EXAMPLES OF ROTATIONAL MOTION
BLOCK ON A FRICTIONLESS HORIZONTAL PLANE
Consider a block of mass m sliding on a horizontal frictionless plane. It is held in a circle
by means of a massless cord, as shown:
We choose an inertial coordinate system with origin at the center. The forces on the block are the
table, gravity, and the tension in the rope. Since there is no friction, gravity and the table cancel
out, leaving only the rope. Since the tension is directed toward the center it produces no torque


( r and F are parallel). Thus:


dJ

0
 J  constant
dt
But
 
J= r  p  rmv
 rv  constant
If the block has initial velocity, v0, at r0, and we pull in the rope slowly (r changes little in a
revolution), then:
r
rv = r0 v0  v  0 v0
r
and the work done in doing so is:

1 2 1 2 1 2  r02
W  KE f  KEi  mv  mv0  mv0   1
2
2
2
 r2

What would happen if we did not pull it in slowly? Then (dv/dt) would not be negligible. This
would require a force along the circumference (the rope would develop a curve). This in turn

would produce a torque which would change J .
We can calculate the work done directly as:
r
 mv 2
 
   F  dr   
dv
r

r
r
0
r0
r

2
2

mv02r02  1  r
mv02r02  1
m  r0  2
1  mv02   r0 









v
dr
1
  0
 
r  r 
2  r 2  r0
2  r02 r 2 
2  r 

r0

 

as before.
TETHER BALL
The tether ball shown consists of a ball of mass m spinning in a horizontal circle by
means of a massless rope.
Initially the ball has velocity v0 and ℓ = ℓ0. The rope is now slowly drawn in (ℓ changes little in
one revolution, as in the first example). What will be  and v when the length is ℓ?
We choose an inertial coordinate system with origin at the top of the pole. We choose the
ball as the system. Then:
T cos   mg  0 (horizontal circle)
v2
T sin   m
(motion in a circle)
 sin 
T 
mg
cos 
Initially
sin 2 0 v02

 0  #
cos 0  0g
There is a torque on the system given by:

 
    mg   mg sin ˆ
where ̂ is horizontal. Thus there are no vertical torques. Thus the vertical component of angular
momentum is constant. But:
 

ˆ  cos  H
ˆ
J    mv  mv  sin V
Hence
J V  mv sin  const   0 mv0 sin   #
Thus we have two equations in two unknowns:
sin 2  v 2

cos  g
v sin   #
we get
v
#
 sin 
sin 2 
#2
sin 4  #2



cos  3g sin 2 
cos  3g
we solve this for θ and then find v from
v
#
 sin 
LOOP THE LOOP
Consider the experiment done in class with a ball rolling down a track and into a loop.
Let r be the radius of the ball and R the radius of the loop. We want to find the minimum h and
maximum  for which the ball will make it over the top always rolling.
First we find the maximum  .
mg sin   Ff  ma
2
a
2
Ff r  mr 2  a  ma
5
r
5
 mg sin  
7
5
ma  a  g sin 
5
7
5
2
 Ff  mg sin   mg sin   mg sin 
7
7
But
Ff max  s N  s mg cos 
2
7
s cos   sin   tan   s
7
2
Now consider the forces when the ball is in the loop. The direction of Ff is set by the fact
that the ball will slow as it rises. This means it must spin less rapidly and therefore the torque
must be out of the board. This requires Ff to have the direction shown:
Then the equations are:
mg cos   N 
mv2
R  r
(1)
Ff  mg sin   ma
2
a
2
Ff r   mr 2  Ff   ma
5
r
5
(- sign because a < 0 but Ff > 0 since we have chosen directions as shown).
7
5
 mg sin   ma  a   g sin 
5
7
But if the ball is rolling without sliding we have:
1
1
mgh  mg  R   R  r  cos   mv 2  I2
2
2
Since
2
I  mr 2
5
This becomes
1
1 2 2 v2
mgh  mg  R   R  r  cos   mv 2 
mr
2
25
r2
 g  h  R   R  r  cos  
7 2
v
10
Putting this result in (1) we find:
 m  10 
mg cos   N  
 g h  R   R  r  cos 
Rr 7
or
10 h  R   R  r  cos 

N  mg 
 cos 
Rr
7

To keep it rolling we find
2
Ff   ma
5
N  
Fmax  s N
2 ma
5 s
But
a
dv d dv
dv v
 
d dt d
d r
Since
1/2
10

v   g h  R   R  r  cos 
7

We get
5 
g R  r  sin 
7
dv

1/2
d 10



 7 g h  R  R  r cos 
5 
g R  r  sin 
7
a  
r
Thus
N
2 g  R  r  m sin 
7
rs
10  h  R   R  r  cos 
 2 mg  R  r  sin 
mg 
 cos  
Rr
rs
7
 7
10  h  R  2  R  r  sin  17
 cos 


7  Rr  7
rs
7
 R  r 2 sin  17
h  R 
  R  r  cos 
5rs
10
We need the minimum h so:
2
dh  R  r  cos  17 
R  r  sin   0


d
5rs
10
 tan  
h  R 
R  r 2

rs 17
 R  r 2

5rs
1   2 1/2

17
1
R  r
10
1   2 1/2
As a check compare this with our previous result where we took S infinite (assumed it
rolled regardless of h).
0
h  R  1.7 R  2.7 R
as before. In general

2 R

  1
17s  r

r  
r


R 1    1   

R
h  R   R    R   1.7 

1   2 1/2  r 50
Suppose R/r = 10 and μs = 0.8. Then

2
(9)  1.324
17  .8


.9

10  .9  1.324

h  R 1 
1.7

   R 1  2.538  3.538 R
1/2 
5  .8
 1  1.3242  

Clearly the requirement of not slipping is more stringent than that of making it over the top if the
coefficient of static friction were infinite.
ANGLES AS VECTORS
I mentioned earlier that finite angles can’t be treated as vectors but infinitesimal ones can.
Now we will see why. A fundamental property of vectors is that they can be added in any order
(at least in flat space – which we are assuming). Hence performing a rotation about the x axis
and then about the y axis should produce the same result and doing y first and then x. Let’s check
this. Perform a 90o ccw rotation about x. This results in:
xˆ  xˆ '  xˆ
yˆ  yˆ '  zˆ
zˆ  zˆ '   yˆ
Now rotate by 90o ccw about y’. This results in:
xˆ '  xˆ ''  yˆ
yˆ '  yˆ ''  zˆ
zˆ '  zˆ ''  xˆ
Now do it in the reverse order:
xˆ  xˆ '  zˆ
yˆ  yˆ '  yˆ
zˆ  zˆ '  xˆ
xˆ '  xˆ ''  zˆ
yˆ '  yˆ ''  xˆ
zˆ '  zˆ ''   yˆ
Clearly this is not the same as our first result. Hence finite angles are NOT vectors.
Now consider infinitesimal rotations by d about x and d about y. Do x first. Consider
an arbitrary point (x,y,z). Rotation about x by d gives:
xˆ  xˆ '  xˆ
yˆ  yˆ '  y cos  d   z sin  d   y  zd
zˆ  zˆ '  z cos  d   y sin  d   z  yd
Now rotate by d around y.
x '  x ''  x 'cos d  z 'sin d  x ' z 'd  x  zd
y '  y ''  y '  y  zd
z '  z ''  z 'cos d  x 'sin d  z  yd  xd
where we have kept first order in d, d.
Now do it in the reverse order. Rotation about y gives:
x  x '  x cos d  z sin d  x  zd
y  y'  y
z  z '  z cos d  x sin d  z  xd
Rotation about x now gives:
x '  x ''  x '  x  zd
y '  y ''  y 'cos d  z 'sin d  y  zd
z '  z ''  z 'cos d  y 'sin d  z  xd  yd
where we have again kept first order in d,d. Now the two results are the same. Thus
infinitesimal angles are vectors. This means that:

 d

dt
is also a vector since dt is merely a scalar. We have already assumed this, but now we see why.
KINETIC ENERGY
Consider the kinetic energy of an object viewed from an inertial coordinate system. We
have:



R i  R cm  ri



Vi  Vcm  vi
  1


1


 mi Vi  Vi   mi  Vcm  vi    Vcm  vi 
2 i
2 i

1
 1
2
 MVcm
 Vcm   mi vi   mi vi2
2
2 i
i
KE 
But


 mi ri  0   mi vi  0
i
Thus
i
KE  KE cm  KE rot
where KECM is the kinetic energy of the center of mass, and KErot is the kinetic energy about the
center of mass. But for an object of fixed shape we know that:
 

vi   ri
 KE rot 
1
 
 
 mi   ri     ri 
2 i
Now at a given instant the situation looks like:
 
 ri  r i sin  ˆ   i ˆ
where ̂ is directed into the board. Thus
 
 
  ri     ri   22i
and
1
1
KE rot    mi 2i  2  I2
2 i
2

where I is the moment of inertia about the instantaneous rotation axis.
ROTATING COORDINATE SYSTEM
We know that for rotations we can use a coordinate system with origin at the center of
mass, but that the orientation of the axis must be fixed – non rotating. The problem is that in
general this causes the moments of inertia to change. It didn’t happen for cylinders or spheres
because of the symmetry, but in general it does (remember the out of balance tire). We now
consider how to fix this problem.
We consider two coordinate systems. Both have origin at the center of mass. One has the
orientation of the axis fixed in space, the other fixed in the object. We need to know how the
time derivatives in the two systems are related.
Consider an arbitrary vector A as seen from the two systems.
Since in flat space (which we are assuming) only the magnitude and direction of the vector
matter, we can move A to the origin:
But now this is exactly the situation we had in deriving the moment of inertia tensor, where now


A plays the role of r i . Recall that:

 dri  
vi 
  ri
dt
Hence

dA
 


A
rot
dt
where

dA
rot
dt

is the derivative of A due to rotation as seen from the system fixed in space. As seen from the
system fixed in the object there is, of course, no change due to rotation. In addition, both systems

see whatever change occurs in A due to changes in its magnitude and orientation in object. Thus

the total derivative of A as seen by the system fixed in space is the sum of that seen by the

system fixed in the object and (d A /dt)rot. Thus:


dA dA  

  A
dt
dt '
where

dA
dt '

is the derivative seen in the system fixed in the object. Since A is arbitrary we have the general
relation:
d
d 

 
dt dt '
Applied to the torque equation we find:


 dJ dJ  


  J
dt dt '
But now we have:
  
J  I 
with I constant. If we now choose principle axis (a set such that the I matrix is diagonal – which
can always be done) we have:
0
I
  11
I   0 I 22
 0
0

0 

0 
I33 
Hence


J  I11x xˆ ' I22y yˆ ' I33z z '

 dJ  

  J
dt '

dJ

 xˆ  y J z  z J y   yˆ  z J x  x J z   zˆ  x J y  y J x 
dt '
  x  I11
dx
  y J z  z J y 
dt '
dx
  y I33z  z I 22y 
dt '
d
 I11 x  yz  I33  I22 
dt '
 I11
Similarly,
 y  I22
z  I33
dy
dt '
 x z  I11  I33 
dz
 x y  I22  I11 
dt '
These equations are known as the Euler equations.
We now use them to understand the tossed book experiment we did in class. Suppose
I11>I22>I33 and that we start the book rotating about the x axis as closely as we can (x>>y,z).
The torque about the center of mass is zero (since it is due only to gravity). Hence:
x 

I I
dx
 yz 33 22
dt '
I11
I I
 y  x z 11 33

I22
I I
 z  x y 22 11

I33
Then
I I
 x    
 yz  y
 z  33 22

I11
I I
I I 
 I  I 
   33 22   x z2 11 33  x 2y 22 11 
I 22
I33 
 I11  
I I I I
 I I
 x  2z 11 33  2y 22 11  33 22
I22
I33  I11

0
 y , z  0 
I I
I I 
 I  I 
 y    11 33   y2z 33 22  2x y 22 11 

I11
I33 
 I22  
I I I I
 I I
 y z2 33 22  2x 22 11  11 33
I11
I33  I 22

Since
I33  I22
0,
I11
I 22  I11
0,
I33
I11  I33
0
I 22
we get
 y    y

with
I I  I I 
 I  I 
  11 33 2z  33 22   2x  22 11  
I22   I11 
 I33  
Hence
y  y0 cos   t  
Thus ωy does not grow!! Similarly
I I
I I 
I I 
 z   22 11  2yz 33 22  2x z 11 33 

I33 
I11
I 22 
I I I I
 I I
 z 2y 33 22  2x 11 33  22 11
I11
I 22  I33

Since ωx2 >> ωy2 we get
 z    z

I I I I
  2x 11 33 22 11
I 22
I33
and
z  z0 cos   t  
and z does not grow. Hence rotation about x is stable. Similarly we can show that rotation about
z is stable.
Now suppose we start with y >> x,z. Then:
I I I I 
 x  x 2y  22 11   33 22   1 

 I33   I11 
With
 I  I  I  I 
  2y  22 11   33 22 
 I33   I11 
x A e  t
and x grows exponentially. Thus rotation about y is NOT stable.