6: Moments and Centroids
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Mass and Slugs
Newton's Second Law states that
F=ma
where F is the force, m is the mass, and a is the acceleration. In the US system, Force is measured in pounds and mass is measures in slugs.
I weigh 165 lbs. What is my mass?
Solution
Since weight corresponds with gravitational force, and the acceleration of gravity is 32 ft/sec2 we have
165=32m
or
m=5.15 slugs.
In the metric system, kg is a mass unit and Newtons is a weight unit.
Moments and Center of Mass for Discrete Mass Points
Suppose that we have a teeter totter and a 10 kg child is on the left 5 meters from the center of the teeter totter and a 15 kg child is on the right 4 meters from the center of the teeter totter. We define the moment as:
10(−5)+15(4)=10.
In general, we define the moment for masses mi at the points xi to be
Moment=∑mixi.
If the moment is 0 then we say that the system is in equilibrium. Otherwise, let x be the value such that
∑mi(xi−x)=0.
Then x is called the center of mass of the system.
Tthe center of mass of the system is given by
x=momenttotal mass
Proof:
∑mi(xi−ˉx)=∑mixi−∑miˉx=∑mixi−ˉx∑mi
so that
x=∑mi∑xi.
Find the center of mass of the teeter totter.
Solution
We have
moment =10
and
total mass=25
hence the
center of mass=1025=0.4.
We can say that if the center of the teeter totter was 0.4 meters from the current center, then the children would be in balance.
For points in the plane, we can find moments and centers of mass coordinate wise. We define:
mx=moment about the x axis =Smixi,
my=moment about the y axis =Smiyi,
Center of Mass =(myM,mxm).
For the points (−3,0) with mass 4, (2,2) with mass 3, and (1,−2) with mass 1 we have
mx=(4)(−3)+(3)(2)+(1)(1)=−5,
my=(4)(0)+(3)(2)+(1)(−2)=8,
Center of Mass=(−58,88)=(−0.625,1).
Center of Mass for a Two Dimensional Plate
First, we recall that for a region of density r bounded by f(x) and g(x)
Mass=(Density)(Area)=ρ∫ba(f(x)−g(x))dx
Mx=ρ∫baf(x)−g(x)2(f(x)−g(x))dx
My=ρ∫bax(f(x)−g(x))dx
and
Center of Mass=(MyM,MxM)
Find the center of mass for the plate of constant density 2 that is bounded by the curves
y=1−x, y=0 and x=0.
We have
Mx=2∫101−x2(1−x)dx,
My=2∫10x(1−x)dx,
M=2∫10(1−x)dx.
Pappus Theorem
Suppose that we revolve a region around the y-axis. Then the volume of revolution is:
V=2prA
where A is the area of the region and r is the distance from the centroid (constant density) to the axis of rotation.
Suppose that we revolve the 4 x4 frame with width 1 centered about (6,2) about the y-axis. Then we have that the Area is
A=4+4+2+2=12.
R=7
so that
V=2p7(12)=168p.
Find the volume of the torus formed by revolving the disk
(x−11)2+y2=4
about the y-axis.
Contributors and Attributions
- Larry Green (Lake Tahoe Community College)
Integrated by Justin Marshall.