Homework 2: Solutions
TM7-12. Put the origin at the bottom of the plane
1
m r& 2 + r 2 θ& 2 − mgr sin θ
2
(
L= T− U=
)
θ = α t ; θ& = α
L=
1
m r& 2 + α 2 r 2 − mgr sin α t
2
(
)
Lagrange’s equation for r gives
mr&& = mα 2 r − mg sin α t
or
&&
r − α 2 r = − g sin α t
(1)
The general solution is of the form r = rp + rh where rh is the general solution of the
homogeneous equation &&
r − α 2 r = 0 and rp is a particular solution of Eq. (1).
So
rh = Aeα t + Be − α t
rp = − C α 2 sin α t . Substituting into
For rp , try a solution of the form rp = C sin α t . Then &&
(1) gives
− C α 2 sin α t − C α 2 sin α t = − g sin α t
C=
g
2α 2
So
r ( t ) = Aeα t + Be − α t +
g
sin α t
2α 2
We can determine A and B from the initial conditions:
Condition (2) implies that r0 = A + B
r ( 0 ) = r0
(2)
r& ( 0 ) = 0
(3)
g
2α 2
Condition (3) implies 0 = A − B +
Solving for A and B gives:
A=
r ( t) =
g 
1
r0 −

2
2α 2 
B=
g 
1
r0 +

2
2α 2 
g  αt 1
g  −αt
g
1
r0 −
e +  r0 +
e +
sin α t
2
2

2
2α 
2
2α 
2α 2
or
r ( t ) = r0 cosh α t +
g
( sin α t − sinh α t )
2α 2
7-15.
b = unextended length of spring
 = variable length of spring
T=
1
m l& 2 + l 2 θ& 2
2
(
)
Placing the origin at the pivot,
U=
1
1
2
2
k ( l − b ) + mgy = k ( l − b ) − mg l cos θ
2
2
L= T− U=
1
1
2
m l& 2 + l 2 θ& 2 − ( l − b ) + mg l cos θ
2
2
(
)
Taking Lagrange’s equations for  and θ gives
l:
d
 ml&  = mlθ& 2 − k ( l − b ) + mg cos θ
dt  
θ :
This reduces to
d
 ml 2θ&  = − mg l sin θ
dt 
&&l − lθ& 2 + k ( l − b ) − g cos θ = 0
m
g
2
θ&& + l& θ& + sin θ = 0
l
l
7-34a. The coordinates of the wedge and the particle are
xM = x
xm = r cos θ + x
yM = 0
ym = − r sin θ
(1)
The Lagrangian is then
L=
M+ m 2 m 2
& & cos θ − 2xr
& θ& sin θ + mgr sin θ
x& +
r& + r 2θ& 2 + 2 xr
2
r
(
)
(2)
Note that we do not take r to be constant since we want the reaction of the wedge on the
particle. The constraint equation is f ( x , θ , r ) = r − R = 0 .
r = 0 to get the equations of motion
a) Right now, however, we may take r = R and r& = &&
for x and θ. Using Lagrange’s equations,
(
&&
x = aR θ&& sin θ + θ& 2 cos θ
)
(3)
&&
x sin θ + g cos θ
θ&& =
R
(4)
where a ≡ m ( M + m) .
Bonus 7-34 b) We can get the reaction of the wedge (normal force) from the Lagrange
equation for r
λ = mx&& cos θ − mRθ& 2 − mg sin θ
(5)
&& in terms of θ and θ& , and substitute the
We can use equations (3) and (4) to express x
resulting expression into (5) to obtain

a − 1  &2
λ = 
 Rθ + g sin θ
2
 1 − a sin θ 
(
)
(6)
To get an expression for θ& , let us use the conservation of energy
H=
M + m 2 m 2 &2
& θ& sin θ − mgR sin θ = − mgR sin θ 0
x& +
R θ − 2 xR
2
2
(
)
(7)
where θ 0 is defined by the initial position of the particle, and − mgR sin θ 0 is the total
energy of the system (assuming we start at rest). We may integrate the expression (3) to
obtain x& = aRθ& sin θ , and substitute this into the energy equation to obtain an expression
for θ&
2 g ( sin θ − sin θ 0 )
θ& 2 =
R ( 1 − a sin 2 θ )
(8)
Finally, we can solve for the reaction in terms of only θ and θ 0
λ = −
(
mMg 3 sin θ − a sin 3 θ − 2 sin θ 0
( M + m) ( 1 − a sin 2 θ )
2
)
(9)