1. KASUS
“Particle sliding on a smooth inclined plane”
-Septiko Aji-
x1
m
M
x2
θ
We see that the square of the speed is given by cosines;
θ
2
2 2
 
v  x1  x 2  2 x1 x 2 cos 
T 
1
2


2 2
 
m x1  x 2  2 x1 x 2 cos  
1
2

M ( x2 )
2
V   mgx 1 sin 
Then, we know;
L  T V
 L
1
2


2 2
 
m x1  x 2  2 x1 x 2 cos  
1
2

M ( x 2 )  mgx 1 sin 
2
Lagrange’s equation;
d  L  L



dt   x   x
d  L

dt   x1
 
 L

 x
1

so, we have
d  L

dt   x1
 
d  L

dt   x1
 
d
 
 L

  x and
1

d  L

dt   x 2
 
 L

 x
1




m x1  x 2 cos   mg sin 
dt
 1  2 cos   g sin 
x
x
 1  g sin   2 cos 
x
x
 L

 x
2


2. d  L

dt   x 2
 
 L

 x
2

d  L

dt   x 2
 
d
dt
 L

 x
2


m x
2




 x1 cos   M x 2  0
 2 ( m  M )  1 cos   0
x
x
Solving for 1 and 2 we find;
x
x
g sin 
1 
x
m cos 
2
1
mM
2 
x
 g sin  cos 
mM
2
 cos 
m