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Nicholas J. Giordano
www.cengage.com/physics/giordano
Nicholas J. 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
Section 8.4
Various Moments of Inertia
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
Section 8.5
Kinematic Relationships
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 = ω R Section 8.6
Rolling Wheel, final
• For the accelerations, a = α R
• 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