ch09: center of mass and linear momentum 9.1: what is physics? · 2015-11-25 · linear momentum...
TRANSCRIPT
The center of mass (com)Motion of a system of particles (Newton’s 2nd
law)Linear momentumImpulseConservation of linear momentumInelastic Collision in one dimensionElastic collision in 1 and 2 dimensions
Ch09: Center Of Mass And Linear Momentum 9.1: What is physics?
Analyzing complicated motion of any sort requires simplification via an understanding of physics. The complicated motion of a system of objects, such as a car or a ballerina or collection of particles, can be simplified if we determine a special point of the system—the center of mass of (com) that system.We can apply equations of motion to the
center of mass to represent the motion of that system
An example for complicate motion is a baseball bat when flipped into air points of bat moves along different path shapes except the center of mass moves a long a simple parabolic path (projectile motion)
9.2: The Center Of MassIf we consider a collection of particles or an extended object with a total mass, M.We can consider the system as a single particle with the total mass, M, concentrated at a single point called the center of mass (com).If a net force external is applied to the system, then you can apply Newton’s second law to the center of mass.
The center of mass of a system of particles is the point that moves as though:
all of the system’s mass were concentrated there all external forces were applied there.
M
xmx
n
iii
com
∑== 1
The position of center of mass xcom is the average position of the system’s masses on x-axis
n
nncom mmm
xmxmxmx++++++
=................
21
2211
Total mass of the system =M
; closer to bigger
9.2: The Center Of Mass: system of particles
For the two particle shown, the xcom is
For n particles along x-axis
For group of particles in space (more than one dimension), we need to find position vector for the center of mass comrr
kzjyixM
rmr CMCMCM
iii
comˆˆˆ ++==
∑ r
r
M
xmx i
ii
com
∑=
M
ymy i
ii
com
∑=
M
zmz i
ii
com
∑=
∑∑
==i
iii
ii
com rmMM
rmr r
r
r 1
9.2: The Center Of Mass: system of particles
A group of particles in two dimensions is shown in the figure. Find the center of mass if m1=1kg, m2=1kg, m3=2kg
m
mmmymymymycom
144
211)2)(2()0)(1()0)(1(
321
332211
==++
++=
++++
=
we have x and y dimensions
mjircom )ˆ1ˆ75.0( +=r
m
mmmxmxmxmxcom
75.043
211)0)(2()2)(1()1)(1(
321
332211
==++
++=
++++
=
9.2: The Center Of Mass: Example
∫= dmrM
rCMrr 1
∫= xdmM
xCM1
∫= ydmM
yCM1
∫= zdmM
zCM1
For an extended solid object (continuous distribution of matter)
M
rmr i
ii
CM
∑∆=
r
r
For ∆mi 0 dm
9.2: The Center Of Mass: Solid Bodiesa) show that the center of mass for the rode of length L lies at the middle b) if the rode is not uniform with λ = αx find center of mass
m=ρV dm= ρdV
m= σA dm=σdA
m=λL dm=λdL
a) Rode lies on x-axis, rode is uniform λ is constant
λdL=λdx because L lies on x-axis
b) λ is variable
9.2: The Center Of Mass: Example
For a system of particles in motion, we usually interesting in the motion of the center of mass that represents the whole system:
9.3: Newton’s Second Law For a System Of Particles
the velocity of the center of mass ( ) of the system is found from
comvr
∑=i
iicom vmM
v rr 1
nni
iicom vm.........vmvmvmvmvM rrrrrr++++== ∑ 332211
Hence, the total mass (M) multiplied by the velocity of center of mass ( ) represents the all system; it is equivalent to multiply each particle velocity with it’s mass and sum them
comvr
9.3: Newton’s Second Law For a System Of Particles
∑=i
iicom amM
a rr 1
netncom
nni
iicom
FFFFFaM
am.........amamamvmaMrrrrrr
rrrrrr
=++++=⇒
++++== ∑...........321
332211
taking the derivative of the , you can get the acceleration of the center of mass ( )
comvr
comar
Newton’s 2nd law
is the net force of all external forces that act on the system; it is equivalent to the sum of all forces act on all particles in the system
in components
9.3: Newton’s Second Law For a System Of Particles: Example
Consider the figure shown. what is the acceleration of the center of mass of the system, and in what direction does it move?
Since we have a system moving under external forces shown, the forces and the masses can be transferred to the center of mass (com) as shown
The com of the system will move as if all the mass were there and the net force acted there
9.3: Newton’s Second Law For a System Of Particles: Example: continued from previous slide
Newton’s 2nd law
x-component
y-component
and
It is a vector quantity has the direction of the velocity For a particle, any change in momentum due to velocity change is related to some external net force
The linear momentum of a particle with mass m and velocity is:
vrvmp rr
= SI unit is kg m/s ≡ N.s
9.4: Linear Momentum
pr
vr
netFr
From Newton’s 2nd law
Since Linear momentum is a vector quantity it may has components in the x-, y-, and z- direction
vmp rr=
zzyyxx mvpmvpmvp ===
dtpdFrr
=∑
kpjpipp zyxˆˆˆ ++=
r
the time rate of change of particle’s linear momentum is equal to the net force acting on the particle (other form of Newton’s 2nd law).
9.4: Linear Momentum
Momentum vector
Momentum components
where
The linear momentum for the center of mass that represents the system particles can be obtained from above equation:
9.5: Linear Momentum For a system of particles
For a system of particles in motion each has its own velocity and mass the velocity of the center of mass ( ) iscomvr
∑=i
iicom vmM
v rr 1
∑
∑++++===⇒
++++===
inicomcom
nni
iicomcom
p.........ppppvMP
vm.........vmvmvmvmvMP
rrrrrrr
rrrrrrr
321
332211wherecomcom vMP rr
= Momentum for system of particles
Hence, the momentum for the system is equivalent to the sum of all individual momentum of particles
9.6: Collision and impulse: Single CollisionMomentum of particle can only changes by a net external force
Consider a collision between a ball and a bat in baseball gameThe ball will experience a force that
varies during collision ball slows down, stop, and reverse direction changes due to that force From Newton’s 2nd law
we can have the change in momentum by integrating both sides over time during collision
A measure for magnitude and duration of a collision force (impulse )
9.6: Collision and impulse: Single Collision
whereIn components (x-components for instant)
the Impulse on the particle (ball in this case):
Since it is difficult to know during collisions we use the average force magnitude and ∆t:
Due to Newton’s 3rd law, the ball make an impulse on the bat of same magnitude but opposite in direction.
Impulse-momentum theorem Impulse is the net change in momentum due to F.
A car made an accident with a wall. Its velocity was 15 m/s to left beforthe accident. After the accident, it is recoil from the wall to the right with velocity 2.6 m/s. If the accident time is 0.15 s, find the impulse caused by the collision and the average force exerted on the car.m = 1500 kg, vi = -15.0 m/svf = 2.6 m/s, Dt = 0.150 sJ = ? Favg = ?
9.6: Collision and impulse: Example
Note: for x-axis
If v to the right v is +ve
If v to the left v is -ve
The impulse J = ∆pJ = pf - pi = mvf – mvi = m(vf - vi)J = 1500(2.6-(-15))J = 1500 (17.6) = 2.64×104 î N.s
Favg = ∆p/∆t = 2.64×104 / 0.15 = 1.76×105 î N
9.6: Collision and impulse: Example: continued from previous slide 9.6: Collision and impulse: Example
Figure below shows the path taken by a race car driver as he collides with the racetrack wall. If driver mass m is 80 kg finda) The impulse on the driver due to collisionb) Average force magnitude on driver if the collision lasts for 14 ms
x-component
y-component
From –ve x-axis
(a) (b)
Consider two particles interact together We can apply Newton’s Third law
constant 21 ==+ ∑ ppp rrr
9.7: Conservation of Linear Momentum
0)( 21 =+ ppdtd rr
Apply Newton’s 2rd law iiff
if
pppp 2121
pprrrr
rr
+=+
= ∑∑
∑∑∑∑∑∑
=
=
=
izfz
iyfy
ixfx
pp
pp
pp
Momentum is vector quantity it has components
conservedismomentumLinear⇒
Take care of velocity v sign (+ve or –ve)
9.7: Conservation of Linear Momentum
m1 = 100 kg, v1 = 5 m/s, m2 = 50 kg
Two skaters recoil each other from rest. After recoil Find a) the total momentum b) the momentum for heavy skater1 and c) the velocity for light skater2
a) ptot = ?, b) p1 = ?, c) v2 = ?Solution: momentum is conserved in x-axis
a) Σpi = ptot = 0 ; they start from rest
b) p1= m1v1= 100(-5)= - 500 kg m/s
c) Σpf = Σpi = ptot
p1f + p2f = 0 p2f = -p1f m2v2 = -p1f
v2=-p1f / m2 v2= -(-500)/50=10 m/s
9.7: Conservation of Linear Momentum: Example
Same with coordinates
9.7: Conservation of Linear Momentum: Example
Two-dimensional explosion: the figure shows a firecracker placed inside a coconut of mass M, initially at rest on a frictionless floor, blows the coconut into three pieces that slide across the floor. Piece C, with mass0.3M, has final speed vfC = 5.0 m/s. (a) What is the speed of piece B, with mass 0.20M? (b) What is the speed of piece A?
0.3M
0.2M
0.5M
= 5m/s
smv
.v.
Mv.Mv.
vmvmvmp
fB
fB
fCfB
fCyCfByBfAyAfy
/67.9
04811530
008sin3005sin200
00
=⇒
=+−
=°+°−
=++⇒=∑)1(0260129050
008cos3005cos2050
00
=++−
=°+°+−
=++⇒=∑
.v.v.
Mv.Mv.Mv.
vmvmvmp
fBfA
fCfBfA
fCxCfBxBfAxAfx
9.7: Conservation of Linear Momentum: Example: continued from previous slide
(a) What is the speed of piece B? (b) What is the speed of piece A?
∑∑∑∑∑∑ ==⇒= iyfyixfxif pppppp andrr
Since coconut was at rest before explosion 0== ∑∑ iyix pp
Sub. in (1) smv fA /3=⇒
p is conserved Inelastic collision
Particles don’t stick together but lose some energy
Completely inelastic collisionParticles stick together and lose some energy (deformation)
Elastic collisionParticles bounce off each other, no loss of energy
Mechanical Energy is conserved ONLY in elastic collisionsMomentum is conserved in both elastic and inelastic collisions
9.8: Momentum and Kinetic Energy in Collisions
ffii vmvmvmvm 22112211rrrr
+=+
( )222
2112
1iii vmvmK += ( )2
222
1121
fff vmvmK +=
ifloss KKE −=
Inelastic collisionMomentum is conserved
Mechanical energy is not conserved
The Energy loss =
∑∑ = fi pp rr
9.9: Inelastic Collision in One dimensions: Inelastic Collision
and
fii vmmvmvm rrr )( 212211 +=+
( )222
2112
1iii vmvmK += ( ) 2
2121
ff vmmK +=
ifloss KKE −=
Completely inelastic collisionMomentum is conserved
Mechanical energy is not conserved
The Energy loss =
∑∑ = fi pp rr
9.9: Inelastic Collision in One dimensions: Completely Inelastic Collision
and
Solution:
Momentum is conserved
9.9: Example: A 900 kg car had a velocity of 20 m/s to the right collides with other 1800kg car that was stopped. If the two car stuck together, find the final velocity after collision
∑∑ = fi pprr
Befor collision After collision
V2i = 0 Vf = ??V1i = 20 m/s
fii mmmm v)(vv 212211rrr
+=+
smmmmm ii
f /67.61800900
0)20(900vvv21
2211 =+
+=
++
=⇒rr
r
9.9: Example: Ballistic pendulum: A bullet (m1) is fired into a large block of wood (m2) suspended from some light wires. The bullet embeds in the block, and the entire system swings through a height h. How can we determine the speed of the projectile from a measurement of h?
Befor collision Immidiatlyafter collision
To find the speed of bullet v1A weneed to find vB
From conservation of energy(only for system after collision from B to C)
righttheto2
221
1
211
2
ghm
mmv
ghvmvmgh
A
BB
+=⇒
=⇒=
rWe have inelastic collision
∑∑ = fi pprr
BA mmm v)(0v 2111rr
+=+
1
211
v)(vmmm B
A
rr +
=⇒
∑∑ = fi pprr
∑∑ = fi KK
)1(vvvv 22112211 ffii mmmm rrrr+=+
=
Momentum is conserved
)2(21
21
21
21 2
222
112
222
11 ffii vmvmvmvm +=+
Energy is also conserved
From equations 1 & 2 )v-v(v-v 2121 ffiirrrr
−=
9.10: Elastic Collision in One dimensions9.10: Elastic Collision in One dimensions: Example
Two metal spheres, suspended by vertical cords, initially just touch. Sphere 1, with mass m1 = 30 g, is pulled to the left to height h1 = 8 cm, and then released from rest. After swinging down, it undergoes an elastic collision with sphere 2, whose mass m2 = 75 g. What is the velocity v1f of sphere 1 just after the collision?
Sphere 1 speed before collision
Ef = Ei
)1(075003000470 21
22112211
ff
ffii
v.v..
vmvmvmvmrr
rrrr
+=+
+=+
)2(02521
)(
21
2121
ff
ffii
vv.
v-vv-vrr
rrrr
+−=−
−= Solve the two equations
9.10 : Example: A block (m1=1.6 kg) initially moving to the right at 4 m/s collides with a spring attached to a second block (m2=2.1 kg) moving to the left with at 2.5 m/s. The spring constant is 600 N/m. Find the velocities of the two blocks after the collision.
Befor collision After collision
Momentum is conserved
(1)
For elastic collision, we have
(2)
From eqn’s (1) and (2)
Conservation of momentum still stands:
Analyze each component separately:
Use conservation of kinetic energy if the collision is elastic:
ffii vmvmvmvm 22112211vrrr
+=+
fyfyiyiyfyiy
fxfxixixfxix
vmvmvmvmpp
vmvmvmvmpp
22112211
22112211
+=+⇒=
+=+⇒=
∑ ∑∑ ∑
( ) ( )222
211
222
211 2
121
iiff
if
vmvmvmvm
KK
+=+
=
∑∑ = fi pp rr
9.11: Collisions In Two Dimensions
9.11:Example: If before collision, m2 was at rest, find magnitude and direction of speed for m2 after collision (m1=1kg, m2=2kg)
v1i = 10 m/s, v1f =5 m/sθ = 30°
v2f = ?φ = ?
momentum is conserved in the two dimensions x and y
smvvv
vmvmvm
vmvmvmvmpp
fx
iff
ixixfxfxixfx
/83.2cos10cos233.4
0cos30cos
222
112211
22112211
==⇒=+
+=+⇒
+=+⇒=∑ ∑
φφ
φ
2-variables 2-eqn’s
m1
m2
x-component smvv
vmvm
vmvmvmvmpp
fyf
ff
iyiyfyfyiyfy
/25.1sin
0sin30sin
22
2211
22112211
−==−⇒
=−⇒
+=+⇒=∑ ∑
φ
φ
°−=−
==+= − 8.2383.225.1tan,/1.3
,
122
222 φandsmvvv
hence
fyfxf
y-component
m1
m2
9.11:Example: continued from previous slide:
Example: Rocket fired vertically, explode ataltitude of 1000 m into three fragements, as shown.The 3 fragements are of equal masses. Find the velocity of the third part after collision
Collision for system of particles
fcomicom vv ,,rr
=fcomicomfcomicom vMvMPP ,,,,rrrr
=⇒=
For isolated system of particles Momentum is conserved
m/sv
vM)(Mvmvmvm
)Mvm
Mvm
MvmM(MvMv
Ppp
Pppconservedismomentum
fx
fxxxx
xxxcom,xix
xcomfxix
comfi
2403
2403
00
0
3
3332211
332211
,
−=⇒
++=++=⇒
++=⇒=
==⇒
==⇒
∑∑∑∑
rrr
rrr
Rocket fired vertically
m/sv
vM)(MM
vmvmvmM
)Mvm
Mvm
Mvm
M(M
MvMv
Ppp
fy
fy
yyy
yyy
ycomiy
ycomfyiy
4503
03
4503
300
300
3
3
332211
332211
,
,
=⇒
++=
++=
++=
=
== ∑∑rrr
Same for momentum in the y-direction
smjiv fy /)ˆ450ˆ240(3 +−=⇒r
Collision for system of particles: Example: continued from previous slide