Download Tutorial 8 Angular Momentum and Planar Kinematics

Tutorial 12
Linear Momentum
Angular Momentum
Zhengjian, XU
Nov 26th, 2008
Conservation of Linear momentum
Newton’s third law:
mv A  mvB  Constant
Central impact
Direct impact:
mAv A  mB vB  mAv A'  mB vB'
Coefficient of restitution

e
t2
tC
tC
Fdt
 Fdt
vB'  v A'

v A  vB
t1
Oblique impact:
 
 
mA v A x  mB vB x  mA v A' x  mB vB'
(vA ) y  (vA ) y
(vB ) y  (vB ) y
x

v   v 
e
'
B x
'
A x
v A x  vB x
Question 16.76

Two small balls, each of mass m = 0.12 kg, hang from
strings of length L = 1 m. The left ball is released from
rest with θ = 30o. As a result of the initial collision, the
right ball swings through a maximum angle of 25o.
Determine the coefficient of restitution.
Question 16.76
Since the right ball can reach a maximum
angle of 25o, After impact, the velocity of B is:
vB '  2 gL(1  cos 25o )
A
B
The impact velocity of Ball A and B:
vA  2 gL(1  cos )
vB  0
Recalling:
vA '  vA  vB '
vB'  v A'
2vB 'v A 2 1  cos 25o  1  cos 30o
e


v A  vB
vA
1  cos 30o
Augular Momentum

Principle of Angular Impulse and momentum
dv
r   F  r  ma  r  m
dt
d (r  mv)
dv dr
 r  m   mv
dt
dt dt
Angular momentum:
rF 
dH
M
dt
Central-Force Motion

If F goes through the center:
r  F  0

HO  constant
In a polar coordinate system:
r  rer
v  vr e r  v θ e θ
H  (rer )  m(v r er  v θ eθ )  mrv θ e z

For central motion:
H  mrvθ e z  constant
rv θ  constant
Question 16.87/16.88




A satellite is in the elliptic earth orbit
shown. Its velocity at perigee A is
8640 m/s. The radius of the earth is
6370 km.
(a) Use conservation of angular
momentum to determine the
magnitude of the satellite’s velocity at
apogee C.
(b) Use conservation of energy to
determine the magnitude of the
velocity at C.
(c) To determine the magnitudes of
the radial velocity vr and transverse
velocity vθ. at B.
(a) : conservation of angular momentum to determine the magnitude of
the satellite’s velocity at apogee C.
For a central-force motion
rv θ  constant
8000  v A  24000  vC
(b):
Use conservation of energy to
determine the magnitude of the velocity
at C.
(c) To determine the magnitudes of the radial velocity vr and transverse
velocity vθ. at B.
Question 16.95

Two gravity research satellites (mA = 250 kg, mB = 50 kg) are
tethered by a cable. The satellite and cable rotate with angular
velocity ,ω0 = 0.25 revolution per minute. Ground controllers order
satellite A to slowly unreel 6 m of additional cable. What is the
angular velocity afterward?
Solution
Consider these two objects as a system, then the angular momentum
of this system relative to its center of mass is conserative.
mArA  mB rB ;
rA  rB  L
mB L
rA 
 2m
m A  mB
mB L
rB 
 10m
m A  mB
The total angular momentum:
H 0  mA rA20  mB rB20
rA
A
rB
O
B
Solution

After the string isunreeled to 18m
mB L '
rA ' 
 3m
m A  mB
mB L '
rB ' 
 15m
m A  mB
The new angular velocity can be obtained by:
H 0  m A rA20  mB rB20
    m r  
' 2
A
 mA r
B
' 2
B
rA
A
rB
O
B
Mass flows

Use the conservation of momentum
mv  (m  m)( v  v)  m( v  vf )
(m - m)v  mvf  0
v m
(m - m)

vf  0
t t
dm
Ff  
vf
dt
Question 104


A nozzle ejects a stream of water horizontally at 40 m/s with a mass
flow rate of 30 kg/s, and the stream is deflected in the horizontal
plane by a plate. Assume that the magnitude of the velocity of the
water when it leaves the plate is approximately equal to the initial
velocity.
Determine the force exerted on the plate by the stream in cases (a),
(b), and (c).

Consider a time interval ∆t, the
change of momentum:
m  v  m  (v0 cos   i  v0 sin   j - v0i)
 m  v0 (cos   1)  i  v0 sin   j

The average force in ∆t:
m  v m

 (v0 cos   i  v0 sin   j - v0 i )
t
t
dm f

 v0 (cos   1)  i  v0 sin   j
dt
Fp 
Solution for Question 16.104
Fp 
dm f
dt
 v0 (cos   1)  i  v0 sin   j
Case (a):


- Fp  30  40(cos 45o 1)  i  v0 sin 45o  j
Case (b):
Case (c):
- Fp  1200  i  1200  j
- Fp  2400i
Question 16.110/16.111


The rocket consists of a 1000-kg payload and a 9000-kg booster.
Eighty percent of the booster’s mass is fuel, and its exhaust
velocity is 1200 m/s. If the rocket starts from rest and external
forces are neglected, what velocity will it attain?
The booster has two stages whose total mass is 9000 kg. Eighty
percent of the mass of each stage is fuel, and the exhaust velocity
of each stage is 1200 m/s. When the fuel of stage 1 is expended,
it is discarded and the motor of stage 2 is ignited. Determine the
velocity attained by the rocket if the masses of the stages are m1
= 6000 kg and m2 = 3000 kg.

The rocket has just one booster
Payload: Mp=1000kg;
Booster: Mb=9000kg
80% of the booster is fuel;
Mf=7200kg
Exhaust velocity: 1200m/s
dm f
dv x
vf  m
dt
dt
dm f
dm

dt
dt
dm
dv x  v f
m

vx
0
dv x  v f 
m
m0
m
dm
 v f ln( 0 )
m
m
After the fuel is exhausted:
m0
 10000 
v x  v f ln ( )  1200  ln 
  1527.6(m / s )
m
 2800 
Question(b)

The rocket has two booster
Payload: Mp=1000kg;
Booster 1: M1=6000kg; Booster 2: M2=3000kg
80% of the booster is fuel;
Exhaust velocity: 1200m/s
(1) After the first booster is discarded:
m0
 10000 
v1  v f ln ( )  1200  ln 
  784.7(m / s )
m
 5200 
(2) After the second booster is discarded:
v2  v1  v f ln (
m0
 4000 
)  1200  ln 
  784.7  1884.2(m / s )
m
 1600 