center of gravity of circular sector
Consider a circular sector with central angle 2α (in radians) and radius R as shown in the diagram below. If we wish to find the distance of the center of gravity
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):
ˉx | = | ∫∫Ax𝑑AA, | ||
ˉy | = | ∫∫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
ˉx | =1αR2∫2α0∫R0xr𝑑r𝑑θ | ||
=1αR2∫2α0∫R0r2cosθdrdθ | |||
=1αR2∫2α0R33cosθdθ | |||
=R3αsin2α |
Now we follow a similar procedure for the y-coordinate:
ˉy | =1αR2∫2α0∫R0yr𝑑r𝑑θ | ||
=1αR2∫2α0∫R0r2sinθdrdθ | |||
=1αR2∫2α0R33sinθdθ | |||
=R3α(1-cos2α) |
The center of gravity is (ˉx,ˉy) and the distance d of the center of gravity from the center of the sector is given by:
d=√ˉx2+ˉy2 |
We substitute for ˉx and ˉy:
d | =√(Rsin2α3α)2+(R(1-cos2α)3α)2 | ||
=R3α√sin22α+(1-cos2α)2 |
From trigonometry, we know that:
2sin2α=1-cos2α
sin2α+cos2α=1
sin2α=2sinαcosα
Keeping these in mind, we substitute:
d | =R3α√4sin2αcos2α+4sin4α | ||
=2R3α√sin2α(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 |
Title | center of gravity of circular sector |
---|---|
Canonical name | CenterOfGravityOfCircularSector |
Date of creation | 2013-03-22 18:06:30 |
Last modified on | 2013-03-22 18:06:30 |
Owner | curious (18562) |
Last modified by | curious (18562) |
Numerical id | 6 |
Author | curious (18562) |
Entry type | Topic |
Classification | msc 44A99 |
Related topic | CentreOfMassOfHalfDisc |