Chapter 9.
Center of Mass and Linear Momentum
Myung-Hwa Jung (Sogang University)
Center Of Mass (COM)
For two particles separated by a distance d, with m1 in origin
If m2 = 0, xcom =0
If m1 = 0, xcom = d
If m1 = m2, xcom =
d
2
1
Center Of Mass (COM)
For two particles, with an arbitrary choice of origin
If x1 = 0,
Center Of Mass (COM)
Two particles in1 dimension (1D)
N particles in1 dimension (1D)
xCOM =
m1 x1 + m2 x2 + × × × + mn -1 xn -1 + mn xn
1
=
m1 + m2 + × × × + mn -1 + mn
M
n
åm x
i
i
i =1
n
M = å mi (총질량 )
i =1
N particles in 3 dimensions (3D)
r
rCOM Þ ( xCOM , yCOM , zCOM )
xCOM =
1
M
r
1
rCOM =
M
n
å mi xi , yCOM =
i =1
n
r
åm r ,
i i
i =1
1
M
n
å mi yi , zCOM =
i =1
1
M
n
åm z
i i
i =1
n
M = å mi (총질량 )
i =1
2
Center Of Mass (COM)
Solid bodies
For uniform density
즉,
Center Of Mass (COM)
Example 9.02 구멍 뚫린 원판의 질량중심
“C의 질량주우심은 P와 S를
합한 것의 질량중심과 같다.”
xC =
mS x S + m P x P - mS R + m P x P
=
=0
mS + m P
mS + m P
mP xP = mS R
®
xP =
mS
R
mP
mP : mS = [p (2 R ) 2 - pR 2 ] : pR 2
mP : mS = 3pR 2 : pR 2
\ xP =
®
mP = 3mS
1
R
3
3
Newton’s Second Law for a System of Particles
r
r
r
r
Mrcm = m1r1 + m2 r2 + m3 r3 + L
r
r
r
r
Mvcm = m1v1 + m2 v2 + m3v3 + L
r
r
r
r
Macm = m1a1 + m2 a2 + m3 a3 + L
r
r
r
= F1 + F2 + F3 + L
Linear Momentum
r
r
r
r
r
r
dp d
dv
dp
= (mv ) = m
= ma Þ Fnet =
dt dt
dt
dt
For system of particles,
If F = 0, P = const.
4
Collision and Impulse
Collision (충돌)
Impulse (충격량)
r
r dP
r r
r
tf r
F=
® dP = Fdt ® DP = ò Fdt
ti
dt
J = Favg Dt
r r r r tf r
DP = Pf - Pi = J = ò Fdt
ti
Linear momentum-impulse theorem
Collision and Impulse
(연속충돌)
f
J = ò Fdt = DP 이므로
i
\ Favg = J / Dt
&
J = - nDp
음의 부호:
J (+)와 Dp(-)가 반대방향
Þ Favg =
J - nDp
Dv
=
= -n m
Dt
Dt
Dt
\ Favg = - n
&
Dm = n × m
Dv Dm
Dm
=Dv
Dt n
Dt
5
Collision and Impulse
Example 9.04 2차원 충격량, 경주차-벽 충돌
r
r r r
(a) J = DP = Pf - Pi = m(vr f - vri )
J x = m[v f cos(-10o ) - vi cos(30o )]
J y = m[v f sin( -10o ) - vi sin(30o )]
r
J = J x iˆ + J y ˆj
J = J x2 + J y2 , q = tan -1
(b) Favg =
Jy
Jx
J
Dt
Conservation of Linear Momentum
r r r r tf r
DP = Pf - Pi = J = ò Fdt
ti
가정: F (and J) = 0 (고립계),
어떤 입자도 계에 출입할 수 없다(닫힌 계)
“Law of Conservation of Linear Momentum”
6
Momentum and Kinetic Energy in Collisions
r
r
Pi = Pf
r
r
r
r
P1i + P2i = P1 f + P2 f
Ki = K f
Ki ¹ K f
One-Dimensional Completely Inelastic Collision
Conservation of linear momentum
m1v1i + m2 v2i = m1v1 f + m2 v2 f
m1v1i = (m1 + m2 )V
V =
m1
vi
m1 + m2
Center of mass velocity :
7
One-Dimensional Completely Inelastic Collision
Example 9.07 운동량 보존, 탄동진자
mv = (m + M )V
V =
m
v
m+M
1
(m + M )V 2 = (m + M ) gh
2
\ v=
m+M
m
2 gh
Elastic Collision (1D): Stationary Target
Conservation of linear momentum
m1v1i + m2 v2i = m1v1 f + m2 v2 f
m1v1i = m1v1 f + m2 v2 f
Conservation of kinetic energy
1
1
1
1
m1v12i + m2 v22i = m1v12f + m2 v22 f
2
2
2
2
1
1
1
m1v12i = m1v12f + m2 v22 f
2
2
2
v1 f =
m1 - m2
v1i
m1 + m2
v2 f =
2m1
v1i
m1 + m2
8
v1 f =
m1 - m2
v1i
m1 + m2
v2 f =
2m1
v1i
m1 + m2
Elastic Collision (1D): Moving Target
Conservation of linear momentum
m1v1i + m2 v2i = m1v1 f + m2 v2 f
Conservation of kinetic energy
1
1
1
1
m1v12i + m2 v22i = m1v12f + m2 v22 f
2
2
2
2
9
Collisions in 2D
x comp.:
y-comp.:
Energy :
Systems with Varying Mass: Rocket
Pi = Pf
Using relative speed,
vrel = (v + dv) - U
Then, dM vrel = M dv
①
dM
dv
vrel = M
dt
dt
R vrel = M a
② dv =
dM
vrel
M
dM ù
é
êë R º dt úû
(R: 연료소비율)
로켓 방정식
10
Summary
• 보기문제
: 9.01, 9.02, 9.03, 9.04, 9.07, 9.08
• 연습문제
11