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Unformatted text preview: mo: two masses
connected by spring.
Why? Newton’s 3rd Law. A fundamental result.
2 1
v
r v
r 2 1 v
r CM O CM
3 v
r 3 Consider a collection of particles or pieces of a continuous
object, with masses m1 , m2 .... and positions r1 , r2 ....
Sum Fi = mi ai over all particles i=1,2,3.....
“Internal Forces” cancel in pairs due to Newton’s 3rd law.
The final result is a simple relation between the net external
force and the acceleration of the center of mass.
(transparency) of mass : rcm = mi ri , M M = tota TEX for keynote a fundamental result...........
et external force : i Fnet,ext = 1
Deﬁne center of mass :
rcm =
mi ri ,
Mi
TEX for keynote
net external force :
Fnet,ext = Fi,ext.
iM = total mass
2 d rcm
= acceleration of center of mass =
i
2
dt
2
d rcm
Fi,ext. acm = acceleration of 1
center of mass =
dt2 mass
Deﬁne center of mass :
rcm =
mi ri ,
M = total
M
net,ext i = M a cm
Then :
Fnet,ext
cm
net external force :
Fnet,ext =
Fi,ext. Then : F =Ma
i d2 rcm
T ∝ L/g
acm = acceleration of center of mass =
dt2
π
Thus: Center of Mass=: 3.141592653589793....
Thenbehaves like= point particle in response
Fnet,ext a M acm T∝ to the net external forces. L/g
r Ramifications........ If there are no external forces, the center of mass is either
stationary or moves with constant velocity.
If there are no external forces, “ total momentum” is conserved.
(we’ll see.)
The cancelation of the effect of internal forces is sometimes
surprising, and powerful for predictions. Intuitions about Center of Mass:
The center of mass is a weighted average of the positions of the
particles, weighted by mass.
For symmetrical objects, like a disk, it is at the geometrical
center.
Later we will see: it is the point where the object can be
“balanced”......(has to do with torque...)
m1 + m2 m1 + m2 Example
Where is the center of mass in the following mass
distribution?
y
2m
(12,12)
m
(0,0) m
(24,0) x Solution: X CM YCM ∑mx
=
i i m0 + (2 m)12 + m24
=
= 12
4m i M ∑my
=
i M i i m0 + (2 m)12 + m0
=
=6
4m y
2m
(12,12)
m
(0,0) € RCM = (12,6) m
(24,0) x of mass : rcm = mi ri , M M = tota TEX for keynote review of THE fundamental result..
et external force : i Fnet,ext = 1
Deﬁne center of mass :
rcm =
mi ri ,
Mi
TEX for keynote
net external force :
Fnet,ext = Fi,ext.
iM = total mass
2 d rcm
= acceleration of center of mass =
i
2
dt
2
d rcm
Fi,ext. acm = acceleration of 1
center of mass =
dt2 mass
Deﬁne center of mass :
rcm =
mi ri ,
M = total
M
net,ext i = M a cm
Then :
Fnet,ext
cm
net external force :
Fnet,ext =
Fi,ext. Then : F =Ma
i d2 rcm
T ∝ L/g
acm = acceleration of center of mass =
dt2
π
Thus: C...
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This note was uploaded on 09/29/2012 for the course MATH 1920 taught by Professor Pantano during the Spring '06 term at Cornell University (Engineering School).
 Spring '06
 PANTANO
 Multivariable Calculus

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