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Physics 409/Classical Mechanics Test #1 17 September 2015
Name ANSWERS 
Answer all parts of each of the two questions. 50 points total. The points value of each part is
indicated. Begin the answer to each question on a new sheet of paper. Clearly define all
coordinates and variables. Be sure to include sufficient detail in each answer so that your logic is
clear to me.
Useful reminders:
One form of Lagrange’s equations of motion is     irF j iij j j
d T T Q
dt q q q
 
  
Another form is where L = T - V0  j j
d L L
dt q q
 
 
(1)(25 pts.) Two points, each of mass m, are connected by
a rigid rod of length 2b whose center is attached to the end
of rigid rod of length a. Each rod is free to move in all
spatial directions Gravity, with constant acceleration g,
acts on each particle.
(a)(10 pts.) Choose a set of suitable generalized
coordinates and find a Lagrangian for this system.
ANS.: There are 2 particles in 3-d space and 2 constraints,
so the system has 4 dof (3x2!2=4). Since the ‘a-rod’ ends
at the COM, the angles 2 and N shown in the figure are a good choice to describe the COM
position. The remaining two coordinates are needed to describe the orientation of the ‘b-rod’
with respect to the inertial Cartesian frame shown. Let angle " be the polar angle between the z
axis and the b-rod, and let angle $ be the azimuthal angle between the x axis and the projection
of the b-rod in the x-y plane. Along with the constant radius r=b, these two angles define a
second set of spherical coordinates. Thus, the kinetic energies for the COM motion and relative
motion are formally similar, and expressions for them can be found in the solutions to HW
problems HW2_prob2_GPS19_G17 and HW2_prob4_dumbbell with suitable adjustments for
changes in notation. The COM motion is equivalent to that of a spherical pendulum of mass 2m
and bob-length a. The potential energy is VCOM = 2mg cos2. The b-rod motion is that of a free
dumbbell with no potential energy. The Lagrangian is L = TCOM + Trel !VCOM , where Trel is the
dumbbell kinetic energy.
(b)(12 pts.) Find the equations of motion for this system and reduce them to a pair of one-
dimensional equations of motion. Obtain a first integral for each of these 1-d equations. 
ANS: See the two HW problems cited in (a) for the EOM, the reduced 1-d EOM, and the first
integrals. When using the dumbbell results of HW2_prob4_dumbbell, you must put f1 = f2 = 0
and d=b.
(c)(3 pts.) Identify three constants of the motion and give a physical interpretation for each one.
ANS: ECOM, Trel, and the two angular momenta pN and p$ are all constants of the motion.
(e)(4 bonus pts.) 
Find a general solution for the center of mass motion of this system in the form of two
quadratures, and , where f(2) and g(2) are explicit
0
( )t f d


   
0
0 ( )g d


      
functions of 2. (You do not have to do the integrals, but you must find the functions f and g.)
ANS.: Both quadratures can be found in part (e) of the results of HW2_prob4_dumbbell,
provided you put d=a and f1 ! f2 = 2mg.
(2)(25 pts.) A block of mass m slides without
friction on a wedge of height h and mass M that
makes an angle 2 with the x-axis. The wedge
slides without friction subject to a constant force
F > 0 that acts in the positive x direction
repelling the wedge from the vertical y axis. 
The small block is not affected by this force, but
its motion is subject to the uniform acceleration
of gravity, g, which acts in the downward y
direction.
(a)(10 pts.) Write the Lagrangian L of this
system in suitable generalized coordinates. 
To aid you with this, note that x1 is the x
displacement of the wedge from the origin, O, of the inertial Cartesian frame, x2 is the x
displacement of the small block from the vertical (left) edge of the wedge, and y2 is the vertical
displacement of the small block from the x axis. Be sure to include the effects of constraints in
the Lagrangian, so that your Lagrangian reflects the correct number of degrees of freedom for
this system.
ANS: L = T !V, where 
2 2 2
1 1 2 2 1 2; ( / 2) ; ( / 2)[( ) ];M m M mT T T T M x T m x x y V Fx mgy           
This form of the Lagrangian contains too many degrees of freedom because x2 and y2 are
constrained by the wedge geometry: h !y2 = x2 tan2 . I chose to eliminate x2 because y2 is the
natural variable in the gravitational potential energy. Upon introducing this constraint, L
becomes 
2 2 2
1 1 2 2 1 2[( ) / 2] cot ( / 2) cscL M m x mx y m y Fx mgy         
(b)(6 pts.) Find the equations of motion (EOM) for this system. 
ANS: 21 2 12 1( ) cot 0 csc cot 0M m x my F and m y mx mg           
(c)(6 pts.) Algebraically manipulate your EOM to find new EOM that independently determine
the horizontal acceleration of the wedge and the vertical acceleration of the small block in terms
of F, M, m, g, and 2. ANS:
2
1
2
12
( sin ) sin cos 0
( sin ) sin cos [ ( ) tan ] 0
M m x F mg and
M m y F M m g
  
   
   
    


(d)(3 pts.) Identify one constant of the motion for this system. ANS: The total energy, T+V
(e)(1 bonus point) For what value of F will the small block remain stationary on the accelerating
wedge? Give your answer in terms of M, m, g, and 2. 
ANS: The small block is stationary on the wedge when Inserting this value into the2 0.y 
decoupled EOM for y2, we find F = (M+m)g tan2.