Transcript Reviewch12-15 1

Review Problems
from chapters
12-15
NA 0
take the torque around B
W  0.50=w  ( x  0.50)
x ?
take torque around P
L
N  L  sin(37 ) W   cos(37o )
2
N ?
o

Mm
F31  F32  G 2
R
Mm 4
F  2 G 2 
R
5
 R  5m 
T  98  60  5880 s
Use third KeplerLaw :
2

 3
4

2
r
T  
 GM 
G  6.67  1011 Nm2 / kg 2
Ei  Ef
1
Mm
Mm
2
mv i  G
 0 G
2
RE
3R E
vi ?
Ei  E f
Ki  U i  K f  U f
1 2
Mm
Mm
mv 0  G
 0G
2
RE
3R
1 2
Mm
v 0  2G
2
3R
4GM
v0 
3R
E  K circular orbit 
W  PA  (  gL )A
R1  R 2
W  30N ,W a ,lwater W  Bwater  20
 Bwater  10N  waterV g
1
1
B liquid   liquidV g  waterV g  Bwater  5
2
2
W a ,liquid W  B liquid  30  5  25N
Av=constant
A1v1= A2v2
A1v1  A2 v2  A1v1  ( A1 / 2)v2  v2  2v1
1 2
1
v 1  P2  gy2  v 2 2
2
2
1
1
P1  v 21  P2  v 2 2
2
2
1
1
2
P1  P2   (v 2 2  v1 )   (3v 21 )
2
2
2P1  P2 
v1 
3
P1  gy1 
1 2
U ( x)  kx
2
compare :
x  xm cos(t   )
x  2 cos(50t )
 xm  2 m,   50 rad / s
k   2 m  2500 * 3  7500 N/m
k
vm  xm  A
m
as xm  A and  
k
m
T  2  0.25  0.50 s
1
f   2 Hz
T
40
xm 
 20 cm
2