7.3 Solutions
337
V =
k
2
(s − l)
2
+ mg(h − s sin α)
(5)
L = T − V =
(M + m)
2
˙
x
2
+
1
2
m ˙
s
2
+ m ˙
x ˙
s cos α −
k
2
(s − l)
2
− mg(h − s sin α)
(6)
(b) The generalized coordinates are q 1 = x and q 2 = s. The Lagrange’s
equations are
d
dt
∂ L
∂ ˙
x
−
∂ L
∂ x
= 0,
d
dt
∂ L
∂ ˙
s
−
∂ L
∂s
= 0
( 7 )
Using (6) in (7), equations of motion become
(M + m) ¨
x + m ¨
s cos α = 0
( 8 )
m ¨
x cos α + m ¨
s + k(s − s 0 ) = 0
( 9 )
where s 0 = l + (mg sin α)/k.
Let x = A sin ωt and s − s 0 = B sin ωt
(10)
¨
x = −ω
2 A sin ωt, ¨
s = −Bω
2 sin ωt
(11)
Substituting (10) and (11) in (8) and (9) we obtain
A(M + m) + B cos α = 0
(12)
Amω
2 cos α + B(mω
2
− k) = 0
(13)
Eliminating A and B, we find
ω =
k(M + m)
m(M + m sin
2
α)
(14)
Components of the velocity of the ball as observed on the table are
7.35 v x = ˙
x + ˙
y cos α
(1)
v y = ˙
y sin α
(2)
v
2
= v
2
x + v
2
y = ˙
x
2
+ ˙
y
2
+ 2 ˙
x ˙
y cos α
(3)
T (ball) =
1
2
mv
2
+
1
2
I ω
2
=
1
2
mv
2
+
1
2
×
2
5
mr
2
ω
2
=
1
2
mv
2
+
1
5
mv
2
=
7
10
mv
2
(4)
T (wedge) =
1
2
(M + m) ˙
x
2
(5)