Transcript Center of Mass notes 1

Center of Mass notes
7-8 Center of Mass
In (a), the diver’s motion is pure translation; in (b) it is
translation plus rotation.
There is one point that moves in the same path a particle
would take if subjected to the same force as the diver. This
point is called the center of mass (CM).
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7-8 Center of Mass
The general motion of an object can be considered as the
sum of the translational motion of the CM, plus
rotational, vibrational, or other forms of motion about
the CM.
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7-8 Center of Mass
For two particles, the center of mass lies closer to the
one with the most mass:
where M is the total mass.
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7-8 Center of Mass
The center of gravity is the point where the gravitational
force can be considered to act. It is the same as the
center of mass as long as the gravitational force does not
vary among different parts of the object.
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7-8 Center of Mass
The center of gravity can be found experimentally by
suspending an object from different points. The CM
need not be within the actual object—a doughnut’s CM
is in the center of the hole.
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7-9 CM for the Human Body
The x’s in the small diagram mark the CM of the listed
body segments.
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7-9 CM for the Human Body
The location of the center
of mass of the leg (circled)
will depend on the position
of the leg.
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7-9 CM for the Human Body
High jumpers have developed a technique where their
CM actually passes under the bar as they go over it. This
allows them to clear higher bars.
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Problem 49.
• The distance between a carbon atom (m= 12 u) and an
oxygen atom (m = 16u) in the CO molecule is 1.13 x
10-10 m. How far from the carbon atom is the center of
mass of the molecule?
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Problem solution
49.Choose the carbon atom as the origin of coordinates. Use Eq. 7–9a.
mC xC + mO xO (12 u)(0)+(16 u)(1.13´10-10 m)
xCM =
=
= 6.5´10-11 m from the C atom
mC + mO
12 u +16 u
Problem 51
• The CM of an empty 1250 kg car is 2.4 m
behind the front of the car. How far from the
front of the car will the CM be when two
people sit in the front sear 2.8 m from the
front of the car and three people sit in the
back seat 3.9 m from the front? Assume that
each person has a mass of 65.o kg.
Solution
51. Find the CM relative to the front of the car. Use Eq. 7–9a.
mcar xcar + mfront xfront + mback xback
xCM =
mcar + mfront + mback
(1250 kg)(2.40 m)+ 2(65.0 kg)(2.80 m)+3(65.0 kg)(3.90 m)
=
= 2.62 m
1250 kg +5(65.0 kg)