# ROTATIONAL AND EQUILIBRIUM OF RIGID BODY by sanmelody

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```									ROTATIONAL AND
EQUILIBRIUM OF RIGID
BODY
Standard Competency
Apply the concept and principle of classical mechanics of continuum
system by problem solving

Base Competency
Formulate the relation of torsion concept, angular momentum and
moment inertia base on Newton’s II Law and their application in
rigid body problem

Learning Objectives
1 Formulate the effect of torsion in an object related to its
rotational motion
2 Reveal the analogy of Newton’s II law about translational and
rotational motions
3 Using the concept of moment of inertia for several shapes of
rigid body
4 Formulate the conservative law of angular momentum on
rotational motion
5 Apply the concept of the center of mass of an object in daily
life
6 Analyze the rotational dynamic of rigid body for several
situations
7 Show the ability of psychomotoric and in experiment groups

References:
[1] John D Cutnell and Kenneth W. Johnson (2002). Physics 5th Ed
with Compliments. John Wiley and Sons, Inc. pp. 212-248
[2] Sunardi dan Etsa Indra Irawan (2007). Fisika Bilingual SMA/MA
untuk SMA/MA Kelas X. CV Yrama Widya pp. 255-298
ROTATIONAL VARIABLES

Consider a rigid body A
surrounding a center of xy-
coordinate. One complete of
revolution object A has elapsed
distance (d) as
d = 2πr
The angle of rotation (θ) equal
s = arc length
or

Base on the picture, θ is equal to distance elapsed by the
rigid body A. It means that θ is equal to distance d in
Kinematics.

If d = v.t or s = v.t,         then θ = (v).t.

The velocity v in rotational motion is known as angular
velocity, simbolized by ω, hence
θ = ω.t

Refer to the scientific terms,
- v denote for linier velocity, whilst
- ω denote for angular velocity

Relation between linier (v) and angular velocity (ω)

s                  θ
s=rθ            ;        v =          ;     ω =
t                  t
rθ
v =         →        v =r ω
t

Angular acceleration (α) is derived from linier acceleration
v
a=
t
ω
hence    α =
t
Relation between linier (a) and angular acceleration (α)

s                         θ
s=rθ             ;         v =                    ;     ω =
t                         t
rθ                                                     ω
v =          →           v =r ω                   ;     α =
t                                                     t
v                        r ω
a=            →             a=
t                         t

Acceleration in rotational motion has
both the radial and the tangential
tangential acceleration (at)

present as long as [omega] is not
equal to zero

The tangential component is only
present if the angular acceleration is
not zero

Tangential acceleration (at)
v             r ω                                           ω
at =        →           at =                 →       at = r
t                        t                             t
at = r α

Every part of the body when a rigid body is rotating around a
fixed axis has:
- the same angular velocity [omega] and the same
angular acceleration
- points that are located at different distances from the
rotation axis have different linear velocities and different
linear accelerations
THE EQUATIONS OF ROTATIONAL KINEMATICS

Basically, rotational variables are similar to translational
(linier) variables described in Kinematics, namely position,
velocity and acceleration. This is apply as well since the
equations of rotational kinematics have similar form with the
equation of translational kinematics.

Linier Motion (Translational)             Circular Motion
(a = constant)                      (Rotational)
(α = constant)
v = vo + a t                        ω = ωo + α t

x = ½ (vo + v)t                     θ = ½ (ωo + ω)t

x = vo t + ½ at2                    θ = ωo t + ½ αt2

v2 = vo2 + 2 a x                    ω2 = ωo2 + 2 α θ

MOMENT OF FORCE (TORQUE)

The measure of a force's
tendency to produce a
called torque.
3.0 m

If a force is used to begin
4.0 N
to spin something, a
torque is generated. A
torque would also be
generated if a force was
used to stop something
from spinning.

Torque is simbolized by T or    τ   (Greek letter tau).

Torque arm or lever arm is defined as distance from applied
force (line of action) to fulcrum or axis of rotation
perpendiculary
Torque Direction of Rotation

Positive (+) torque direction   Negative (−) torque direction
as it rotates counter           as it rotates clockwise
clockwise

Torque Equation

Torque angle φ is measured counter clockwise from the
radial line, see the picture below.

Pivot point = titik tumpu
ccw = counter clockwise
Radial line is drawn from pivot
point to the point where force
is applied
Only the tangential component of force causes a torque

The torque increases as the
force increases and also as
the distance increases

The same force will be much
more effective at rotating an
object such as as nut or a
door if our hand is not too
close to the axis.

This is why we have long handled wrench, and why doorknobs
are not next to hinges.

Torque is not a type of energy eventhough its unit is newton-
meter, therefore it should not have a unit of joule.
EQUILIBRIUM OF RIGID BODY

Body Equilibrium
It’s a condition of a body (an object) which still remain as
in initial condition after external forces applied vanished.

According to Newton’s I law, an object to said as equlibry if
resultant force applies on object is equal to zero.

Rigid Body
A characteristic of an object when one or more forces
apply, it will remain as it was (no volume or shape
change)

Equilibrium
An object is in equilibrium if the linear momentum of its
center of mass is constant and if its angular momentum about
its center of mass is constant:
p = constant
L = constant

An object is in static equilibrium if its linear momentum and
angular momentum is equal to zero:
p = 0 kg m/s
L = 0 kg m2/s

Requirements for Equilibrium

If a body is in translational equilibrium then dp/dt = 0, or

If a body is in rotational equilibrium then dL/dt = 0, or
In summary, the following equations must be satisfied for an
object in static equilibrium

If we restrict ourselves to two dimensions (the x-y plane) the
following equations must be satisfied:

Equilibrium and the Force of Gravity

Figure left. Weight of an
object balanced by a single
force. It shows a body of
arbitrary shape balanced by a
single force. The origin of the
coordinate system is defined
such that it coincides with
the center of gravity of the
object, which is the point
upon which the balancing force acts.

An object that is supported at its center of gravity will be in
static equilibrium, independent of the orientation of the
object. If the body is in equilibrium, the net force acting on it
must be zero. It shows that

Since the body is in equilibrium

and therefore
In obtaining this result we have assumed that the
gravitational acceleration is the same for every point of
the body. The net torque acting on the body is given by

Since the body is in static equilibrium

and therefore

This shows that rcm = 0 or rcm is parallel to g.

We conclude that for a body to be in equilibrium, its
center of mass must coincide with its center of gravity.

Example

A uniform beam of length L whose mass is m, rest with its
ends on two digital scales. A block whose mass is M rests on
the beam, its center one-fourth away from the beam's left
end. What do the scales read ?

For the system to be in equilibrium, the net force and net
torque must be zero.
Here we have replaced the force acting on the beam with a
single force acting on its center of gravity. The net torque of
the system, with respect to the left scale, is

This shows immediately that

From the equation of the net force we obtain

Example
L and mass m rests
against a wall. Its
upper end is a
distance h above the
ground. The center
of gravity of the
the way up the

A firefighter with
mass M climbs
halfway up the
the wall, but not the ground, is frictionless. What is the force
exerted on the ladder by the wall and by the ground ?
The wall exerts a horizontal force FW on the ladder (the
normal force); it exerts no vertical force. The ground exerts a
force Fg on the ladder with a horizontal component Fgx and a
vertical component Fgy.

If these two components were not present, the system would
not be in equilibrium. The net force in the x and y directions is
given by

and

The net torque, with respect to O (which is the contact point
between the ladder and the ground), is given by

This immediately shows that

We can now calculate the force Fg:

and

We observe that Fgx depends on the position of the firefighter.
Suppose that the firefighter is a distance f L up the ladder. In
this case Fgx is given by
If the coefficient of static friction between the ladder and the
ground is us, than the maximum distance the firefighter can
climb is reached when

or

This shows that

Stacking Blocks

Two bricks of length L
and mass m are
stacked. Using
conditions of static
equilibrium we can
determine the
maximum overhang of
the top brick

Two bricks of length L and mass m are stacked. Using
conditions of static equilibrium we can determine the
maximum overhang of the top brick.

The two forces acting on the top brick are the gravitational
force Fg and the normal force N, exerted by the bottom brick
on the top brick. Both forces are directed along the y-axis.
Since the system is in equilibrium, the net force acting along
the y-axis must be zero. We conclude that
The torque of the normal force and the gravitational force
with respect to O is given by

The net torque acting on the top brick is given by

If the system is in equilibrium, then the net torque acting on
the top brick with respect to O must be zero.
This implies that

or

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