Transcript Physics 141 Mechanics Yongli Gao Lecture 4 Motion in 3-D

Physics 141
Mechanics
Lecture 11
CM of Rigid Body
Yongli Gao
• To find the center of mass of an object of finite size,
we first concentrate on one whose shape and size are
fixed, a rigid body, which is an idealized model of a
solid object. We further assume that the object is
uniform and of density r=M/V. For a small volume
dV at r, the mass is dm=rdV, and the CM of a rigid
body is 1
1
1 M
1
rCM 
rdm

M
x CM 
1

xdV ,, y

V
M
CM
 r r dV


M V
1
ydV , , z

V
CM
 r dV


r dV

V
1
zdV

V
• We usually take advantage of the symmetry of the
object to reduce the actual calculation. If the object
is not uniform, we’d have to find the position
dependent density r(r), and keep it as part of the
integrand.
Example: CM of a Half Sphere
Find the CM of a uniform half sphere of radius R0.
Solution:
Assume that the z-axis is along the center of the half
sphere and perpendicular to the flat bottom, the
origion is at the center of the bottom. From the
symmetry of the sphere, we have xCM=yCM=0.

R0
zdV 
0

R0
z  R dz 
2
0

R0
0
z  ( R 0 2  z 2 )dz
1 2 2 1 4  R 0  R 0



R 0 z  z  |0 
2
4 
4
z CM 

R0
zdV
0
V

 R0 4 / 4
2  R0 / 3
3

4
3
8
R0
CM of a Composite Object
• If an object is composed of a number of pieces, and
the CM of each is known, then the CM of the
composite object can be obtained by treating each
piece as a particle.
m 11 , m12 , m 13 ,..., m 1n 1 , m 21 , m 22 , m 23 ,..., m 2 n 2 , m 31 , m 32 , m 32 , ..., m 3 n 3 ,...
rCM 1 
r m
m
1i
1i
1i
1i
m 1i
M1
, , rCM 2 
r
2i
m2 i
M2
,, rCM 3 
r
3i
m3 i
M3
,, ...
r11 m 11  r12 m 12  ...  r11 n m 1n  r21 m 21  r22 m 22 ...  r2 n m 2 n 
rCM 
1
1
2
2
m 11  m 12  m 13  ...  m 1n 1  m 21  m 22  m 23  ...  m 2 n 2  ...
r


r
CMj
Mj
j
M
j
j
Example: CM of a Right Angle Ruler
Find the CM of a right angle ruler of uniform arms
of length l1 and l2, respectively.
y
Solution:
For each arm, the CM is at its center.
x CM 1 
1
2
x CM 2  0, , y CM 2 
x CM 
y CM 
l2
l1 , , y CM 1  0, , M 1  r l1
1
2
l1
l2 ,, M 2  r l2
x CM 1 M 1  x CM 2 M 2
M1  M 2
y CM 1 M 1  y CM 2 M 2
M1  M 2
+


l1 r l1 / 2  0
r l1  r l2
0  l2 r l2 / 2
r l1  r l2


l1 2
2 (l1  l2 )
l2
2
2 (l1  l2 )
x
Example: CM of a New Moon
Sometime we can use the composite rule to find the
CM of each piece. Find the CM of disk of radius R
with a hole of radius r cut from its one edge.
Solution:
y
The piece cut from the hole is a disk
of radius r. The new moon and
the disk form a complete disk of
x
radius R, whose CM is at the center.
x CM 2  R  r , , M 2  r r 2 , , M 1  r ( R 2  r 2 )
x CM 
x CM 1 M 1  x CM 2 M 2
x CM 1  
M1  M 2
( R  r )r  r
2
r (R  r )
2
2

 
x CM 1 r ( R 2  r 2 )  ( R  r ) r r 2
r R
r
2
Rr
2
 0
Motion of CM
• The motion of the CM of a particle system or an
object of finite size can be described by Newton’s
2nd law as if all the mass is centered at the CM,
Fext 
F
 M a CM
i ,ext
where  Fi ,ext is the vector sum of all the external
forces acting on the system. The forces exerted by
one part of the system on another is called internal
forces, aCM is the acceleration of the CM.
M rCM 
mr
i i
i
d
dt
d
dt
M rCM  M
d rCM
dt
d
mr

dt
 M v CM 
M v CM  M a CM 

i i
i
d
mv

dt
i
i
i

ma
i
i
m
d ri
i
i
i

F
i
i
dt

mv
i
i
i
• Fi is the total force acting on the ith part of the
system, composed of external force Fi,ext that is
exerted by sources external to the system, and
internal force Fi,int, that is by the other parts of the
system. From the Newton’s 3rd law,  Fi ,int  0
F
i
i

F
i, ext
i
 Fext  M a CM

F
i ,int
i

F
i ,ext
i
 Fext
i
• The motion of each part of the system, however, is
determined by all the forces on it, including both
Fi,ext and Fi,int.
Example: Walking on Fishing Boat
If you walk slowly from one end of a fishing boat of
mass M and length L, how much will the boat
move? Suppose your mass is m.
Solution:
Walking slowly means we can ignore friction by
water. Now there’s no net force on the system
formed by you and the boat, and the CM remains
L
stationary. 0 m  LM / 2
ML
+
before :: x CM 0 

m M
after :: x CM 1 
2( m  M )
( x boat  L / 2 )m  x boat M
m M
m
M
x
 x CM 0
 ML / 2  ( x boat  L / 2 )m  x boat M
M  m
The center of the x

L
boat
2(m  M )
boat is now at
+
M
m
x
Example: Motion of Biatomic Molecule
A diatomic molecule can be approximated as two
particles of masses m1 and m2, linked by a rigid but
massless rod of length l. At a given moment, m1 is
moving up with velocity v10, and m2 horizontally
with v20, what will be the motion of the molecule as
a whole.
Solution:
0 m  lm
m
x CM 0 
v CMx 0 
v CMy 0 
1
2
m1  m 2

0 m 1  v 20 m 2
m1  m 2
v 10 m 1  0 m 2
m1  m 2
2
m1  m 2


l,, y CM 0  0
m2
m1  m 2
m1
m1  m 2
y
v10
v 20
m1
v10
l m
2
v20
x
The motion of the CM is a projectile of initial
velocity of vCM0. The total external force is
(m1+m2)g.
x CM  x CM 0  v CMx 0 t 
y CM  y CM 0  v CMy 0 t 
m2
m1  m 2
1
2
gt
2

(l  v 20 t )
m1
m1  m 2
v10 t 
1
2
gt
2
The motion of m1 is determined by the vector sum of
m1g+F21, and that of m2 by m2g+F12=m2g-F21