Transcript 幻灯片 1 - Yangtze University

Mechanics Exercise Class Ⅲ
Brief Review
Ⅰ Rotation and Angular Momentum
1 Angular Quantities
Angular displacement   2  1
avg  
Angular velocity
ins  d 
2 Rotational Inertia
t Angular acceleration
dt
 avg  
ins  d 
I   mi ri 2   r 2 dm
i
3 The Parallel-Axis Theorem
m
I  Icom  Mh2
t
dt
4 The Kinetic Energy of the rolling
1
1
2
2
K  I com   Mvcom
2
2
5 Torque
  r  F ;   rF sin   rF  r F
6 Angular Momentum
Angular Momentum of a Particle
 r  mv  rmv sin   rmv  rmv
Angular Momentum of a System of Particles
L
1

2
 .... 
n
n

i 1
i
Angular Momentum of a Rigid Body L  I 
7 Work and Rotational Kinetic Energy
W 

f
i
d   K 
1
I ( 2f  2i )
2
8 Newton’s Second Law in Angular Form
Particle:
 net
d

dt
System of Particles:
A Rigid Body With fixed axis:  net  I
9 Conservation of Angular Momentum
0
L  a constant
 net
dL

dt
Ⅱ Gravitation
1 The Law of Gravitation
m1m2
F G 2
r
2 Gravitational Potential Energy
GMm
U 
r
3 Escape Speed
2GM
v
R
1. A wheel rotating about a fixed axis through its center has a
constant angular acceleration of 4.0rad/s2. In a certain 4.0s
interval the wheel turns through an angle of 80rad. (a) What is
the angular velocity of the wheel at the start of the 4.0s interval?
(b) Assuming that the wheel starts from rest, how long is it in
motion at the start of the 4.0 interval?
Solution :
(a)The key idea here is that the angular acceleration is
constant so we can use the rotation equation:
   0  0 
1 2
t
2
Substituting the given data and solving for 0 we find,
1 2
1
  0   t 80   4  42
2
2
0 

 12rad / s
t
4
(b) If the wheel starts from rest, the key idea here is that the
initial angular velocity is 0, so we can use the rotation
equation:
1
  t2
2
Inserting the given data and solving for 
1
   4  42  32rad
2
,we can obtain,
2. Each of the three helicopter rotor blades shown in the figure is
5.20 m long and has a mass of 240 kg. The rotor is rotating at
350 rev/min. (a) What is the rotational inertia of the rotor
assembly about the axis of rotation ? (Each blade can be
considered to be a thin rod rotated about one end).
(b) What is the total kinetic energy of rotation?
Solution :
(a)The key idea here is each blade can be
considered to be a thin rod rotated about one
end, so the total rotational inertia of the rotor
blades can be written as :
I  I1  I 2  I 3  3I1 ( I1  I 2  I 3 ) (1)
5.20m
The second key idea is the rotational inertia of a thin rod rotated
about its end is
1
I1 
ml 2
3
(2)
Combine the Eq (1) and (2) and substitute the given data ,we can
get the total rotational inertia
I  3I1  ml 2  240  5.22  6489.6kg.m2
(b) Substituting I and

into Eq 11-27, we find
1
K  I 2
2
1
350  2
  6489.6 
2
60
 1188678.4J
3. (P206.43) In Fig.6-79, a small 50g block slides down a
frictionless surface through height h=20 cm and then
sticks to a uniform rod of mass 100 g and length 40 cm.
The rod pivots about point O through angle θ before
momentarily stopping. Find θ .
Solution : The whole process can be divided
into three parts :
(1) The small block slides down the
frictionless surface through height h, in this
part only the gravitational force being a
conservative force, does work, so the law of
conservation of mechanic energy holds
m1 gh 
1
m1v12
2
Where v1 is the speed of the block before it collides with the rod
(2) The small block collides with the rod and sticks to it. During
this interaction there is no net torque acting on the block–rod
system relative to the point O, the angular momentum of the
system is conserved (Note: since there is a net force acting on the
rod at point by the pivot, the law of conservation of linear
momentum does not hold! )
m1v1 L  I
Where ωis the angular speed of the system about point O just after
the collision. I is the rotational inertia of the block-rod system about
point O, which is
1
I  m2 L2  m1 L2
3
(3) The block-rod system swings up until it momentarily stops .
During this process the mechanic energy of the system is
conserved , we thus write
1 2
I  (m1  m2 ) ghcom
2
Where Δh is the height change of the center of mass of the blockrod system in the swing up process. In the vertical pisition the
center of mass of the system is below point O at
m1L  m2 (0.5L) (0.05kg )(0.4m)  (0.1kg )(0.2m)
Lcom 

 0.2667m
m1  m2
0.05kg  0.10kg
So
hcom  Lcom (1  cos )
We have
cos   0.85
Thus the angle θ we look for is
  31.8
4. A 150.0 kg rocket moving radially outward from Earth has a
speed of 3.70 km/s. When its engine shuts off 200 km above
Earth’s surface. (a) Assuming negligible air drag , find the rocket’s
kinetic energy of the rocket 1000 km above the Earth’s surface. (b)
What maximum height above the surface is reached by the rocket?
solution
(a)The key idea is that the mechanical energy of
the rocket is constant
Ki  Ui  K f  U f
(1)
When the rocket is 200km above the Earth’s
surface, the energy is
Ei 
1
GMm
mv 21 
2
( R  h1 )
(2)
When the rocket is 1000km above the Earth’s surface, the energy is
Ef  K f 
GMm
( R  h2 )
(3)
Combining the Eq(1)-(3), we can get
1 2
1
1
K f  mv 1  GMm (

)  3.8 107 J
2
R  h1 R  h2
(4)
(b) The key idea is that when the rocket reaches the maximum height ,
the kinetic energy of it is zero , the Eq (4) is changed as
1
1
1
2
K f  mv 1  GMm (

)0
2
R  h1 R  hmax
Rewriting the above equation , we obtain
1
1
GM
2
mv 1  GMm

2
R  h1
R  hmax
Substituting the known data, we finally get
hmax  1034.9km
5. A geosynchronous satellite is one that stays above the same
point on the Earth, which is possible only if it is above a point on
the equator. Such satellites are used for such purpose as cable TV
transmission, for weather forecasting, and as communication
relays. Determine (a) the height above the Earth’s surface such a
satellite must orbit and (b) such a satellite’s speed.
solution
(a) The only force on the satellite is gravity, assuming the satellite
moves in a circle, then
mSat mE
v2
G
 mSat
2
r
r
And the speed of the satellite must be
v
2r
T
where T=1 day=(24h)(3600s/h)=86400 s. We put this into the first
equation above and obtain (after canceling mSat on both sides):
2r 
m
G 2E 
r
rT 2
2
We solve for r:




11
2
2
24
Gm
T
6
.
67

10
N

m
/
kg
5
.
98

10
kg 864000s 
3
E
r 

2
4
4 2
 7.54  1022 m3
and, taking the cube root, r=4.23107 m, or 42300 km from the
Earth’s center. We substrate the Earth’s radius of 6380 km to
find that the satellite must orbit about 36000 km (about 6 rE )
above the Earth’s surface.
(b)
GmE
v

r
6.67 10
11


N  m2 / kg 2 5.98 1024 kg
 3070m / s
7
4.23 10 m


We get the same result if we use v  2r / T