Solutions - University of Illinois at Chicago

January 4, 2012
University of Illinois at Chicago
Department of Physics
Classical Mechanics Qualifying Exam Solutions
Problem 1.
A cylinder of a non-uniform radial density with mass M,
length l and radius R rolls without slipping from rest down
a ramp and onto a circular loop of radius a. The cylinder
is initially at a height h above the bottom of the loop. At
the bottom of the loop, the normal force on the cylinder is
twice its weight.
h
a
a) Expressing the rotational inertia of the non-uniform
cylinder in the general form (I=βMR2), express the β
in terms of h and a.
2
b) Find numerical value of β if the radial density profile for the cylinder is given by ρ (r ) = ρ 2 r ;
c) If for the cylinder of the same total mass M the radial density profile is given by ρ n ( r ) = ρ n r n ,
where n ∈ 0,1,2,3... , describe qualitatively how do you expect the value of β to change with increasing
n. Explain your reasoning.
Solution:
a) Centripetal acceleration as the ball rolls around the circular loop at the bottom of the track is ac
= v2/a, and could be expressed from free body diagram equation:
N-W = Mv2/a, where N is the normal force and the W is the weight. We are given that N=2Mg, so
2Mg – Mg – Mv2/a, i. e. v2 = ga
Relating the angular and translational velocities by v=aω, we next use the expression for the total
kinetic energy of rolling object (no slipping)
K = ½ Mv2 + ½ Iω2
1
And apply energy conservation for the ball between its initial position at rest and its position at the
bottom of the loop:
Mgh = ½ Mv2 + ½ Iω2
Substituting for v and for ω (from above) and using I = βMR2 expression we find
h = ½a +½ β a
Rearranging: β = 2h/a -1
b) The moment of inertial about the rotational axis of the cylinder is given by I = ∫ r 2dm . For the
quadratic density profile dm = ρ 2 r 2l 2πrdr
R
R
R6 2
πlR 4 ρ 2
, so
I = ∫ r ρ 2 r l 2πrdr = 2πρ 2l
= MR 2 , with M = ∫ ρ (r )dV = ∫ 2πlρ 2 r 3dr =
6
3
2
0
0
2
2
β =
2
3
c) For an arbitrary value of n:
R
R
I = 2πlρ n ∫ r n+3 dr = 2πlρ n
0
I=
n+2
MR 2
n+4
,
β=
2πlρ n R n+2
R n+ 4
, with M = 2πlρ n ∫ r n+1dr =
, so
n+4
n+2
0
n+2
n+4,
e.g.
n = 0, β =
2
1
; n → ∞, β → 1
2
Problem 2.
A particle of unit mass is projected with a velocity v0 at right angles to the radius vector at a distance a
from the origin of a center of attractive force, given by
⎛ 4 a 2 ⎞
f (r ) = −k ⎜⎜ 3 + 5 ⎟⎟
r ⎠
⎝ r
For initial velocity value given by v02 =
9k
, find the polar equation of the resulting orbit.
2a 2
Solution: Calculating the potential energy
⎛ 4 a 2 ⎞
dv
− = f ( r ) = −k ⎜ 3 + 5 ⎟
dr
⎝ r r ⎠
⎛ 2 a 2 ⎞
Thus, V = −k ⎜ 2 + 4 ⎟
⎝ r 4r ⎠
The total energy is …
1
1 ⎞ 1 ⎛ 9k ⎞ 9k
⎛ 2
E = To + Vo = vo2 − k ⎜ 2 + 2 ⎟ = ⎜ 2 ⎟ − 2 = 0
2
⎝ a 4a ⎠ 2 ⎝ 2a ⎠ 4a
Its angular momentum is …
9k
l 2 = a 2vo2 =
= constant = r 4θ&2
2
Its KE is …
2
2
⎤
⎤ l 2
1
1 ⎡⎛ dr ⎞
1 ⎡⎛ dr ⎞
T = r&2 + r 2θ&2 = ⎢⎜ ⎟ + r 2 ⎥ θ&2 = ⎢⎜ ⎟ + r 2 ⎥ 4
2
2 ⎣⎢⎝ dθ ⎠
2 ⎣⎢⎝ dθ ⎠
⎦⎥
⎦⎥ r
The energy equation of the orbit is …
2
⎤ l 2
⎛ 2 a 2 ⎞
1 ⎡⎛ dr ⎞
T + V = 0 = ⎢⎜ ⎟ + r 2 ⎥ 4 − k ⎜ 2 + 4 ⎟
2 ⎢⎣⎝ dθ ⎠
⎥⎦ r
⎝ r 4r ⎠
⎡⎛ dr ⎞2 2 ⎤ 9k
⎛ 2 a 2 ⎞
= ⎢⎜ ⎟ + r ⎥ 4 − k ⎜ 2 + 4 ⎟
⎢⎣⎝ dθ ⎠
⎥⎦ 4r
⎝ r 4r ⎠
2
⎛ dr ⎞ 1 2 2
or
⎜
⎟ = a − r
⎝ dθ ⎠ 9
dr
dφ
Letting
then
= −a sin φ
r = a cos φ
dθ
dθ
2
1
⎛ dφ ⎞ 1
So
∴φ = θ
⎜
⎟ =
3
⎝ dθ ⎠ 9
1
Thus r = a cos θ
r = a @θ = 0o
3
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3
Problem 3.
A simple pendulum of length b and mass m is suspended from a point on the circumference of a thin
massless disc of radius a that rotates with a constant angular velocity ω about its central axis. Using
Lagrangian formalism, find
a) the equation of motion of the mass m;
b) the solution for the equation of motion for small oscillations.
Solution:
y
a)
ωt
Coordinates:
x = a cos ω t + b sin θ
y = a sin ω t − b cosθ
x
x&= −aω sin ω t + bθ&cosθ
y&= aω cos ωt + bθ&sin θ
θ
m
1
L = T − V = m ( x&2 + y&2 ) − mgy
2
m
= ⎡⎣ a 2ω 2 + b2θ&2 + 2bθ&aω sin (θ − ω t )⎤⎦ − mg ( a sin ω t − b cosθ )
2
d ∂L
= mb 2θ&&+ mbaω θ&− ω cos (θ − ω t )
&
dt ∂θ
∂L
= mbθ&aω cos (θ − ω t ) − mgb sin θ
∂θ
∂L d ∂L
The equation of motion
−
= 0 is
∂θ dt ∂θ&
(
ω 2a
)
g
sin θ = 0
b
b
(Note – the equation reduces to equation of simple pendulum if ω → 0 .)
θ −
cos(θ − ωt ) +
b) For small θ the equation of motion reduces to that of a constant driving force harmonic
oscillator :
g
ω 2a
θ + θ =
cos(ωt )
b
b
4
The general solution for this equation consists of two parts, a complementary function u(t) and
a particular solution v(t). Complementary solution comes from the simple harmonic oscillator
⎛ g
⎞
g
⎟ .
equation: θ + θ = 0 , and could be immediately written as u (t ) = A cos⎜
t
−
ϕ
⎜ b
⎟
b
⎝
⎠
Given the form of the driving force, the particular solution could be defined as
v(t ) = B cos(ωt ) .
Taking the derivatives and plugging back in the equation of motion, one finds the expression
ω 2a
for B as B =
, and the general solution in form:
g − ω 2b
⎛ g
⎞
ω 2a
θ (t ) = u(t ) + v(t ) = A cos⎜⎜
t − φ ⎟⎟ +
cos(ωt )
2
b
g
−
ω
b
⎝
⎠
5
Problem 4.
A rigid body consists of six particles, each of mass m, fixed to the ends of three light rods of length 2a,
2b, and 2c respectively, the rods being held mutually perpendicular to one another at their midpoints.
a) Write down the inertia tensor for the system in the coordinate axes defined by the rods;
b) Find angular momentum and the kinetic energy of the system when it is rotating with an
angular velocity ω about an axis passing through the origin and the point (a,b,c).
Solution:
a)
I xy = ∑ mi xi yi = 0 since either xi or yi is zero for all six
i
particles. Similarly, all the other products of inertia are zero.
Therefore the coordinate axes are principle axes.
2
2
I xx = ∑ mi yi2 + zi2 = m ⎡0 + 0 + b2 + ( −b ) + c 2 + ( −c ) ⎤
⎣
⎦
i
(0,0,c)
(
)
I xx = 2m (b2 + c2 )
I yy = 2m ( a2 + c2 )
(0,b,0)
I zz = 2m ( a 2 + b2 )
(a,0,0)
b)
r
ω
ω=
(a
2
1
+ b2 + c 2 ) 2
⎡b2 + c 2
t
⎢
I = 2m ⎢ 0
⎢ 0
⎣
⎡ a ⎤
ω
⎢b ⎥
( aeˆ1 + beˆ2 + ceˆ3 ) =
1 ⎢ ⎥
( a2 + b2 + c2 ) 2 ⎢⎣ c ⎥⎦
r tr
L
Using = I ω ,
r
L=
1
⎡b2 + c 2
⎢
⎢ 0
⎢ 0
⎣
1
⎡ a ( b 2 + c 2 )⎤
⎢
⎥
⎢b ( a 2 + c 2 )⎥
⎢
⎥
⎢c ( a 2 + b 2 )⎥
⎣
⎦
2mω
( a 2 + b2 + c 2 ) 2
r
L=
2mω
(a
2
+ b2 + c2 ) 2
0
a2 + c2
0
0 ⎤ ⎡ a ⎤
⎥
0 ⎥ ⎢⎢b ⎥⎥
a 2 + b 2 ⎥⎦ ⎢⎣ c ⎥⎦
6
0
a + c2
0
2
0 ⎤
⎥
0 ⎥
a 2 + b2 ⎥⎦
1 r r
Using T = ω ⋅ L
2
⎡ a ( b 2 + c 2 )⎤
⎢
⎥
1
2mω 2
2
2
⎢
T=
a
b
c
b
a
+
c
[
] (
)⎥⎥
2 ( a 2 + b2 + c2 )
⎢
⎢c ( a 2 + b 2 )⎥
⎣
⎦
2
mω
T = 2 2 2 ⎡⎣a 2 b2 + c 2 + b2 a 2 + c 2 + c 2 a 2 + b2 ⎤⎦
a +b +c
2mω 2
T = 2 2 2 a 2b 2 + a 2 c 2 + b 2 c 2
a +b +c
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)
Problem 5.


The force of a charged particle in an inertial reference frame in electric field E and magnetic field B

  

is given by F = q E + v × B , where q is the particle charge and v is the velocity of the particle in the
inertial system.
a) Prove that the transformation from a fixed frame to a rotating frame is given by
r r& r& r
r r r r r
&
r&= &
r ʹ′ + ω × r ʹ′ + 2ω × vʹ′ + ω × (ω × r ʹ′ )
(
)
b) Find the differential equation of motion referred to a non-inertial coordinate system



rotating with angular velocity ω = −⎛⎜ q ⎞⎟ B , for small B (neglect B2 and higher order
⎝ 2m ⎠
terms).
Solution:
  
a) Let’s consider a coordinate system (i ʹ′, j ʹ′, k ʹ′) , rotating about the axis defined by the unit vector
  
 

n with respect to the (i , j , k ) system with angular velocity ω = n ω . Position of any point P
in space can be expressed in two systems (in case of common origin) as
  



 
r = i x + j y + k z = r ʹ′ = i ʹ′xʹ′ + j ʹ′y ʹ′ + k ʹ′z ʹ′
The velocity then can be written


 dxʹ′
  dx

  
di ʹ′
di ʹ′
v =i
+ ... = i ʹ′
+ ... + xʹ′
+ ... = vʹ′ + xʹ′
+ ... = vʹ′ + ω × r ʹ′
dt
dt
dt
dt
This finding is generally true for any vector, e.g. for derivative of the velocity vectors:


 
⎛ dv ⎞
⎛ dv ⎞
= ⎜ ⎟
+ω×v
⎜ ⎟
⎝ dt ⎠ fixed ⎝ dt ⎠ rotating

  
   
⎛ dv ⎞
⎛ d (vʹ′ + ω × r ʹ′) ⎞
= ⎜
+ ω × (vʹ′ + ω × r ʹ′) =
⎜ ⎟
⎟
dt
⎝ dt ⎠ fixed ⎝
⎠ rotating
 

    
⎛ dvʹ′ ⎞
⎛ d (ω × r ʹ′) ⎞
+ ⎜
+ ω × vʹ′ + ω × ω × r ʹ′ =
⎜
⎟
⎟
dt
⎝ dt ⎠ rotating ⎝
⎠ rotating



    
  ⎛ dr ʹ′ ⎞
⎛ dvʹ′ ⎞
⎛ dω ⎞
+ ⎜
× r ʹ′ + ω × ⎜
+ ω × vʹ′ + ω × ω × r ʹ′
⎜
⎟
⎟
⎟
⎝ dt ⎠ rotating ⎝ dt ⎠ rotating
⎝ dt ⎠ rotating
Changing notations, one arrives at the transformation equations from a fixed to a rotating frame:
8
r r r r
v = vʹ′ + ω × r ʹ′
r r& r& r
r r r r r
&
r&= &
r ʹ′ + ω × r ʹ′ + 2ω × vʹ′ + ω × (ω × r ʹ′ )
b) The equation of motion:
r
r
r r
mr&&= qE + q v × B
(
)
Transforming from a fixed frame to a moving rotating frame:
q r
r
r
ω =−
B so ω&= 0
2m
r r q r r r
r
r
r
r r r
mr&&ʹ′ − q B × vʹ′ − B × (ω × r ʹ′ ) = qE + q ⎡⎣( vʹ′ + ω × r ʹ′ ) × B ⎤⎦
2
r
r
r
r
q r r
r
r
r r
r r
mr&&ʹ′ + q vʹ′ × B + (ω × r ʹ′ ) × B = qE + q vʹ′ × B + q (ω × r ʹ′ ) × B
2
r
r&& r q r r
mr ʹ′ = qE + (ω × r ʹ′ ) × B
2
r q qB
q r r
(ω × r ʹ′) × B = ⎛⎜ ⎞⎟ ( r ʹ′)(sin θ )( B ) ∝ B 2
2
2 ⎝ 2m ⎠
r
r
Neglecting terms in B 2 , mr&&ʹ′ = qE (Larmor’s theorem)
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