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单词 CenterOfGravityOfCircularSector
释义

center of gravity of circular sector


Consider a circular sector with central angleMathworldPlanetmath 2α (in radians) and radius R as shown in the diagram below. If we wish to find the distance of the center of gravityPlanetmathPlanetmath from the center of the sector, we divide the sector into elements of area dA as illustrated.

In general, the mass of a lamina element is given by dm=δ2dA and the coordinates of the centers of mass are (given that the mass is evenly distributed over the area):

x¯=Ax𝑑AA,
y¯=Ay𝑑AA

In this case, we will use polar coordinates as it would be much easier to carry out the integration, and the boundaries can be defined easily. In polar coordinates dA=rdrdθ. The area of the sector is A=12R2(2α)=αR2.Now

x¯=1αR202α0Rxr𝑑r𝑑θ
=1αR202α0Rr2cosθdrdθ
=1αR202αR33cosθdθ
=R3αsin2α

Now we follow a similarMathworldPlanetmath procedure for the y-coordinate:

y¯=1αR202α0Ryr𝑑r𝑑θ
=1αR202α0Rr2sinθdrdθ
=1αR202αR33sinθdθ
=R3α(1-cos2α)

The center of gravity is (x¯,y¯) and the distance d of the center of gravity from the center of the sector is given by:

d=x¯2+y¯2

We substitute for x¯ and y¯:

d=(Rsin2α3α)2+(R(1-cos2α)3α)2
=R3αsin22α+(1-cos2α)2

From trigonometryMathworldPlanetmath, we know that:

2sin2α=1-cos2α

sin2α+cos2α=1

sin2α=2sinαcosα

Keeping these in mind, we substitute:

d=R3α4sin2αcos2α+4sin4α
=2R3αsin2α(cos2α+sin2α)
=2Rsinα3α

In conclusion, the distance of the center of gravity of a circular sector with radius R and angle 2α from the center of the sector is given by:

d=2Rsinα3α
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更新时间:2025/5/5 2:31:07