Lecture-XI
Angular momentum and Fixed axis
rotation
How to study rotational motion of an object?
We need to develop techniques to handle the rotational motion of solid bodies.
For example, consider the common Yo-Yo running up and down its
string as the spool winds and unwinds. Each particle of the Yo-Yo
moves according to Newton's laws.
dpi
Fi =
, i = 1, 2,3, , N . N = ?
dt
Unfortunately, analyzing rotational problems on a particle-by-particle
basis is an impossible task. One needs to introduce the rotational
analog of the above equation.
In the case of translational motion, one needed the concepts of force, linear
momentum, and center of mass. For rotational motion the analogous concepts are
torque, angular momentum, and moment of inertia.
Leaving aside extended bodies for a time being, let us start with a particle. Since a
particle has no size, its orientation in space is of no consequence, and one is only
concerned with translational motion. In spite of this, particle motion is useful for
introducing the concepts of angular momentum and torque.
Angular Momentum of a Particle
The angular momentum L of a particle with linear momentum p and position vector r
with respect to a given coordinate system is given by
The unit of angular momentum is kg.m2/s in the SI system.
The direction of angular momentum L is defined by the
properties of the cross product of r and p . The vectors r
and p determine a plane, plane of motion, and L is
perpendicular to this plane as shown in the figure.
There are various methods for visualizing and calculating angular momentum. We will
demonstrate a general one.
Consider motion is in the xy plane. Say, r = (x,y,0) and
p = (px,py,0). The L is given by
Since motion is in the xy plane, L lies
entirely along the z axis. For a general
case, L has components along all three
axes.
Angular Momentum of a Sliding Block
Origin at B:
Origin at A:
L depends on the choice of origin.
The Conical Pendulum
Mass M hangs by a massless rod of length l which rotates at
constant angular frequency ω. The mass moves with steady
speed in a circular path of constant radius r. Evaluate L about A
and about B.
ˆ
r
=
rr
and
p
=
Mv
= Mrωθˆ
About A:
LA = r × p = Mr 2ω rˆ × θˆ = Mr 2ω kˆ
(
)
Note that LA is constant, both in magnitude and direction.
About B:
LB = r '× p,
r ' = l sin α rˆ − l cos α kˆ, p = Mrωθˆ
rˆ
LB = r '× p = l sin α
0
θˆ
0
Mrω
kˆ
−l cos α = Mrωl cos α rˆ + Mrωl sin α kˆ
0
Again, L depends on the choice of origin.
But unlike LA, the direction of LB is not constant. LB is perpendicular to
both r' and p. As the bob swings around, LB sweeps out the shaded cone
as shown. The z-component of LB is constant, but the horizontal
component travels around the circle with the bob.
Torque
The torque due to force F which acts on a particle at position r is defined by
If r=(x,y,z) and F=(Fx,Fy,Fz), then τ is given by
A "sense of rotation" can be associated using r and F. Assume in
the sketch that all the vectors are in the xy plane. The torque on
m1 due to F1 is along the positive z-axis (out of the paper) and
the torque on m2 due to F2 is along the negative z-axis (into the
paper).
Few important things to notice:
1. Torque depends on the origin but force
does not.
2. τ and F are always mutually perpendicular.
3. There can be a torque on a system with
zero net force, and there can be force with
zero net torque.
Torque due to Gravity
The problem is to find the torque on a body of mass M about origin A when the
applied force is due to a uniform gravitational field g. We can regard the body as a
collection of particles. The torque τj on the jth particle is
where ri is the position vector of the jth particle from
origin A, and mj is its mass.
Torque as rate of change of angular momentum:
If the torque is zero, L = constant and the angular momentum is conserved.
The change in angular momentum is zero,
∆L = L f − Li = 0 or L f = Li
Thus when the total torque about a point is zero, the final angular momentum is equal
to the initial angular momentum about the same point.
Conservation angular momentum even for a single particle leads to important
consequences.
Central Force Motion and the Law of Equal Areas
Consider a particle moving under a central force,
The torque on the particle about the origin is
Hence, the angular momentum of the particle
magnitude and direction.
is constant both in
Kepler's Laws
Johannes Kepler, working with data painstakingly collected by Tycho Brahe without the
aid of a telescope, developed three laws which described the motion of the planets
across the sky.
1. The Law of Orbits: All planets move in elliptical orbits, with the sun at one focus.
2. The Law of Areas: A line that connects a planet to the sun sweeps out equal areas in
equal times.
3. The Law of Periods: The square of the period of any planet is proportional to the
cube of the semimajor axis of its orbit.
Kepler's laws were derived for orbits around the sun, but they apply to satellite orbits
as well.
Torque on a Sliding Block
dLB
τ = 0, and
= 0, LB = constant
dt
Torque on the Conical Pendulum
There is no vertical acceleration,
Conservation laws: An example
Mass m is attached to a post of radius R by a string. Initially it is distance r from the
center of the post and is moving tangentially with speed vo. In case (a) the string
passes through a hole in the center of the post at the top. The string is gradually
shortened by drawing it through the hole. In case (b) the string wraps around the
outside of the post. What quantities are conserved in each case? Find the final speed
of the mass when it hits the post for each case.
(a) Since
∫ T ⋅ dr ≠ 0
energy is not conserved.
momentum is not conserved.
The force is radial, hence τ=0 , so dL/dt=0.
v0 r
mv0 r = mv f R, So, v f =
R
(b) The force is NOT central, hence angular momentum is NOT conserved.
momentum is not conserved.
But energy is conserved, because ∫ T ⋅ dr = ∫ T ⋅ vdr = 0
Hence,
1 2 1 2
mv f = mv0 , so v f = v0 .
2
2