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