrix}2\\-2\\\end{matrix}\right.=\delta\left[32-\frac{64}{10}\right]=25.6\delta

My=Ī“āˆ«āˆ’22x(4āˆ’x2)dx=Ī“[2x2āˆ’14x4]∣2āˆ’2=0M_y=\delta\int_{-2}^{2}x\left(4-x^2\right)dx=\delta\left[2x^2-\frac{1}{4}x^4\right]\left|\begin{matrix}2\\-2\\\end{matrix}\right.=0

Center of Mass Coordinates

xC=MyM=0x_C=\frac{M_y}{M}=0

yC=MxM=256āˆ™3320=2.4y_C=\frac{M_x}{M}=\frac{256\bullet3}{320}=2.4



LATEST TUTORIALS
APPROVED BY CLIENTS