AP Physics “C” – Unit 4/Chapter 10 & 11 Notes – Yockers Rotational Motion Angular Position, Speed, and Acceleration - rotational motion – motion of a body that spins about an axis - axis of rotation – the line about which rotation occurs - circular motion – occurs when any single point on an object travels in a circle around an axis

of rotation →r – radius →s – arc length – distance moved along the circumference of a circle

- angles can be measured in radians →radian – an angle whose arc length is equal to its radius

→rad = radian(s); 1 rad = arc length/length of radius when s=r;

3.572

3601

rad

→ the radian is a pure number with no dimensions (ratio of lengths)

→in general, any angle, , measured in radians, is defined by the relation

rsorr

s

→conversions

deg180

rad

- angular displacement, , describes how much an object has rotated →the angle through which a point, line, or body is rotated in a specified direction and

about a specified axis

→ - angular displacement

r

s

r

ss

r

s

r

s ff

f

00

0

- s is positive (+) when rotation is counterclockwise (ccw)

- s is negative (-) when rotation is clockwise (cw)

- angular speed, , describes rate of rotation →the rate at which a body rotates about an axis, usually expressed in rad/s (or s-1

because radians are not dimensional)

→average angular speed = avg

0

0

ttt f

f

avg

unit is rad/s or revolutions/unit time

- instantaneous angular speed

dt

d

tt

0lim

- angular acceleration, , occurs when angular speed changes →the time rate of change of angular speed, expressed in rad/s/s or rad/s2 (or s-2)

→average angular acceleration = avg

0

0

ttt f

f

avg

- instantaneous angular acceleration

dt

d

tt

0lim

- rigid body model - all points on a rotating rigid object have the same angular acceleration and angular speed →if not, the object’s shape would change (it would not be rigid)

- similarities of angular and linear quantities

linear angular

displacement s speed v acceleration a

- direction of and

→the direction of is along the axis of rotation

- if rotation is counterclockwise then the direction of is out of the plane of the diagram

- if rotation is clockwise then the direction of is into the plane of the diagram

→the direction of follows from its vector definition of dt

d

(along the same axis as )

- for rotation about a fixed axis

→the direction of is the same as if the angular speed (the magnitude

of ) is increasing in time

→the direction of is antiparallel to if the angular speed is decreasing in

time Rotational Kinematics: The Rigid Body Under Constant Angular Acceleration - for rotational motion about a fixed axis, the simplest accelerated motion to analyze is motion

under constant angular acceleration

→manipulating the equation for instantaneous angular acceleration results in

tdtd f

tf

0

00

→manipulating the equation for instantaneous angular speed results in

2

21

000

000

ttdttdtd f

ttf

→eliminating results in

tff 021

0

→and eliminating t results in

0

2

0

2 2 ff

Tangential and centripetal acceleration - objects in circular motion have a tangential speed - an object farther from the axis of rotation must travel at a higher tangential speed around the

circular path, s, to travel the same angular displacement as would an object closer to the axis

- tangential speed

rv

dt

dr

dt

dsv

- tangential acceleration is tangent to the circular path →the instantaneous linear acceleration of an object directed along the tangent to the

object’s circular path

ra

dt

dr

dt

dva

t

t

- centripetal acceleration

→remember 0

0

tt

vva

f

f

and that velocity can be changed either by magnitude or

direction; a change in either produces acceleration →for an object moving in a circular path with a constant speed, acceleration is due to a

change in direction →centripetal acceleration (ac) – acceleration directed toward the center of a circular

path - magnitude of ac is given by

r

va t

c

2

- because tangential speed is related to the angular speed by rvt , the

centripetal acceleration can also be found using angular speed 2rac

- http://www.phy.ntnu.edu.tw/oldjava/circularMotion/circular3D_e.html - http://www.walter-fendt.de/ph14e/carousel.htm - tangential and centripetal accelerations are perpendicular →tangential component of acceleration is due to changing speed →centripetal component of acceleration is due to changing direction - total acceleration is found using the Pythagorean theorem

42422222 rrraaa ct

- direction of the acceleration can be found using

t

c

a

a1tan

Moment of Inertia and Rotational Kinetic Energy - if a rigid object is considered to be a collection of particles that rotates about a fixed axis with

an angular speed

→each particle has some kinetic energy determined by its mass and tangential speed →if the mass of the ith particle is mi and its tangential speed is vi, the kinetic energy of

the particle is 2

21

iii vmK

→the total kinetic energy KR of the rotating object is

22

2122

212

21

i

ii

i

iiii

ii

iR rmrmvmKK 2

21 mvK

- the quantity in parentheses above is the moment of inertia, I, of the rigid object

i

iirmI 2

→moment of inertia is the rotational analog of mass →the greater the moment of inertia of a body, the greater its resistance to change in its

angular speed - therefore, the kinetic energy of the rotating rigid object can be expressed as

2

21 IKR

- on the storage side of the continuity equation for energy (Unit 3 Notes), the kinetic energy term should now be considered the sum of the changes in both translational and rotational kinetic energy

- for an extended continuous object, the moment of inertia can be calculated by dividing the object into many small elements

dmrmrIi

imi

22

0lim

- it is usually easier to calculate the moments of inertia in terms of the volume of the elements

rather than their mass (especially if is uniform)

dVrI

V

m

2

- examples (moment of inertia problems)

Torque - (tau) – SI derived unit is N·m - point mass vs. extended object →extended object – has a definite, finite size and shape →point mass – assuming all of an object’s mass is concentrated at its center of mass - rotational and translational motion can be separated →translational motion – motion along a path →rotational motion – the motion of a body that spins about an axis - net torque produces rotation →torque – a quantity that measures the ability of a force to rotate an object about some

axis (torque measures the effectiveness of a given force in producing rotation of a body about a specified axis)

→torque is not work: work requires motion and torque does not; work is a dot product (the product of two vectors results in a scalar quantity) whereas torque is a cross product (the product of two vectors results in a third vector quantity)

- torque depends on a force and a lever arm →how easily an object rotates depends not only on how much force is applied but also

on where the force is applied →lever arm – the ┴ distance from the axis of rotation to a line drawn along the direction

of the force - torque also depends on the angle between a force and a lever arm →forces do not have to be ┴ to cause rotation

dF

rF

)(sin

- two equal but opposite forces can produce a rotational acceleration if they do not act along

the same line - if torques are equal and opposite, there will be no rotational acceleration →seesaw – momentary torque produced by pushing with legs Rotational Equilibrium - equilibrium requires zero net force and zero net torque →if the net force on an object is zero, the object is in translational equilibrium →if the net torque on an object is zero, the object is in rotational equilibrium - the dependence of equilibrium on the absence of net torque is called the second condition of

equilibrium →the resultant torque acting on an object in rotational equilibrium is independent of

where the axis is placed - an unknown force that acts along a line passing through this axis of rotation will

not produce any torque - beginning a diagram by arbitrarily setting an axis where a force acts can

eliminate an unknown in a problem - conditions for equilibrium

Type of Equation Symbolic Eq. Meaning

translational Fnet=0 Fnet on an object must be zero

rotational net=0 net on an object must be zero

- unstable equilibrium – center of mass is above support point - stable equilibrium – center of mass is below support point The Rigid Object Under a Net Torque - only the tangential component of an applied force contributes to the torque of a rotating rigid

object - the radial component of an applied force provides no torque because its line of action is

through the axis of rotation - the torque on each particle in a rigid rotating object

iiii

tiiitii

tiiti

rm

amrFr

amF

2

- the sum of the torque on all particles in a rigid rotating object

i

ii

i

iii

i

i

rm

rm

2

2

- recognizing moment of inertia in the second equation above

maF

I

- the net torque on a rigid body is proportional to the body’s angular acceleration Work and Energy in Rotational Motion - the work done by external forces on an object will equal the change in rotational kinetic

energy as long as energy is not stored by any other means - work done by a torque

drFsdFdW sin

→the radial component of F does no work because it is perpendicular to the

displacement of the point of application of the force →the above equation can be written as

ddW this is analogous to the work done in translational motion (the product

of force and translational displacement) →combining this result with the rotational form of Newton’s second law allows us to

express the torque as

d

dI

dt

d

d

dI

dt

dII which leads to dIdWd

→integrating the above gives the total work done by the torque

Rf KIIdIdWf f

2

0212

21

0 0

- if a system consists of components that are both translating and rotating, the work-kinetic

energy theorem generalizes to RKKW

- the rate at which work is being done by F on an object rotating about a fixed axis (instantaneous power, P)

FvP

dt

dWP

dt

d

dt

dW

Angular Momentum - for a rotating object (with no translational motion), each particle in the object is moving in a

circular path, so momentum is associated with the motion of each particle - although the object has no linear momentum (why not?), a “quantity of motion” is associated

with its rotation (angular momentum) - the instantaneous angular momentum, L, of the particle relative to the axis of rotation is

defined by the vector product of its instantaneous position vector r and the instantaneous linear momentum p

prL

→SI units of angular momentum are kg·m2/s →the sense of L is governed by the right-hand rule →because vmp

, the magnitude of L is

sinmvrL

- L is zero when r is parallel to p ( = 0° or 180°)

- L is a maximum equal to mvr when r is perpendicular to p ( = 90° or 270°) - the net torque acting on a particle equals the time rate of change of the particle’s angular

Momentum

dt

dprFr

p

dt

dr

dt

dprpr

dt

d

dt

dL

prL

(product rule for differentiation; cross product is not commutative)

→the last term on the right is zero because v=dr/dt is parallel to p, so

dt

dpr

dt

dL substituting from the the box above,

dt

dL which is the rotational analog of

dt

dpF

- the total angular momentum L of a system of particles about some point is defined as the vector sum of the angular momenta of the individual particles

i

in LLLLL 21

- the time rate of change of the total angular momentum of the system equals the vector sum of all torques, including those associated with internal forces between particles and those associated with external forces

- the total angular momentum can vary with time only if there is a net external torque on the system

i

toti

i

iext

dt

dLL

dt

d

dt

dL

- determining the rotational analog of mvp

→each particle of mass mi in a rigid object moves in a circular path of radius ri, with a tangential speed vi

IrmrrmrvmLi

ii

i

iii

i

iii

2

IL

Conservation of Angular Momentum - the total angular momentum of a system is conserved if the net torque acting on the system

is zero

ftotitottot

totext

LLconL

dt

dL

,,

0

→the total energy, linear momentum, and angular momentum of an isolated system are all conserved

→the angular momentum of an isolated system is conserved whether the system is a rigid object or not

Parallel axis theorem - the total kinetic energy of a rolling object of mass M and moment of inertia I is the

combination of the rotational kinetic energy around the center of mass plus the translational kinetic energy of the center of mass

2

CM212

CM21 MvIK

- the parallel axis theorem permits the expression of the equation above in terms of the moment of inertia Ip through any axis parallel to the axis through the center of mass of the object

2

CM MDII p

→D is the distance from the center-of-mass axis to the parallel axis →M is the total mass of the object →so, the moment of inertia around an axis passing through the contact point P between

a rolling object and the surface would be expressed as 2

CM MRII p

→using the equation for total kinetic energy of a rolling object (above) and expressing the center of mass’ translational speed in terms of angular speed gives

2

21

22

CM21

22

212

CM21

2

CM212

CM21

pIK

MRIK

MRIK

MvIK

→thus, the kinetic energy of the rolling object can be considered as equivalent to a purely rotational kinetic energy of the object rotating around its contact point

→depending on the situation, the equation for total kinetic energy of a rolling object can also be written as

2

CM2

CM

21

2

CM21

2

CM

CM21

2

CM212

CM21

vMR

IK

MvR

vIK

MvIK

- examples in book! Kepler’s Laws - First - all planets move in elliptical orbits with the Sun at one of the focal points →any object bound to another by a force that varies as 1/r2 will move in an elliptical

orbit - Second - a line drawn from the sun to any planet sweeps out equal areas in equal time

intervals

→conservation of angular momentum

aapp

aapp

ap

rvrv

rmvrmv

LL

- Third - the square of the orbital period of any planet is proportional to the cube of the average distance from the planet to the Sun

→period of a circular orbit (derive this!)

33

22 4

rKrGM

T S

S

32 rT

→period of a planet’s orbit around the Sun - for elliptical orbits, r is replaced with the length of the semimajor axis, a

33

22 4

aKaGM

T S

S

and 3219

3

2

m/s1097.2 SKa

T which is nearly constant

for all the solar system’s planets! →period of a satellite’s orbit around the Earth

33

22 4

aKaGM

T E

E

and 3219 m/s1097.2 SE KK

- Conservation concepts and orbits →escape velocity (conservation of energy)

fE

Ei

f

E

E

Ei

fgfigi

rRGMv

r

mGM

R

mGMmv

UKUK

1122

2

21

,,

R

GMvescape

2

AP Physics “C” – Unit 4/Chapter 12 Notes – Yockers Oscillatory Motion Motion of a Particle Attached to a Spring

- Hooke’s law

xkFs

→Fs = force of a spring (may be referred to as a “restoring force”) →k = spring constant (reflects the stiffness of the spring) →x = displacement (x = 0 is the equilibrium position) - the direction of the restoring force (Fs) is such that the object is being either pushed or pulled

toward the equilibrium position - what is the description of an object’s motion in a system governed by Hooke’s law?

→Fs →v and p →a

- simple harmonic motion occurs when the net force along the direction of motion is a Hooke’s law type of force – that is, when the net force is proportional to the displacement and in the opposite direction

- related terms

→amplitude (A) ≡ the magnitude of the maximum position of the object relative to its

equilibrium position

→period (T) ≡ the time required for one complete vibration

→frequency (f) ≡ the number of vibrations (oscillations) per second (also the inverse of

T)

- acceleration of an object undergoing SHM as a function of its position

x

m

ka

xkam

since the maximum value of x is ±A, the acceleration ranges over the values –kA/m to +kA/m, so we also know that an object is undergoing SHM if its acceleration is proportional to its displacement and is in the opposite direction to it

- in the absence of friction, the motion will continue forever because the force exerted by the spring is conservative

Mathematical Representation of Simple Harmonic Motion - position, velocity, and acceleration

→remember that 2

2

dt

xd

dt

dva , so

xm

k

dt

xda

2

2

→the ratio m

k is denoted (explained later in notes) with the symbol 2 which results in

2m

k

xdt

xd 2

2

2

→deriving the above from an equation describing the position of a particle in simple

harmonic motion where A, , and are constants of the motion - A (the amplitude of the motion) is the maximum value of the position of

the particle in either the positive or negative direction

- is the angular frequency found by m

k and has units of radians

per second

- is the phase constant (phase angle) and is determined uniquely by the position and velocity of the particle at t = 0. →if the particle is at its maximum position A at t = 0, the phase

constant is = 0

- position

tAtx cos

- velocity

tAtdt

dA

dt

dxv sincos

- acceleration

xtAtdt

dA

dt

xda 22

2

2

cossin

- time

→period (T) of the motion is the time interval required for a particle to go through one full cycle of its motion →the values of x and v at time t equal the values of x and v at time t + T →angular frequency is related to the period by noting that the phase increases

by 2 radians in a time interval of T

2 tTt

fT

12

Tf

22

k

mT

2

2

m

k

Tf

2

11

- maximum values of speed and acceleration

Am

kAv max and A

m

kAa 2

max

- evaluating the constants of the motion

→the graphs above show that the phase of the velocity differs from the phase of the

position by /2 rad or 90° - when x is a maximum or a minimum, the velocity is zero - when x is zero, the speed is a maximum

→the phase of acceleration differs from the phase of position by or 180° - when x is a maximum, a has a maximum value in the opposite direction

→ is determined by k and m, T, or f (see the equations above)

→the constants A and are evaluated from the initial conditions (the state of the oscillator at t=0) using the equations for position, velocity and/or acceleration

- examples Energy Considerations in Simple Harmonic Motion - kinetic energy of a simple harmonic oscillator

tAmtAmmvK 222

212

212

21 sinsin

- potential energy of a simple harmonic oscillator

tkAtAkkxU 22

212

212

21 coscos

- total energy of a simple harmonic oscillator

2

21

222

21

22

2122

21

22

21222

21

cossin

cossin

cossin

kAE

ttkAE

tkAtAm

kmE

tkAtAmE

UKE

→the total energy of an isolated simple harmonic oscillator is a constant of the motion

and proportional to the square of the amplitude - the total energy is equal to the maximum potential energy stored in the spring

when x ± A - at these points, v = 0 and there is no K

→at the equilibrium position, x = 0 and U = 0, so the total energy is all in the form of kinetic energy of the particle

2

212

max21

max kAmvK

→the above results are appropriate for the simplification model in which the spring is considered to be massless

- the conservation of kinetic energy for an isolated system can be used to obtain the velocity

for an arbitrary position x of the particle

2222

2

212

212

21

xAxAm

kv

kAkxmvUKE

The Simple Pendulum - the motion of a pendulum is not SHM even though it is periodic - if the angular displacement from equilibrium is small (less than about 10 degrees or 0.2 rad,

so sin=) then a pendulum follows the general equation form of Hooke’s law, and it is justified to say there is SHM as long as the size of the object is small relative to the length of the string

→oscillation is about the lowest point which is the equilibrium position →the motion occurs in a vertical plane and is driven by the gravitational force (restoring

force) which is always in the opposite direction of displacement →using Newton’s second law to write the equation of motion in the tangential direction

2

2

sindt

sdmmamaF gtt

→because Ls and L is constant, the above equation can be written as

sin2

2

L

a

dt

d g

→for small angles sin , so the equation of motion becomes

L

a

dt

d g

2

2

- from this equation we can conclude that the motion is approximately simple

harmonic for small amplitudes because it contains the expression Lag

(remember Lag2 )

- using the equation for displacement for a particle-spring system can be

written as tcosmax where max is the maximum angular position

and the angular frequency for a pendulum is

L

ag which gives the period as

ga

LT 2

→L = fixed length of the pendulum →ag = the acceleration due to gravity (not necessarily the Earth’s!) →period does not depend upon the mass of the bob →period does not depend upon amplitude - how can a time piece using a pendulum be adjusted if it runs fast/slow? - examples The Physical Pendulum - if a hanging object (that cannot be modeled as a particle) oscillates about a fixed axis that

does not pass through its center of mass, it must be treated as a physical, or compound, pendulum

- for a physical pendulum it is necessary to use a rigid object under a net torque

→the torque about O is provided by the gravitational force, and its magnitude is

magdsin →applying Newton’s second law for rotation I

2

2

sindt

dIdmag

- the negative sign on the left indicates that the torque about O tends to decrease

(the gravitational force produces a restoring force) →using the small angle approximation

I

dma

dt

d g

2

2

- tcosmax where max is the maximum angular position

→with the substitution Idmag2 , the motion of the object is approximately simple

harmonic for small amplitudes because the equation will now take the form of

xdt

xd 2

2

2

→angular frequency and period for a physical pendulum

I

dmag and

dma

IT

g

2

2

Note: d is the distance from the axis of rotation to the center of mass (CM) as shown in the diagram above