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
 6 Analyze the rotational dynamic of rigid body for several
 7 Show the ability of psychomotoric and in experiment groups

[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

                                          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
                                          to 2π radian,
   s = arc length
   r = radius                                θ = 2π (rad)

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

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
                   v =         →        v =r ω

Angular acceleration (α) is derived from linier acceleration
hence    α =
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
                                     components, namely radial (ar) and
                                     tangential acceleration (at)

                                     The radial component is always
                                     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

Radial acceleration (ar)

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

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


                                         The measure of a force's
                                         tendency to produce a
                                         rotation about an axis is
                                         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
Torque Direction of Rotation

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

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

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.


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

                                           A ladder with length
                                           L and mass m rests
                                           against a wall. Its
                                           upper end is a
                                           distance h above the
                                           ground. The center
                                           of gravity of the
                                           ladder is one-third of
                                           the way up the

                                            A firefighter with
                                            mass M climbs
                                            halfway up the
                                            ladder. Assume that
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


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:


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


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


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