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Nicholas J. Giordano
Nicholas J. Giordano
www.cengage.com/physics/giordano
Rotational Motion
Introduction
• Until now, objects have been treated as point
particles
• The point particle treatment is the correct way to
describe translational motion
• The point particle approximation does not allow us to
analyze rotational motion
• Need to learn how to use Newton’s Laws to
describe and analyze rotational motion
• The size and shape of the object will have to be taken
into account
Introduction
Describing Rotational Motion
• To deal with rotational motion, rotational quantities
need to be defined
• Angular position
• Angular velocity
• Angular acceleration
• These will be analogous to the translational (linear)
quantities of position, velocity and acceleration
Section 8.1
Coordinate System
• Need to identify the
•
•
•
•
rotation axis
In A, the rod is fixed at
one end and the rotational
axis is the z- axis
The angle θ is measured
with respect to the x-axis
For the CD in B, the z-axis
is the axis of rotation
The angular position, θ, is
specified by the angle the
reference line makes with
the x-axis
Section 8.1
Radian
• The end of the rod
sweeps out a circle of
radius r
• Assume the end of the
rod travels a distance s
along the circular path
• At the same time, the
rod sweeps out an
angle θ
Section 8.1
Radian, cont.
• The distance s and angle θ are related by
• θ is measured in radians
• Angles can also be measured in degrees
• 360°= 2 π rad
•
Both measure one complete circle
Section 8.1
Angular Velocity
• Angular velocity describes how the angular position
is changing with time
• Denoted by ω
• For some time interval, Δt, the average angular
velocity is
Section 8.1
Instantaneous Angular Velocity
• The instantaneous angular velocity is
• The instantaneous angular velocity equals the
average angular velocity when ω is constant
• SI unit is rad/s
• May also see rpm (revolutions / minute)
Section 8.1
Angular Velocity, Direction
• Since angular velocity is
a vector quantity, it must
have a direction
• If θ increases with time,
then ω is positive
• Therefore, a
counterclockwise
rotation corresponds to
a positive angular
velocity
• Clockwise would be
negative
Section 8.1
Angular Acceleration
• Angular acceleration is the rate of change of the
angular velocity
• Denoted by α
• The average angular acceleration is
• The instantaneous angular acceleration is
• SI units is rad/s²
• The instantaneous angular acceleration equals the
average angular acceleration when α is constant
Section 8.1
Angular Acceleration and Centripetal
Acceleration
• Angular acceleration and centripetal acceleration are different
• As an example, assume a particle is moving in a circle with a
constant linear velocity
• The particle’s angular position increases at a constant rate,
therefore its angular velocity is constant
• Its angular acceleration is 0
• Since it is moving in a circle, it experiences a centripetal
acceleration of ac = v2/ r
• This is not zero, even though the angular acceleration is zero
• The centripetal acceleration refers to the linear motion of the
particle
• The angular acceleration is concerned with the related angular
motion
Section 8.1
Angular and Linear Velocities
Compared
• When an object is
rotating, all the points
on the object have the
same angular velocity
• Makes ω a useful
quantity for describing
the motion
• The linear velocity is not
the same for all points
• It depends on the
distance from the
rotational axis
Section 8.1
Period of Rotational Motion
• One revolution of an object corresponds to 2 π
radians
• The object will move through ω / 2π complete
revolutions each second
• The time required to complete one revolution is the
period of the motion
• Denoted by T with
Section 8.1
Connection Between Linear and
Angular Velocities
• The linear velocity of any point on a rotating object is
related to its angular velocity by v = ωr
• r is the distance from the rotational axis to the point
• Technically, this is the relationship between the
speeds, since direction has not been taken into
account
• When a point is farther from the axis of rotation (rA >
rB , for example), then its linear velocity will be
greater
• vA > vB
• The angular velocities of both points are the same
Section 8.1
Connection Between Linear and
Angular Accelerations
• The relationship between linear and angular
velocities can be used to determine the relationship
between linear and angular accelerations
• Analysis indicates the angular acceleration and the
linear acceleration of a point a distance r from the
rotation axis is given by a = α r
Section 8.1
Torque and Newton’s Laws for
Rotational Motion
• A connection between force and rotational motion is
needed
• Specifically, how forces give rise to angular
accelerations
• The approach will be similar to looking at forces and
linear motion
Section 8.2
Torque
• Torque is the product of
an applied force and the
distance it is applied
from the support point
• Denoted τ
• The point P is called
the pivot point
•
Since the object can
rotate around that point
Section 8.2
Lever Arm and Torque
• The lever arm is the distance between the pivot point
and where the force acts
• When the force is perpendicular to a line connecting
its point of application to the pivot point, the torque is
given by τ = F r
• Torque in rotational motion is analogous to force in
translational motion
• Torque is the product of the force and the distance to
the pivot point
• SI unit is N. m
Section 8.2
Torque and Angular Acceleration
• Only forces with a
component
perpendicular to the rod
can contribute to the
angular acceleration
• Newton’s Second Law
for rotational motion is
written as Στ = I α
• I is called the moment
of inertia of the object
Section 8.2
Newton’s Second Law – Rotational
• The Στ is the vector sum of all the torques acting on
the object
• The moment of inertia, I, plays the same role that
mass did in translational motion
• For a ball and massless hinged rod, I = m r2
Section 8.2
Analogy with Translational Motion
• Torque plays the role of force in rotational motion
• Torque depends on the magnitude of the force and
where the force is applied relative to the pivot point
• There may be multiple forces acting on the system, all
involving a single pivot point
Section 8.2
Analogy with Translational Motion, 2
 The motion of inertia enters into rotational motion in
the same way that mass enters into translational
motion
 For an object composed of many pieces of mass
located at various distances from the pivot point, the
moment of inertia of the object is
 The moment of inertia depends on the mass and on
how that mass is distributed relative to the axis of
rotation
Section 8.2
Analogy with Translational Motion, 3
• Newton’s Second Law for translational motion can
be used to derive Newton’s Second Law for
rotational motion
• For rotational motion, Newton’s Second Law
becomes Στ = I α
Section 8.2
Linear and Rotational Comparison
Section 8.2
Torque and Lever Arm: Generalized
• In general, the force
that produces a torque
does not have to be
applied in a
perpendicular direction
• Assume the force acts
at an angle ϕ with
respect to the rod
• Only the perpendicular
component contributes
to the torque
Section 8.2
Torque and Lever Arm: Generalized, cont.
• The perpendicular component of the applied force is
Fapplied sin ϕ
• The torque is Fapplied r sin ϕ
• This is a general definition of torque
• It can be used when the force is not directed
perpendicular to the lever arm
• When ϕ is 90°, sin ϕ = 1 and Fapplied = F|
• When ϕ is 0°, sin ϕ = 0 and Fapplied = 0
• A force applied parallel to the lever arm cannot cause
an object to rotate
Section 8.2
One Way to Think About Torque
• Since the perpendicular
component of the force is
Fapplied sin ϕ, the torque
can be expressed as
Fapplied r sin ϕ
• The angle can be found
by extending the radius
line beyond the point
where the force acts
• The angle is between the
force and this radius line
Section 8.2
Another Way to Think About Torque
• You can also use the
perpendicular distance
from the pivot point to
where the force is acting
to calculate the torque
• rperpendicular = r sin ϕ
• This gives the general
expression for the lever
arm
• Therefore,
τ = Fapplied r sin ϕ
Section 8.2
Ways to Think About Torques, Final
• Using the idea of the
perpendicular force or
the perpendicular
distance (lever arm) will
give the same results
• For example:
• ϕ = 90° gives
maximum torque
• ϕ = 0° give zero
torque
Section 8.2
Torque and Direction – Mass at End
• For a single rotation axis, the direction of the torque
is specified by its sign
• A positive torque is one that would produce a
counterclockwise rotation
• A negative torque would produce a clockwise
rotation
Section 8.2
Torque and Direction – Distributed Mass
 We could imagine
breaking the clock hand
up into many
infinitesimally small
pieces and finding the
torques on each piece
 A more convenient
approach is to use the
center of gravity of the
hand
Section 8.2
Center of Gravity
• For the purposes of calculating the torque due to the
gravitational force, you can assume all the force acts
at a single location
• The location is called the center of gravity of the
object
• The center of gravity and the center of mass of an
object are usually the same point
Section 8.2
Rotational Equilibrium
• Equilibrium may include rotational equilibrium
• An object can be in equilibrium with regard to both its
translation and its rotational motion
• Its linear acceleration must be zero and its angular
acceleration must be zero
• The total force being zero is not sufficient to ensure
both accelerations are zero
Section 8.3
Example: Equilibrium
• The applied forces are
equal in magnitude, but
opposite in direction
• Therefore, ΣF = 0
• However, the object is
not in equilibrium
• The forces produce a
net torque on the object
• There will be an
angular acceleration in
the clockwise direction
Section 8.3
Rotational Equilibrium, cont.
• For an object to be in complete equilibrium, the
angular acceleration is required to be zero
• Στ = 0
• This is a necessary condition for rotational
equilibrium
• All the torques will be considered to refer to a single
axis of rotation
•
The same ideas can also be applied to multiple axes
Section 8.3
Problem Solving Strategy – Equilibrium
• Recognize the principle
• The object is in translational static equilibrium if its
linear acceleration and linear velocity are both zero
• The object is in rotational static equilibrium if its
angular acceleration and angular velocity are both
zero
• Sketch the problem
• Show the object of interest along with all the forces
that act on it
• Include a set of coordinate axes
Section 8.3
Problem Solving Strategy, 2
• Identify the relationships
• Find the rotation axis and the pivot point
• Calculate the torque from each force
•
Determine the lever arm for each force
• Calculate the magnitude of the force using τ = F r sin ϕ
• Determine the sign of the torque
•
•
If the force acting alone would produce a counterclockwise
rotation, the torque is positive
If the force acting alone would produce a clockwise rotation,
the torque is negative
• Add the torques from each force to get the total force
•
Be sure to include the sign of each torque
Section 8.3
Problem Solving Strategy, 3
• Solve
• Solve for the unknowns by applying the condition for
rotational equilibrium
•
Στ = 0
• If necessary, apply the condition for translational
equilibrium
•
ΣF = 0
• Check
• Consider what your answer means
• Check that your answer makes sense
Section 8.3
Problem Solving Reminders
• There are some new considerations to remember
• Draw the entire object to show where the forces are
acting on it
•
You can no longer draw the object as a point
• Decide where to put the pivot point
•
•
•
•
There is often a natural choice for the pivot point
Sometimes there can be more than one plausible choice
For an object in rotational equilibrium, any spot may be
chosen to be the pivot point without affecting the final
answer
To simplify the equations, remember that forces whose lever
arms are zero will not contribute to the torque
Section 8.3
Rotational Equilibrium Example: Lever
 Use rotational
equilibrium to find the
force needed to just lift
the rock
 We can assume that
the acceleration and
the angular
acceleration are zero
 Also ignore the mass of
the lever

mlever << mrock
Section 8.3
Lever Example, cont.
• Three forces are present
• The force from the rock on the lever
• The force the person applied to the lever
• The force the of support on the lever
•
Choosing this point to be the pivot point produces a zero
torque from this force
• The force exerted by the person can be less than the
weight of the rock
• If Lperson > Lrock
• The lever will amplify the force exerted by the person
Section 8.3
Amplification of Forces in the Ear
• The incus is supported
by a hinge that acts like
a lever
• The ratio of the lever
arms is about 3
• The lever amplifies the
forces associated with a
sound vibration by the
same factor
Section 8.3
Example: Tipping a Crate
• We can calculate the
force that will just cause
the crate to tip
• When on the verge of
tipping, static
equilibrium applies
• If the person can exert
about half the weight of
the crate, it will tip
Section 8.3
Moment of Inertia
• The moment of inertia of an object composed of
many pieces of mass is
• The moment of inertia of an object depends on its
mass and on how this mass is distributed with
respect to the rotation axis
• The definition can be applied to find the moment of
inertia of various objects for any rotational axis
• SI unit of moment of inertia is kg · m2
Section 8.4
Moment of Inertia, cont.
• The value of I depends
on the choice of rotation
axis
• In the two examples, m
and L are the same
• Their moments of inertia
are different due to the
difference in rotation
axes
Section 8.4
Various Moments of Inertia
Section 8.4
Rotational Dynamics
• Newton’s Second Law for a rotating system states
Στ = Iα
• Once the total torque and moment of inertia are
found, the angular acceleration can be calculated
• Then rotational motion equations can be applied
• For constant angular acceleration:
Section 8.5
Kinematic Relationships
Section 8.5
Example: Real Pulley with Mass
• Up to now, we have
assumed a massless
pulley
• Using rotational dynamics,
we can deal with real
pulleys
• The torque on the pulley
is due to the tension in the
rope
• Apply Newton’s Second
Laws for translational
motion and for rotational
motion
Section 8.5
Problem Solving Strategy – Rotational
Dynamics
• Recognize the principle
• Find the total torque
• Find the moment of inertia
• Use Newton’s Second Law to find the angular
acceleration
• Sketch the problem
• Show all the objects of interest
• Include all the forces that act on the objects
• Include coordinate axes for translational motion
Section 8.5
Problem Solving Strategy – Rotational
Dynamics, 2
• Identify the relationships
• Determine the rotation axis and the pivot point for
calculating torques on any object that rotates
• Find the total torque on the objects that are
undergoing rotational motion
•
These torques will be used in Newton’s Second Law: Στ =
Iα
• Calculate the sum of the forces acting on the objects
that are undergoing linear motion
•
These will be used in Newton’s Second Law: ΣF = ma
• Check for the relationship between linear and
rotational accelerations
Section 8.5
Problem Solving Strategy, 3
• Solve
• Use both forms of Newton’s Second Law to solve for
the unknown(s)
• Check
• Consider what your answer means
• Be sure your answer makes sense
Section 8.5
Example: Motion of a Crate
 The crate undergoes
translational motion
 The pulley undergoes
rotational motion
 For the pulley:
 The tension in the rope
supplies the torque
 The pulley rotates
around its center, so
that is a logical axis of
rotation
Section 8.5
Motion of a Crate Example, cont.
• Pulley equation
• Στ = - T Rpulley = Ipulley α
• The pulley is a disc, so I = ½ mpulley R²pulley
• For the crate
• Take the +y direction as +
• Equation: ΣF = T – mcrate g = mcrate a
• Relating the accelerations
• a = α Rpulley
• Combine the equations and solve
Section 8.5
Combined Motions
• Many common situations include a combination of
rotational and translational motion
• Two examples are
• A rolling wheel
• The motion of a baseball bat
Section 8.6
Rolling Wheel
 Rolling motion combines
translational and rotational
motions
 Assume the center of the
wheel is moving at a
constant linear speed v
 The point on the edge of
the wheel does not move
with constant velocity
 The point of the wheel in
contact with the ground is
at rest during the instant it
is in contact
Rolling Wheel, 2
• The wheel undergoes
rotational motion about
an axis through its
center
• The axle
• The rotational motion is
described by an angular
velocity, ω
• The wheel starts with
point 1 in contact with
the ground
Section 8.6
Rolling Wheel, cont.
• When it completes one
rotation it has traveled a
distance along the
ground equal to the
circumference of the
wheel
•
• Combining gives
v=ωR
Section 8.6
Rolling Wheel, final
• For the accelerations,
a=αR
• Follows the same
argument as for the
velocity
• Friction is essential for
rolling motion
• Usually, the wheels do
not slip so static friction
is involved
Section 8.6
Sweet Spot of a Baseball Bat
• Newton’s Second Law
for the linear motion is
applied to the center of
mass of the bat
• As the batter swings,
the bat undergoes
rotational motion about
the batter’s hands
• We will approximate P
as being fixed
Section 8.6
Sweet Spot of a Baseball Bat
• If the force acts at the
sweet spot, there is no
recoil at your hands
• If the bat was uniform,
the sweet spot would be
L/6
• Bats are not actually
uniform, but the sweet
spot of a real bat can
be found in a similar
way
Section 8.6