Transcript Finding the Center

Finding the Center
Discrete Center of Mass


Discrete masses are treated
as separate objects.
They can lie on a line or in a
plane.


m
r
 i i
N

rCM 

rCM
i 1
M

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
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m1r1  m2 r2  m3r3  m4 r4

m1  m2  m3  m4
m2
m1
m3
rCM
m4
Discrete Problem
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Two trucks are on a barge.
Find the center of mass.
• m1 = 11 Mg, d1 = -12 m
• m2 = 23 Mg, d2 = +12 m
• mb = 35 Mg
m1
m2
d CM 
•
•
•
•
m1d1  m2 d 2  mb d b
m1  m2  mb
m1 d1 = -132 Mg m
m2 d2 = +276 Mg m
mb db = 0
M = 69 Mg
• dCM = 2.1 m
mb
d1 0
d2
Center of Gravity
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All forces act as if at the center of mass.
• This includes gravity
• Sometimes called the center of gravity
m1
m2
Symmetry
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Continuous objects have a center of mass.
• In general the calculus is needed
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If an object is regular, the symmetry tells us the
center along that direction.
Break Up
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Before the split, momentum
is P = MV
• M total mass
• V center of mass velocity

After the split, the sum of
momentum is conserved.
• P = m1v1 + m2v2
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
Center of mass velocity
remains the same.
The kinetic energy is not
conserved.
v1
M
V
V
v2
Explosions
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A 325 kg booster rocket and
732 kg satellite coast at 5.22
km/s.
Explosive bolts cause a
separation in the direction of
motion.
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• K = (1/2)(1057 kg)(5.22 x
103 m/s)2 = 14.4 GJ
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Find the kinetic energy
before separation, and the
energy of the explosion.
Kinetic energy after
separation
• K1 = 16.4 GJ
• K2 = 0.592 GJ
• Satellite moves at 6.69 km/s
• Booster moves at 1.91 km/s
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Kinetic energy before
separation is (1/2)MV2
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The difference is the energy
of the explosion.
• Kint = 2.6 GJ
Thrust
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If there is no external force
the force to be applied must
be proportional to the time
rate of change in mass.
• The mass changes by Dm
• The velocity changes by Dv
• The mass added or
removed had a velocity u
compared to the object
Dv
Dm
m
u
Dt
Dt
The force u(Dm/Dt) is the thrust
Water Force
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Thrust can be used to find
the force of a stream of
water.
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A hose provides a flow of 4.4
kg/s at a speed of 20. m/s.
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The momentum loss is
Dp
Dm
u
Dt
Dt
• (20. m/s)(4.4 kg/s) = 88 N
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The momentum loss is the
force.
Heavy Water
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Water is poured into a
beaker from a height of 2 m
at a rate of 4 g/s, into a
beaker with a 100. g mass.
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There is extra momentum
from the falling water.
v  2 gh
  Dm / Dt
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What does the scale read
when the water is at 200. ml
in the beaker (1 ml is 1 g)?
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Answer: 302 g
Dp
Dm
v
  2 gh
Dt
Dt
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This is about 0.024 N or an
equivalent mass of 2.4 g.
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