PHYS1110H, 2011 Fall.
Shijie Zhong
Rotation, Angular Momentum, Torque, and Conservation of Angular
Momentum.
Rotational Motion about a Fixed–axis (kinematics)
Angular velocity ω:
Consider a rotating disk with a fixed axis. An effective way to describe the
motion is to introduce a coordinate or angle θ (in radian). θ increases in
counter-clockwise direction, and decreases in clockwise direction. Angular
velocity ω (rad/second) is the rate of change of angle θ.
€
ω = dθ dt , or ω = Δθ Δt ,
(1)
for instantaneous and average angular velocities, respectively. For counterclockwise rotation, ω is positive (a right-hand rule), while it is negative for
clockwise rotation.
€
It is useful to make an analogy between 1-D motion and fixed-axis rotation
at this point. The coordinate x and v for 1-D motion are analogous to θ and
ω here. Of course, their units are different. It’s also interesting to point out
the difference in directions. Velocity v is a vector, and is in either positive x
axis or negative axis. For angular velocity ω, we may also assign it a
direction – for positive (i.e., counter-clockwise rotation) angular velocity, ω
points to z axis direction that is perpendicular to the rotational plane, and
negative ω points to negative z axis direction.
Angular velocity ω is also called angular frequency that can be related to the
usual definition of frequency f, the number of revolution per second. We
may define period T as the time it takes for a full revolution or 2π radian.
T=1/f. Therefore, we have ωΤ=2π or
Note
8
1
T=2π/ω,
and
ω=2πf.
(2)
Angular acceleration α:
We may further define angular acceleration α as the rate of change in ω:
2
α = dω dt = d θ 2 ,
(3)
dt
with direction in either positive z or negative z directions (but not dependent
on clockwise or counter-clockwise rotation anymore).
€
€
€
€
For constant angular acceleration α, the following relations hold
(4)
ω = ω 0 + αt ,
1
θ = θ0 + ω 0 t + αt 2 ,
(5)
2
(6)
ω 2 = ω 0 2 + 2α (θ − θ0 ) ,
which are similar to what we learned for 1-D motion, not surprising, given
that the mathematical forms are the same.
Angular acceleration α differs from the centripetal acceleration ar=v2/r for
circular motion in direction and meanings, but it may be related to tangential
acceleration atan, that is the rate of change in magnitude of velocity.
atan = dv dt = d(rω ) dt = r dω dt = rα ,
(7)
The total acceleration is a vector sum:
€
a = atan + ar or | a |= atan 2 + a r 2 . (8)
Polar coordinate system
€
At this point, it is useful to discuss velocity and acceleration vectors in polar
coordinates. So far, we have limited ourselves in Cartesian coordinate
system. Sometimes, it is convenient to use polar coordinate system where
particle’s position is described in terms of r and θ.
Note
8
2
 
In polar coordinates, position vector r is r = rer , where er is unit vector in r
direction.
€
The velocity vector is:


der
 dr d(rer ) dr 
v=
=
= er + r
,
dt
dt
dt
dt
€
€
€
€
€
€
(9)
where notice that both er and eθ change with time (see the middle figure
above) as a particle moves from positions r(t) to r(t+Δt), and we will need to
compute der/dt and deθ/dt using the right figure above.




der
Δer
er (t + Δt) − er (t)
Δθ 
dθ 

= lim
= lim
= lim
eθ =
eθ = ωeθ , (10)
dt
Δt
Δt 
Δt
dt


deθ
Δe
e (t + Δt) − eθ (t)
Δθ 
dθ 

= lim θ = lim θ
= lim(−
er ) = − er = −ωer , (11)
dt
Δt
Δt
Δt
dt
Substituting (10) to (9), we have velocity in polar coordinates:

 dr dr 

v=
= er + rωeθ .
(12)
dt dt
We can now derive an expression for acceleration.



deθ
 dv d 2 r  dr der d(rω ) 
,
a=
=
er +
+
eθ + (rω )
dt dt 2
dt dt
dt
dt
d 2 r  dr 
dr 
dω 

=
er + ωeθ + ωeθ + r
eθ − rω 2 er .
dt
dt
dt
dt 2
dr
 d 2r


(13)
a=(
− rω 2 )er + (2 ω + rα )eθ .
2
dt
dt
€
For circular motion with r as a constant, (12) and (13) become





v = rωeθ , and a = −rω 2 er + rαeθ .
€
Note
8
€
3
There are two components in acceleration -- the radial component is our
usual centripetal acceleration, while the θ-component is tangential
acceleration. They can be related back to (7) and (8). For circular motion at a


constant rate, tangential component vanishes, and a = −rω 2 er is our usual
centripetal acceleration.
Torque.
€
For a particle with position vector r in a given coordinate system with force
F, its torque τ is given as cross product of r and F,
  
τ = r ×F.
€
(14)
The unit of τ is kg m2/s2. Recall all the properties we discussed about the
cross product.
If r and F are in xy plane, τ is parallel
to z axis,
  

τ = r × F = rF sin θez .
€
(15)
where θ is the angle between vectors r
and F when they share the same origin
(remember that we can move a vector
with no rotation).
However, vector r often starts from the
origin of coordinates or some reference point (e.g., hinge of rotation) and
ends at the location of a particle, while vector F starts at the particle
location. τ points to either positive or negative z direction that would bring
in either counter-clockwise (positive z) or clockwise (negative z) rotation of
the particle around the reference point.


If vectors r and F are given in 3-D as r = (x, y,z) and F = (Fx ,Fy ,Fz ) , we
can use the general form -- determinant form for cross product.



e x e y ez

τ= x
y
z .
€(16)
€
Fx Fy Fz
€
Note
8
4
Notice that torque τ is always perpendicular to vectors r and F.
If several forces acting on the same object (e.g., a wheel or a bin), the total
or net torque is


 
τ net = ∑ τ j = ∑ r j × F j ,
(17)
where j is index for each torque or position and force vector pair.
€
€
Net force and net torque are not closely related to each other. The following

4 scenarios are all possible. 1) τ net = Fnet = 0 (in equilibrium), 2)




τ net = 0, Fnet ≠ 0 , 3) τ net ≠ 0, Fnet = 0 , and 4) they are both nonzero.
Moment of Inertia about a rotation axis.
€
€ particles with mass m and
For a system of
j
distance to a rotation axis rj, the moment of
inertia of the particles about the rotation axis:
I = ∑ m j rj 2 ,
(18)
j
For an object (e.g., a ring or sphere) with
density ρ, (18) can be generalized:
€
€
I = ∫ r 2 dm = ∫ ρr 2 dV ,
(19)
where r is the distance of the element of volume to the rotation axis. If the
rotation axis is along z axis, r2=x2+y2.
Note
8
5
As we will see below, the torque is analogous to force in linear motion, and
moment of inertia is analogous to mass.
Rotational Dynamics about a Fixed-axis.
We drew comparisons between 1-D linear motion kinematics and fixed-axis
rotational motion. The analogy can be further extended in dynamics as well
with torque, moment of inertia, and angular acceleration in rotational motion
analogous to a force, mass, and linear acceleration in 1-D linear motion.
Apply force F to a block of mass m that is attached to a rod with the other
end fixed at a free-to-rotate axis (see figure). The force F is always
perpendicular to the rod and causes the mass to accelerate in tangential
direction, according to the Newton’s 2nd law:
F = matan = mrα or
rF = mr 2" .
(20)
€
!
I = mr 2 is the moment of inertia.
(20) becomes
!
(21)
" = I# .
(21) is analogous to the Newton’s 2nd law F=ma.
!
Example 1. I for a uniform ring around axis passing the center.
For a ring with mass M and radius R, the perimeter of the ring is 2πR, and
the linear density λ=M/(2πR).
2π
I = ∫ λR2 dl = ∫ λR2 Rdθ = 2 πλR 3 = MR 2 .
0
€
Example 2. I for a uniform disk around axis passing the center.
For a disk with mass M and radius R, the area of the disk is πR2, and the
surface density λ=M/(πR2). We may divide the disk into concentric rings,
Note
8
6
and for a given ring with radius r and thickness dr, the surface area of this
ring is 2πrdr. The moment of inertia I for the disk is
R
1
1
2
I = ∫ λr dS = ∫ r 2 λ 2 πrdr = 2 πλ R 4 = MR2 .
4
2
0
Notice from examples 1 and 2 that for a given mass M, the more the mass is
near the center or rotational axis, the smaller the moment of inertia.
€
€
€
€
€
1
ML2 , around the axis
12
passing the center of mass. For a uniform sphere of mass M and radius R
2
around the axis passing the center of mass, I = MR2 .
5
€
Example 3. I for a uniform disk around an axis at a distance l to the
center – parallel axis theorem.
Without loss of generality, let the rotation axis to be on positive x axis at a
distance of l from the center of the€disk which is also used as the origin of
the coordinates (see figure below). Consider a small element with surface
area dS and mass λdS, with λ=M/(πR2) as in example 2. The distance
between this element and the rotation axis
is r, and its distance to the origin is r’. For
the given triangle in the figure below, the
law of cosine is
r 2 = l 2 + r'2 −2r'l cos θ .
The moment of inertia with respect to the
rotation axis:
I = ∫ λr 2 dS = λ ∫ (l 2 + r'2 −2lr'cos θ )dS
For a uniform rod of mass M and length L, I =
= λ ∫ l 2 dS + ∫ λr'2 dS − 2lλ ∫ r'cos θdS
= πR2 λl 2 + ICM = Ml 2 + ICM ,
where ICM is the moment of inertia relative
to the axis passing through the center of
mass (i.e., 1/2MR2 from example 2), and the integral with cosθ is zero,
because of the definition
of CM R=0.
  
 

∫ λr'cos θdS = ∫ λr '⋅l dS = ∫ λr ' dS ⋅ l = R ⋅ l = 0 .
I = Ml 2 + ICM ,
(22)
€
€
Note
8
7
(22) is called parallel axis theorem. Although we showed this for a special
case of a disk, this theorem is true for all the cases.
€
€
€
€
€
€
Example 4. Atwood’s machine with a massive pulley.
Two masses M1 and M2 hang on two sides of a pulley with mass Mp and
radius R. Find acceleration for the masses.
We solved similar problems before with massless pulley, but we now need
to consider mass of pulley. While the acceleration for the two masses
remains the same, the tensional force on the string T1 and T2 differ now.
Newton’s 2nd law for the two masses:
M1g − T1 = M1a , (i)
(ii)
T2 − M 2 g = M 2 a ,
For the pulley, we may use (14) that
relates net torque to tangential
acceleration and also notice
atan = a = Rα ,
T1 R − T2 R = Iα , or
1
a
T1 R − T2 R = M p R 2 ,
2
R
1
T1 − T2 = M p a . (iii)
2
Equations (i), (ii), and (iii) can be
solved for a, T1 and T2.
(M1 − M 2 )g
a=
.
M1 + M 2 + M p /2
Rotational kinetic energy.
€
If we place a rotating disk with zero translational velocity on a ramp, the
disk may rotate and climb the ramp. This indicates that a rotating object has
rotational kinetic energy. For a pure rotational motion, the kinetic energy can
be considered as sum for small elements of masses that are all rotated around
a fixed axis at constant angular velocity ω.
1
1
1
1
KE = ∑ m j v j 2 = ∑ m j r j 2ω 2 = ω 2 ∑ m j r j 2 = Iω 2 ,
2
2
2
2
(23)
€
Note
8
8
For an object that rotates around an axis passing the center of mass and that
the center of mass has a translational motion vCM (e.g., a rolling wheel), the
total kinetic energy for this object is
1
1
KE total = KE trans + KE rot = mvCM 2 + Iω 2 .
(24)
2
2
€
Example 5. Down a ramp, which goes faster? a sphere, a hoop and a
cylinder, each with mass M and radius R and starts at rest at a height h.
€
€
€
Conservation of energy:
1
1
1
1 vf
Mgh = Mv f 2 + Iω 2 = Mv f 2 + I( )2 .
2
2
2
2 R
v f = 2Mgh /(M + I / R 2 ) .
Substituting I for a sphere, a hoop, and a cylinder, we found that vf is
v f = gh , 4gh / 3 , and 10gh / 7 for a hoop, cylinder, and sphere,
respectively, in the order from slower to faster.
The sphere has a smaller moment of inertia, thus taking less rotational
kinetic energy, and hence moving faster. One can easily show that for a
€
€
block sliding down the ramp with no rotation and no friction, v f = 2gh is
the largest among them all.
€
Note
8
9
Angular momentum of a particle.
For a particle with position vector r in a given coordinate system with
momentum p, its angular momentum L is given as cross product of r and p,
  
L = r × p.
(25)
The unit of L is kg m2/s. If r
and p are in xy plane, L is
parallel to z axis, or
€
  

L = r × p = rp sin φez , (26)
where φ is the angle between
vectors r and p when they share the same origin.
€
However, often vector r starts from the origin of coordinates and ends at the
location of a particle, while vector p starts at the particle location and points
to the direction of motion or v. L points to either positive or negative z
direction that represents counter-clockwise (positive z) or clockwise
(negative z) rotation of the particle.


Suppose that r = (x, y) and p = (p x , p y ) , we can use the determinant form
for cross product.



e x e y ez
  

(27)
L =€r × p = x €y 0 = (xp y − yp x )ez .
px py 0
Angular momentum depends on the reference point where vector r starts, as
in the following example.
€
Torque can be related to the rate of change of angular momentum.
€
€
€





dL d   dr   dp    dp  dp
= (r × p) =
× p+r ×
=v × p+r ×
=r ×
dt dt   dt
dt
dt
dt
=r ×F =τ,
where we
 considered that v x p = 0 as they are parallel to each other.
 dL
.
(28)
τ=
dt
Note
8
10
Example 6. Angular momentum of a sliding block.
A block with mass m slides along x axis at a velocity v.
€
€
€
€
First,
angular momentum relative to the origin or point A,

  

L A = r × p = re x × mve x = 0 .
We did not specify any force on the block, but we can infer based on motion
along x axis, if there is a net force, the force must be also along x axis.
Therefore, the torque with respect to point A would also be zero:




τ A = r × F = re x × Fe x = 0 .
The torque and angular momentum satisfy (28).
Now let’s compute the angular momentum relative to point B along y axis.



e x e y ez

 

L B = r × p = x −l 0 = mlvez .
mv 0 0
We can compute the torque due to a net force F along x direction, given the
motion of the block is along x direction.



e x e y ez

 

τ B = r × F = x −l 0 = Flez .
F 0 0
Again, the net torque and angular momentum satisfy (28), as demonstrated
below:



dL B d(mlvez )
dv 

=
= lm ez = lFez = τ B .
dt
dt
dt
€
Note
8
11
Example 7. Angular momentum of a conical pendulum (the particle
mass M, velocity v, and length of the string l).
First, let’s compute angular momentum relative
to the center of the circle on the plane of motion,
or point A. Since r and p are perpendicular to
each other in this case, and the motion is
counter-clockwise,

 

L A = r × p = Mvrez ,
which is along z direction.
We can also compute the torque with respect to
€
point A caused by the net force.
The net force acting on the pendulum is the
sum of the tension along the string and gravity, and should point to A or
parallel to vector r. Therefore, the net torque relative to point A is ZERO.
Again, this is consistent with (28), because vector LA is a constant, and its
time derivative is also zero.
€
Now, let’s compute angular momentum relative to the hinge point at B, and
its magnitude is given as below,
 
| L B |=| r '× p |= Mvl,
where vector r’ starts from B and ends at the particle with length of l, and is
perpendicular to vector v or p. The direction of vector LB, as shown, changes
with time (around z axis). To show (28) is more tedious for this case, and we
skip it.
Angular momentum for a fixed
rotation axis
Equation (28) is equivalent to
Newton’s second law, and can be
reduced to (21) for a fixed axis
rotation. Without loss of
generality, we may consider an
object rotating around z axis with
angular velocity ω also in z
direction, and the object may be
divided into many small elements
Note
8
12
of
The total angular momentum for the rotating object
 volumes.
 


L = ∑ rj × p j = ∑ rj × m j v j .
€
Let vector r’j be position vector to j-th element from the axis, we have



v j = r' j ωevj where evj is the unit vector in xy plane in the direction of
velocity (see figure above). This is because the velocity must be
perpendicular to z axis and in xy plane.
€
€
The z-component
of the angular momentum,
  



L z = ∑ [(r j × p j ) ⋅ ez ] = ∑ [(r j × m j r' j ωevj ) ⋅ ez ] = ∑ m j r' j 2 ω = Iω .
 

where we considered (r j × evj ) ⋅ ez = r' j , and ∑ m j r' j 2 = I .
€
€
L z = Iω .
(29)
Substituting (29) into (28) leads to (21). We should point out that equation
€
€
(29) is the angular momentum of a rotating object with a fixed axis in the
direction of angular velocity.
Motion with both translation and rotation
A rolling disk on a plane or a spinning planet orbiting a star has both
translation and rotation motion. Here we only consider the rotation motion
with rotation axis parallel to z axis (i.e., having a fixed direction in space).
Note
8
13
First, let’s consider the z component angular momentum and state that


(30)
L z = I0ω + ( R × MVcm ) z ,
where I0 is the moment of inertia of the body relative to a rotation axis
passing the center of mass CM, ω is the angular velocity of the rotation
relative to the same axis, M is the mass of the body, R is the position vector
for CM and Vcm is CM’s velocity, both relative to an inertial reference
frame. Equation (30) can be proved as following.
€
Let’s divide the object into many small elements, and for j-th element with
position vector rj, in an inertial reference frame, the angular momentum is





L = ∑ r j × m j v j = ∑ r j × m j r˙j .
(31)


CM’s position vector is R = ∑ m j r j / M . Suppose that r’j is the position
vector for j-th element relative to the CM, we have


rj = R + r ' j ,
and substituting
it to (31), we have

  €
 
L = ∑ ( R + r ' j ) × m j ( R˙ + r˙ ' j )
€


 


 
= R × ∑ m j R˙ + ∑ m j r ' j ×R + R × ∑ m j r˙ ' j + ∑ m j r ' j ×r˙ ' j .
€
(32)
The 2nd and 3rd terms in
(32) are zero, because

 

∑ m j r ' j = ∑ m j (r j − R) = ∑ m j r j − MR = 0 .
we have
€ Therefore,
 



€
€
 
 
L = R × ∑ m j R˙ + ∑ m j r ' j ×r˙ ' j = R × MR˙ + ∑ m j r ' j ×r˙ ' j .


 
(33)
L z = ( R × MVcm ) z + (∑ m j r ' j ×r˙ ' j ) z .
The 2nd term in (33) is for the rotational motion about the axis that is parallel
to z axis and passes the CM, and it can be written as (29) or I0ω, according
to derivation there. Therefore, we proved equation (30).
€
€
Similarly, we may write total torque acting on the body in the same inertial
frame as



 
  
 
(34)
τ = ∑ ri × fi = ∑ (r 'i + R) × fi = ∑ (r 'i × fi ) + R × ∑ fi .


Let F = ∑ fi be the total force on the body. The z component of the total
torque is
 
 
 
τ z = ∑ (r 'i × fi ) z + ( R × F ) z = τ 0 + ( R × F ) z ,
(35)
€
€
€
Note
8
14
where τ0 is the z component of the torque relative to the CM.
Clearly, we should have τ z = dL z / dt , with z-component of torque and
angular momentum defined in (35) and (30), respectively. More specifically,
we have
(36)
τ 0 = d(I0ω )/ dt = I0α ,
€
where α is angular acceleration. Equation (36) states that the rotation about
the axis passing the CM only depends on the torque relative to the CM.
€
Recall that equation (24) for total kinetic energy with both rotational and
translational motions.
Example 8. A uniform disk with radius R and mass M rolls with no slipping
down a ramp with angle θ. Find acceleration.
€
€
€
€
Method 1: See the left diagram, using Newton’s 2nd law and also (36),
W sin θ − f = Ma , (W=Mg)
fR = I0α , (Only frictional force contributes to torque relative to the CM)
a = Rα . (because of no slipping)
Considering I0=1/2MR2, solve these three equations for three unknowns, f,
a, and α, and we have a = 2 / 3g sin θ .
Method 2: See the right diagram, and use a coordinate system with origin at
the contact point. Use (30) to compute the angular momentum:
€


1
3
L z = I0ω + ( R × MVcm ) z = MR2ω + MRVcm = MRVcm .
2
2
where we consider no slipping condition or Vcm=Rω.
With the contact point as the origin, only gravity contributes to the torque,
3 
τ z = ( ∑ ri × Fi ) z = MgR sin θ .
i=1
τ z = dL z / dt , or MgR sin θ = 3/2MRdVcm / dt , or a = dVcm / dt = 2 / 3g sin θ .
€
€
Note
8
€
€
15
Conservation of Angular momentum.
From (28), if the net torque on a particle is zero, then its angular momentum
L is conserved or constant with time. Conservation of angular momentum is
an important conservation law, similar to conservation laws of momentum
and energy.
Example 9. Proof of Kepler’s 2nd law – equal area.
The gravitational force between a planet and its star always points to the
star, or F(r)=f(r)r, therefore with the gravity as the only force on the planet,
 

the torque on the planet is zero -- τ = r × ( fr ) = 0 . Hence, the angular
momentum of a planet is conserved or constant with time.
€
€
€
€
€
Consider the planet at times t and t+Δt, at coordinates (r,θ) and (r+Δr,θ+Δθ).
The area swept out is
ΔA ≈ 0.5(r + Δr)(rΔθ ) = 0.5r 2 Δθ + 0.5rΔrΔθ .
The rate of sweeping the area is
dA
ΔA
dθ
1 dθ
= limΔt→0
= 0.5r 2
+ 0 = r2
.
dt
Δt
dt
2
dt
The velocity of the planet can be written as

de
dr 
dθ 
 dr d(rer ) dr 
v=
=
= er + r r = er + r eθ ,
dt
dt
dt
dt
dt
dt
which is the velocity vector expressed in a polar coordinate system, and in
deriving
it we considered

der dθ 
=
eθ ,
dt
dt
in the polar coordinate system, as in (10).
Then the angular momentum of the planet is
 
dr 
dθ 
dθ 


L = r × mv = mrer × ( er + r eθ ) = mr 2
ez .
dt
dt
dt
Since L is constant with time, dA/dt must be a constant.
€
Note
8
16
Example 10. Capture cross section of a planet.
Suppose that we send a spacecraft to a planet with radius R. If gravity of the
planet plays no role, then the spacecraft has to aim at a cross section area
πR2 to land the spacecraft to the planet. However, when the planet’s gravity
field is considered, then the cross section area to capture the spacecraft
would be larger. Here we determine this cross section area.
€
€
€
Suppose the distance between where the spacecraft is launched and the
planet is much larger than R. The distance between the initial trajectory of
spacecraft and the line aa is b, and is called impact parameter of the
trajectory. We need to determine the largest b (let’s call it b’) for which the
spacecraft can still be captured by the planet. The capture cross section area
is πb’2.
Because the position vector and gravitational force on the spacecraft are
parallel, the net torque on the spacecraft is zero, and the angular momentum
of the spacecraft is conserved (is the linear momentum of the spacecraft
conserved?). The total energy of the spacecraft is E=K+U and is
1
GmM
E = mv2 −
.
2
r
The angular momentum of the spacecraft about the center of the planet is
L = −mvr sin φ .
Initially, at very large distance r, the spacecraft moves at a speed of v0 and is
1
a trajectory with impact parameter b’. E = mv0 2 and L= -mv0b’.
2
The point of collision occurs at a distance of closest approach of the orbit,
that is r=R. And at this point, vectors r and v are perpendicular to each other,
i.e., L= -mvR. Therefore, conservation of angular momentum and energy
€
lead to:
1
1
GmM
mv0 2 = mv2 −
,
2
2
R
Note
8
17
€
−mv0 b'= −mvR .
Two unknowns are v and b’ that can be solved from these two equations.
GmM / R
b'2 = R 2 (1 +
) , or
2
mv0 /2
GmM / R
U (R)
πb'2 = πR2 (1 +
) = πR2 (1 −
).
E
mv0 2 /2
€
€
Note
8
18