Transcript 幻灯片 1

Mechanics Exercise Class Ⅰ
英文数字、算式表达法
= is equal to / equals
< is less than
a=b+c
a=b-c
a=bxc
a = b/c
3
> is larger than
>> is much greater than
a is equal to b plus c
a is equal to b minus c
a is equal to b times c / a equals to b
multiplied by c
a is equal to b divided by c / a equals
to b over c
x (square) root x / the square root of x
x cube root (of) x ; 4 x fourth root (of) x ; n x nth root (of) x
英文数字、算式表达法
f( x )
fx / f of x / the function f of x
lim
the limit as x approaches zero
x  0


0
the integral from zero to infinity
x2
x square / x to the second power / x to the power two
x3
x cube / x to the third power / x to the power three
x7
the seventh power of x（x to the seventh power）
lognX
log x to the base n
Brief Review
Displacement

r

 dr
v
Velocity
dt 
2
 dv d r
Acceleration a 
 2
dt dt

v
Projectile Motion  v cos iˆ  v sin ˆj
gx 2
y  tan  0 x 
2
2v0 cos  0 
Centripetal Acceleration
(trajectory)
v2
a
R
Newton’s Second Law F net  ma Drag Force D  1 CAv 2
2
1. A model rocket fired vertically from the ground ascends
with a constant vertical acceleration of the 4.00m/s2for 6.00
s. Its fuel is then exhausted ,so it continues upward as a
free-fall particle and then falls back down. (a) What is the
maximum altitude reached? (b) What is the total time
elapsed from the takeoff until the rocket strikes the ground?
Solution:
(a) One key idea is since the fuel is
exhausted and before the rocket strikes the
ground, its acceleration is g of magnitude.
And when the fuel is exhausted ,the velocity
is
v1  a1t1  4  6  24m / s (upward)
y
then we can get the position of the rocket
0
s1 
1 2 1
a1t1  4  62  72m
2
2
Second key idea is the velocity equals zero, when the
rocket is at maximum altitude. So
v22  v12 0  242
s2 

 29.39m
g
9.8
So the maximum altitude is
s  s1  s2  72  29.4  101.39m
(b) Since the fuel is exhausted and before the rocket returns,
the time interval is
t2 
v2  v1 0  24

 2.45s
a2
9.8
Then the rocket from the maximum altitude falls back to the
ground. From the Eq. 2-50, we can obtain
0  101.39 
1
( 9.8 )t32
2
Work out this equation, yielding
t3  4.55s
So the total time elapsed from the takeoff until the rocket
strikes the ground is
t  t1  t2  t3  6  2.45  4.55  13.00s
2. A block of mass m1 on a frictionless inclined plane of
angle is connected by a cord over a massless , frictionless
pulley to a second block of mass m2 hanging vertically.
What (a) the magnitude of the acceleration of each block
and (b) The direction of the acceleration of the hanging
block? (c) what is the tension of the cord?
Solution:
Choose m2 to be a system and draw it’s free-body diagram
T
m2
m2 g
With Newton’s second law
applied to the m2 system, we
can obtain
m2 g  T  m2 a2
(1)
m
1
m2
Choose m1 to be a system and draw it’s
free-body diagram and we can write
N
T
Parallel direction T  m1 g sin   m1a1
Perpendicular direction m1 g cos   N

m1g
The acceleration components a1 and a2, have the same
value since the string does not stretch, thus
T  m1 g sin   m1a2 (2)
Now we can solve equation (1) and (2) simultaneously for T
and a2. First solve equation (1) for T
T  m2 ( g  a2 )
(3)
Then substitute for T into equation (2):
m2 ( g  a2 )  m1 g sin   m2 a2
m2  m1 sin 
g
Solving for a2, we find a2 
m1  m2
Discussion : If m2  m1 sin   0 the direction of the
acceleration of the hanging block is vertically downward.
On the other hand if m2  m1 sin   0 the direction is vertically
upward.
Substituting for a2 for Eq. (3), we find that tension in the
rope is of magnitude
T
m1m2 ( 1  sin  )
g
m1  m2
3 A truck moves along a straight road with a constant
speed of 30m/s, and a projectile is launched from it. What
is the (a) magnitude and (b) direction of the projectile’s
initial velocity relative to the truck if it can fall back to the
launching point on the truck after the truck moves 60m.
The drag force is negligible.
Solution:
Choose the truck as reference. Set the launching point as
origin, and construct coordinates O-xy as shown.
Set the launching time as starting point of time.
a x  0,
ay  g
vx  v0 x
v y  v0 y  gt,
v0 x  v0 cos  ,
v0 y  v0 sin 
1 2
y

v
t

gt
x  v0 xt
0y
2
1 2
x  v0 cos t ,
y  v0 sin t  gt
2
The loading time:
t
60
 2s
30
And at the loading time x  0,
so
0  v0 cos   2
(1)
1
0  v0 sin   2  g  2 2
2
(2)
Eq (1) yields cos   0,

y0

2
Eq (2) yields v0  9.8m / s
So, if the projectile is thrown vertically relative to the truck,
it will fall back to the launching point after the truck moves
60m.
4 An inclined plane of mass m2 can slide on a smooth
horizontal surface, with the inclined angle of α. An athlete of
mass m1 slides on the inclined plane frictionlessly, what is (a)
the relative acceleration of the athlete with respect to the
inclined plane and (b) the pressure force on the plane from
the athlete?
Solution：Isolating the bodies m1 , m 2 ,
N sin   m a
For m 2
2
1
R  m g  N cos   0
2
N cos   m g  m a sin 
For m1
1
1
2
 N sin   m a  m a cos 
1
1
1
2
Solving the equations for a2 and N:
(m  m )g sin 
m 1m 2 g cos 
a 
N
m  m sin 
m 2  m 1 sin 2 
1
2
2
2
2
1
5 A parachute athlete dives with an initial speed of v0, and
the drag force is proportional to the square of the speed as:
av2. The total mass of the parachute and the athlete is m.
What is the function of v=v(t)?
Hint: this can be utilized in the integration
1
1
1


1  v 2 2(1  v ) 2(1  v )
Solution:
dv
mg  av  m
,
dt
2
dv
dt

,
2
mg  av
m
Transferring to
dv
dt

,
2
mg (1  av / mg ) m
dv
 gdt ,
2
1  av / mg
a
, 
set  
mg
2
a
C
mg
the above equation can be written as
dv
 gdt ,
2
2
1 v
dv
dv

 gdt ,
（
2 1   v） （
2 1   v）
d（1   v） d（1   v）

 gdt ,
2 （1   v） 2 （1   v）
Quadrature ln( 1   v )  2 gt  C
1  v
1  v 0
1  v
)  2 gt  ln(
)
Then ln(
1  v
1  v 0
1  v 0
 1)
1  v 0
1
v
 2 gt 1   v 0
(e
 1)
1  v 0
(e 2 gt
ln(
1  v 0
)C
1  v 0