Download torque and circular motion - PHYSICS I PRE-AP

TORQUE AND CIRCULAR
MOTION
CHAPTER 8
TORQUE
A PHYSICAL QUANTITY RESPONSIBLE
TO CREATE ANGULAR ACCELERATION
FO A BODY.
IT STARTS, STOPS, OR CHANGES THE
DIRECTION OF THE MOTION.
IT IS PRODUCED BY A FORCE APPLIED
ON A BODY.
Ł= rFsinθ
TANGENTIAL AND RADIAL
COMPONENTS
A FORCE HAS A TANGENTIAL AND A
RADIAL COMPONENT
THE RADIAL COMPONENT GOES
THROUGH THE AXIS OF ROTATION
AND SO DOES NOT CAUSE ROTATION
THE TANGENTIAL COMPONENT IS THE
ONE THAT CAUSES ROTATION
DIAGRAM OF TORQUE
PHI IS THE ANGLE F MAKES THE ANGLE OF ROTATION
τ=rFsin Φ
F
Φ
MOMENT ARM – PERPENDICULAR
DISTANCE OF THE LINE OF ACTION OF
THE FORCE FROM THE AXIS TO
ROTATION. (MOMENT ARM IS ALSO
CALLED LEVER ARM)
FR
FT
P
O
MOMENT
ARM
τ=Fperpendicular
OR FIND THE COMPONENT
OF THE FORCE CAUSING
THE TORQUE
TORQUE
F – FORCE ACTING ON THE BODY.
R – LENGTH OF THE POSITION
VECTOR FROM 0 TO ANY POINT ON
THE FORCE.
Θ – THE ANGLE BETWEEN r AND F
F
θ
r
Ł= rFsinθ
RESOLVE THE FORCE INTO X & Y
WHEN TORQUE IS APPLIED, ROTATION
OCCURS AROUND PIVOT POINT OR
FULCRUM
F
F sin θ
θ
F cos θ
PIVOT POINT(FULCRUM)
DIRECTION OF ROTATION
ROTATION IS TAKEN IN A SINGLE
PLANE
ROTATION CAN BE CLOCKWISE
ROTATION CAN BE
COUNTERCLOCKWISE
SIGN CONVENTION FOR TORQUE
TORQUE –
CLOCKWISE IS NEGATIVE
COUNTERCLOCKWISE IS POSITIVE
NET TORQUE
 THE SUM OF ALL TORQUES ACTING ON THE
BODY. THE SUM IS CARRIED OUT TAKING
INTO ACCOUNT THE SIGN CONVENTION.
 WHEN MORE THAN ONE TORQUE ACTS ON
A BODY, THE ACCELERATION PRODUCED IS
PROPORTIONAL TO THE NET TORQUE.
CENTER OF GRAVITY
THE POINT AT WHICH THE ENTIRE
WEIGHT OF AN OBJECT SEEMS TO
ACT.
SEE-SAW TORQUE
UNIFORM
IF AN OBJECT IS UNIFORM ITS
CENTER OF GRAVITY IS AT ITS
GEOMETRIC CENTER.
CENTER OF MASS
THE POINT WHERE ALL THE MASS
SEEMS TO ACT.
AN OBJECT IS IN EQUILIBRIUM AS
LONG AS THE CM STAYS OVER ITS
BASE LEVEL
AN OBJECT IS CONSIDERED UNIFORM
WHEN THE CM IS ITS GEOMETRIC
CENTER
XCM=1/M Σ mi xi
M = TOTAL MASS
Mi = MASS OF PARTCLE
Xi = DISTANCE FROM THE ORIGIN
MECHANICAL EQUILIBRIUM
ANY MOTION OF A RIGID BODY IS A
CONBINATION OF ROTATIONAL AND
TRANSLATIONAL MOTION
IT IS SAID TO BE IN MECHANICAL
EQUILIBRIUM IF THE TRANSLATION
AND ROTATIONAL MOTION ARE IN
EQUILIBRIUM
CONDITIONS
ROTATIONAL EQUILIBRIUM
Σ τ(ABOUT ANY POINT) = 0
(ANGULAR MOMENTUM IS CONSTANT)
TRANLATIONAL EQUILIBRIUM
ΣFx = 0 AND ΣFy = O AND ΣFz = O
ELASTICITY
THE BRANCH OF PHYSICS WHICH
DEALS WITH HOW OBJECTS DEORM
WHEN FORCES ARE APPLIED TO
THEM.
ELASTIC LIMIT- WHEN PERMANENT
DEFORMATION OCCURS AND THE
OBJECT WILL NOT RETURN TO ITS
ORIGINAL SHAPE.
THREE WAYS AN OBJECT CHANGES
ITS SHAPE
SHEARING FORCES – CAUSE
MOVEMENT SIMILAR TO THE PAGES
OF A BOOK. EX: EARTH LAYERS
DURING A QUAKE.
STRETCHING OR COMPRESSING
FORCES. EX: BALLOON, STRING
BULK FORCES – APPLIES TO FLUIDS.
PRESSURE FROM ALL SIDES CAN
CAUSE A VOLUME CHANGE.
STRESS AND STRAIN
STRESS(CAUSED BY FORCES)
PRODUCE STRAIN.
STRESS IS PROPORTIONAL TO
STRAIN
THAT PROPORTIONALITY CONSTANT
IS CALLED THE MODULUS(E)
(YOUNG’S MODULUS)
EQUATIONS
STRETCHING
STRESS = FORCE/AREA OVER WHICH
THE FORCE IS APPLIED
STRAIN = ratio of the change.
The ratio could be length, volume, height
STRAIN = ∆L/L OR ∆H/H OR ∆V/V
STRESS = E x STRAIN
EQUATIONS
 STRETCHING FORCES
 F/A=E∆L/L
 SHEARING FORCES
 F/A=G∆H/H
G = SHEARING MODULUS
 BULK FORCES
 F/A=B∆V/V
B = BULK MODULUS
REMEMBER: PRESSURE = F/A
UNIT FOR THE MODULUS IS PASCALS
ROTATIONAL MOTION
PURE ROTATIONAL MOTION IS
MOTION WHERE ALL POINTS OF THE
OBJECT MOVE IN A CIRCULAR PATH
AROUND AN AXIS OF ROTATION.
WE ARE MOST CONCERNED WITH
RIGID BODIES – BODIES THAT DO NOT
DEFORM WITH MOTION.
TWO TYPES OF VELOCITY –RIGID
OBJECTS
LINEAR VELOCITY – GREATER THE
FURTHER FROM THE AXIS OF
ROTATION.
M/S
ANGULAR VELOCITY IS THE SAME
FOR EVERY POINT IN THE ROTATING
BODY AT ANY GIVEN INSTANT.
RAD/S OR REV/S
PICTURE THIS
AN OBJECT ROTATES FROM POINT A
TO B IN THE SAME ANGLE AS FROM
POINT a TO b.
A
BOTH TRAVEL
THE SAME
ANGULAR
DISTANCE – θ
a
θ
B
b
TANGNTIAL DISTANCE
THEY TRAVELED DIFFERENT
TANGENTIAL DISTANCES IN THE SAME
TIME PERIOD.
ARC LENGTH AB > ARC LENGTH ab
A
a
θ
B
b
ANGULAR DISPLACEMENT
THE ANGLE THROUGH WHICH AN
OBJECT TURNS AROUND AN AXIS OF
ROTATION.
MEASURED IN DEGREES, RADIANS,
OR REVOLUTIONS
1 REV = 360° = 2π RADIANS
Θ = D/R
D – TANGENTIAL DISTANCE
R - RADIUS
CONVERSION BETWEEN METERS AND
RADIANS (TRANSLATIONAL TO
ROTATIONAL)
DISTANCE IN METERS = RADIANS
RADIUS IN METERS
A RADIAN IS A UNIT THAT SERVES AS A
PLACE HOLDER
ANGULAR POSITION
GIVEN AS THE OBJECT HAS ROTAED
THROUGH SOME ANGLE WHEN IT
TRAVELS THE DISTANCE “L”
MEASURED ALONG THE
CIRCUMFERENCE OF ITS CIRCULAR
PATH.
WHAT IS A RADIAN
ONE RADIAN (RAD) IS THE ANGLE
SUBTENDED BY AN ARC WHOSE
LENGTH IS EQUAL TO THE RADIUS.
IN OTHER WORDS, IF L=R, THEN θ IS
EXACTLY EQUAL TO ONE RADIAN
ANGULAR SPEED – VELOCITY ώ
THE RATE AT WHICH AN OBJECT
ROTATES
RADIANS/SECOND
ω = θ/t
ω = V/R (V IS TANGENTIAL V)
ω = ∆θ/t
ANGULAR SPEED AND FREQUENCY
FREQUENCY IS THE NUMBER OF
COPELTE REVOLUTIONS PER SECOND
1 REV/S = 2πRADIANS/S
f = ω / 2π

ANGULAR ACCELERATION (α)
 THE RATE AT WHICH A ROTATING OBJECT
CHANGES ANGULAR SPEED
 RADIANS/S2
 α = (ωf - ωi)/T
 α = A/R
 A IS TANGENTIAL ACCELERATION
 R = RADIUS
 AVERAGE ANGULAR ACCELERATION
 α = ∆ω/t
TOTAL LINEAR ACCELERATION
THE VECTOR SUM OF THE
CENTRIPETAL ACCELERATION (THE
RADIAL COMPONENT OF THE ACCELERATION)
aR = ω2r
AND THE TANGENTIAL COMPONENT
OF THE ACCLERATION.
DIAGRAM
BECAUSE OF THIS THE CENTRIPETAL
ACCELERATION INCREASES THE
FARTHER YOU ARE FROM THE AXIS
OF ROTATION.
a tan
aR
EQUATION COMPARISON
 EQUATIONS FOR LINEARMOTION CAN
BE TRANFORMED INTO ROTATIONAL
FORMS
D=V/T
D=Do+Vi + ½ at2
Vf=Vi + at
Vf2 = Vi2 + 2ad
θ = ωt
θ = θo + ωi t + ½ αt2
ωf = ωi+ αT
ωf 2= ωi2+ 2αθ
NEWTON’S 2ND LAW IN ROTATIONAL
FORM
F = MA …… Ł = Iα
Ł - TORQUE
I – MOMENT OF INERTIA
α – ANGULAR ACCELERATION
ROLLING WHEEL
TANGENTIAL VELOCITY OF THE
WHEEL AT THE AXEL = V
TANGENTIAL VELOCITY OF THE
WHEEL AT THE EDGE = 2V
MOMENT OF INERTIA, I
 ANALAGOUS TO MASS
 DEPENDS UPON THE ROTATING BODY’S
MASS
 AND THE DISTRIBUTION OF THE MASS
 I = ΣM1R12 + M2R22 + M3R32 ….
HOW UNITS COMPARE
VARIABLE
TRANSLATIONAL
ROTATIONAL
DISTANCE
d – METERS
Θ - RADIANS
VELOCITY
V – M/S
ω – RADIANS/S
ACCLERATION
a – M/S2
Α – RADIANS/S2
FORCE
F – NEWTONS
Ł–Nm
MASS
m – KILOGRAMS
I – kg m2
MOMENT OF INERTIA
 A FORCE IS REQUIRED TO START AN
OBJECT ROTATING ABOUT AN AXIS.
 TORQUE PRODUCES ANGULAR
ACCELERATION.
 TORQUE IS REQUIRED TO START THE
ROTATION
 THE MOMENT OF INERTIA TELL HOW THE
MASS OF THE BODY IS DISTRIBUTED ABOUT
THE AXIS.
Ł = Iα
ROTATIONAL MOMENT OF INERTIA
FOR BODIES
THIN HOOP OF RADIUS R
I = mr2
SOLID DISK OF RADIUS R
I = 1/2 mr2
UNIFORM SPHERE OF RADIUS R
I = 2/5 mr2
LONG UNIFORM ROD OF LENGTH L WITH ITS I = 1/12 mr2
AXIS OF ROTATION THROUGH ITS CENTER
LONG UNIFORM ROD OF LENGTH L WITH ITS I = 1/3 mr2
AXIS OF ROTATION THROUGH ONE END
ROTATIONAL K.E.
ROTATION AROUND THE CM
MEASURED IN JOULES
KE = ½ Iω2
ROTATION OF OBJECT NOT AROUND
THE CENTER OF MASS HAS
ROTATIONAL AND TRANSLATIONAL
K.E.
KE = ½ mv2 + ½ Iω2
SLIDING VS ROLLING
OBJECTS THAT SLIDE CONVERT ALL
OF THEIR POTENTIAL ENERGY INTO
TRANSLATIONAL KINETIC ENERGY.
OBJECTS THAT ROLL CONVERT MOST
OF THEIR POTENTIAL ENERGY INTO
ROTATIONAL K.E. AND THE REST INTO
TRANSLATIONAL K.E. SO IT WILL NOT
MOVE AS QUICKLY AS A SLIDING
OBJECT.
ANGULAR MOMENTUM
p=mv – LINEAR
L = Iω – ANGULAR
ANGULAR MOMENTUM REMAINS
CONSTANT IF NO EXTERNAL
TORQUES ARE ACTING
IF TORQUE NET IS ZERO AND
ROTATION IS AROUND A FIXED AXIS
THE MOMENTUM IS CONSERVED
Iω = CONSTANT
EXAMPLE
ICE SKATER SPINNING ON ICE –
EXTENDED ARMS INCREASE I AND
GIVE A SMALL ω
AS SHE BRINGS HER ARMS IN THE I
DECREASES SO THE ω INCREASES
BECAUSE ANGULAR MOMENTUM IS
CONSERVED.