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RotationChapters 8 and 9
Rotational motion can be most easily approached by considering how it is analogous to linear motion. For each concept in rotational motion, there exists a linear analog. Combined with a few "bridge relationships“ between rotational and linear motion, a table of analogies presents a powerful tool for conceptual understanding.
Axis of Rotation
Rigid Body RotationRIGID BODY is an extended object whose size and shape do NOT change as it moves.
Rotational Kinematics
Chapter 8
Kinematics is a complete description of how motion occurs without consideration of the causes of motion
Some terms to Describe Rotational Motion
Angular Position, q in radiansDirection is + when measured CCW from x-
axis - when measured CW from x-axis
Angular Displacement, D , q change in angular position
0
q(rad) = s/rArc length, s
Since for a full circle, s=2pr, 1 rev = 2p rad = 360o 1 rad =57.3o
ExampleWhich particle has angular position 5p/2?
y
A.
x
y
B.
x
y
C.
x
y
D.
x
A
Some terms to Describe Rotational Motion
Angular Velocity, w in rad/s, is the velocity of rotation. Direction is along axis of rotation
given by the right hand rule (+ is CCW, - is CW)
fTt
22
Angular
velocity
Tangential
velocity
w
wvT
vT
Only for rad
r
v
tr
s T
T
rvT
2
Using Equations as Guide to Thinking
How do the variables relate to each other?
T
2
A BC D If A is 1m from the center and
D is 4 m from the center and all take 10 sec to complete a revolution, what are the velocities at A and D
vA = ______m/s
vD = ______m/s
wA = ______rad/s
wD = ______rad/s
TANGENTIAL ANGULAR Speed Speed
0.63
2.5
0.63
0.63
rv
A bike wheel with diameter 80cm rotates at a rate of 45rpm a)What is the angular velocity of the bike tire (in rad/s)
? b)What is the tangential speed of the tire in m/s? c)At 90rpm, what is the tangential velocity of the tire?
sm
rvT/88.1
)71.4)(4.0(
vTwsradsrad
sx
rev
radxrpm
/71.4/5.160
min1
1
245
2 x 1.88 =3.76m/s
Some terms to Describe Rotational Motion
Angular Acceleration, a in rad/s2, is the rate of change of the angular velocity.Direction is along axis of rotation, parallel or anti-parallel to w
r
a
tr
v
tT
Angularaccelerati
on
Tangentialacceleratio
n
Only for rad
Some terms to Describe Rotational Motion
Rotational variables
Translational or Linear variables
ra
rv
rs
T
T
Only if angle unit
is in radians
wf wi
wiwf
SPEEDING UPw and a are in same
direction
wfwi
wiwf
rotating CCW (+): w +, a +
SLOWING DOWNw and a are in opposite
direction
rotating CW (-): w -, a -
rotating CCW (+): w +, a -
rotating CW (-): w -, a +
12
The Vector Nature of Angular Variables
Right-Hand Rule Grasp the axis of rotation with your right hand, so that your fingers circle the axis in the same sense as the rotation. Your extended thumb points along the axis in the direction of the angular velocity vector.
Angular acceleration arises when the angular velocity changes, and the acceleration vector also points along the axis of rotation. The acceleration vector has the same direction as the change in the angular velocity.
Problem: Assume the particle is speeding up.a) What is the direction of the instantaneous velocity, v?
b) What is the direction of the angular velocity, w?c) What is the direction of the tangential acceleration, aT?
d) What is the direction of the angular acceleration a?
e) What is the direction of the centripetal acceleration, ac?f) What is the direction of the overall acceleration, a, of the
particle?g) What changes if the particle is slowing down?w
v
Tangent to circleOut of page
w
Same direction as v, tangent to circle
aT aC
Same direction as w, out of page
a
Into center of circle
a
aT and a would point anti- parallel to w and a
A jet awaiting clearance for takeoff is momentarily stopped on the runway. As seen from the front of one engine, the fan blades are rotating with an angular velocity of -110 rad/s, where the – sign indicates a CW rotation. As the plane takes off, the angular velocity of the blades reaches -330 rad/s in 14 s. Find the angular acceleration of the blades assuming it to be constant.
-16 rad/s2
Linear and Rotational KinematicsEquations for Constant Angular Acceleration
2.5
.4
.3)(.2
.1
20
2
221
0
0
021
ttt
t
xavv
attvxatvv
vvvtvx
2.5
.4
.3)(.2
.1
20
2
221
0
0
021
Rotational Motion
(a = constant)
Linear Motion(a = constant)
Applying the Equations of Rotational Kinematics
1.Draw a picture to represent the system if necessary.
2. Select a coordinate system to analyze your system with the x-axis along the initial angular position. Decide which direction of rotation will be + and which – (convention is + for CCW and – for CW). Keep directions consistent for the problem.
3. List the variables that are given
4. Write down the kinematic equation that will be used to solve the problem.
5. Isolate the unknown variable.
6. Solve for relevant unknowns by putting in numbers. Keep in mind that you may sometimes have more than one unknown and therefore need the same number of equations as unknowns.
17
tradrev
sradsrev
42
//5.000
Example: The blades of a ceiling fan start from rest and, after two revolutions, have an angular speed of 0.50 rev/s. The angular acceleration of the blades is constant. What is the angular acceleration of the fan?What is the angular speed after eight revolutions?
28
2
20
2
/39.0
)4(20
2
srad
t
radrev
168
?0
8
0
sradsrevsrad /28.6/1/2
)16(20
2
82
20
2
18
sradsradrev
radsrev /8.106/34
217/17
mrrrvT
30.0)8.106(32
The blade of a lawn mower is rotating at an angular speed of 17 rev/s. The tangential speed of the outer edge of the blade is 32 m/s. What is the radius of the blade?
NONuniform Circular MotionMagnitude of the velocity (speed) is NOT constant and the direction of the velocity continuously changes as object moves around in a circle
v1v2
r
w
Dq a
aT = ra
ac
22
rr
va Tc
aT and a due to change of velocity magnitude or speed. They are due to forces acting in the line of motion.
ac is due to change in direction of velocity. It is due centripetal force.
rt
va TT
NONuniform Circular MotionMagnitude of the velocity (speed) is NOT constant and the direction of the velocity continuously changes as object moves around in a circle
v1
v2
r
w
Dq
a
aT = raac
22
rr
vac r
t
vaT
c
T
Tc
a
a
aaa
1
22
tanj
Acceleration and net force vectors do not point to the center in nonuniform circular motion. They have both radial (centripetal) and tangential components.
21
Example: A 220kg speedboat is making a circular turn (radius 32m) around a buoy. During the turn, the engine applies a net tangential force of 550N to the boat. The initial tangential speed of the boat is 5m/s. a) Find aT. b) After 2s into turn, find ac.
st
sradra
sradrv
T
T
0.2
/078.032/5.2/
?
/156.032/5/
2
0
22
rr
vac
v
v0
rDq
ac0
F
2/5.2
220550
sma
a
maF
T
T
TT
ac
aT
aT
ac=3.1 m/s2
Rotational Dynamics
Chapter 9
Dynamics is a complete description of WHY motion occurs
Rotation of a Rigid BodyRIGID BODY is an extended object whose size and shape do NOT change as it moves.
All points in object move along parallel paths. Object can be treated as a point.
All points in object move along different paths around rotation axis. One point – center of mass – follows the path of a point particle. Object must be treated as an EXTENDED object
Rotational DynamicsWhat causes rotational motion to change? What causes angular acceleration?
Linear MotionFORCE, F, causes accelerationF
rFrF sinAngular MotionTORQUE, t, causes angular acceleration
F F
No torque
No torqueF
2F
q
FF=Fsinq
Rotational DynamicsWhat causes rotational motion to change? What causes angular acceleration?
Linear Motion - FORCE, F, causes acceleration
rFrF sin
Angular Motion - TORQUE, t, causes angular acceleration
Distance to axis of rotation
Magnitude of
Force
Angle between r and
F
Units of Torque: N
m
TORQUE is a VECTOR+ when it produces CCW rotation about the axis- when it produces CW rotation about the axis
r
r/22F 2F
F F
F
AB
C D
E
A wheel turns freely on an axle at the center. Which one of the forces shown will provide the largest positive torque? Negative torque?E A
F
r=20cmq
F=Fsinq
F//
Problem: Luis uses a 20cm long wrench to tighten a nut. The wrench handle is tilted 30o above the horizontal and Luis pulls straight down on the end with a force of 100 N. How much torque does Luis exert on the nut?
30o
Nm
rFrF
3.17)60sin100)(2.0(
sin
Static Equilibrium(Translational AND
Rotational)
Torque and Static EquilibriumObject at rest is in static equilibrium
Linear Motion
0F
Rotational Motion 0
Static Equilibrium of a Rigid BodyA rigid object at rest is in static equilibrium – it has
zero translational acceleration and zero angular acceleration.
0
0
x
x
F
F
0The net torque about ANY AXIS is zero if a rigid object is in rotational equilibrium.
Applying Equilibrium Conditions to a Rigid Body
1.Select object.
2. Draw a Free Body Diagram. We are now dealing with extended objects and the position of the forces are important.
3. Choose a convenient x- and y- axis and resolve forces into their x- and y-components
4. Apply Newton’s 2nd Law for equilibrium to each direction: Fx = 0 and Fy = 0 .
5. Select a convenient axis of rotation or pivot point. The rotation axis is arbitrary since object is in equilibrium with respect to ANY axis
6. Identify the point where each force acts and calculate the torque produced about the axis of rotation. Set t = 0.
\
6. Solve equations in 4 and 6 to determine unknowns
32
SEESAW or PLANK PROBLEM: A woman whose weight is 530 N is poised at the right end of a diving board with a length of 3.90 m. The board has negligible weight and is bolted down at the left end, while being supported 1.40 m away by a fulcrum. Find the forces F1 and F2 that the bolt and the fulcrum, respectively, exert on the board
F1
F2
W
1.40m
3.90m
530
0
0
21
21
FF
WFF
Fy
NF
F
mgrrF W
4.946
1325)5.2)(530(4.1
0
0
1
1
11
NF 14762 Choose Axis
(r=0)
Center of GravityThe center of mass of an object or system is the point at which all the WEIGHT can be assumed to reside.
Sometimes the system is an assortment of particles and sometimes it is a solid object.
Mathematically, you can think of the center of mass as a “weighted average”.
CMn
nnCG x
mggmgm
xgmxgmxgmx
...
)(...)()(
21
2211
If gravitational field is uniform throughout object, then CM and CG are the same
Center of Gravity, TORQUE and EquilibriumIn extended objects or a group of objects, force of gravity acts at the CG so the weight of an object can produce a torque
CG plays important role in determining whether an object or group of objects remains in equilibrium. WHEN THE CG is in line with a pivot point, then the object or system is in equilibrium.
STABLE
CG
W
FN
0yF 0
CG
W
FN
0yF 0UNSTABLE
Center of Gravity, TORQUE and EquilibriumIn extended objects, force of gravity acts at the CG so the weight of an object can produce a torque
CG plays important role in determining whether an object or group of objects remains in equilibrium. WHEN THE CG of a system is in line with a pivot point, then the system is in equilibrium.
STABLE
CG
W
FN
0yF 0
CG
W
FN
0yF 0STABLE
Wb
CGCG
CG of the system of boy and seesaw is supported at fulcrum
1. Plank or See Saw problems
2. Shelf or Crane problems: beam sticking out of the wall
3. Ladder Problems
Some Common Equilibrium Problems
15 m
60.0
4.0 m
37
LADDER PROBLEM: An 8.00-m ladder of weight WL = 355 N leans against a smooth vertical wall. The term “smooth” means that the wall can exert only a normal force directed perpendicular to the wall and cannot exert a frictional force parallel to it. A firefighter, whose weight is WF = 875 N, stands 6.30 m from the bottom of the ladder. Assume that the ladder’s weight acts at the ladder’s center and neglect the hose’s weight. Find the forces that the wall and the ground exert on the ladder.
38
Pivotr=0
r F=6
.3m NF
WWF
F
N
FLN
y
1230875355
0
0
2
2
sN
sN
x
fF
fF
F
1
1 0
0
FN1
WFWL
fS
FN2
fsNF
F
rWrWrF
N
N
LLFFN
727
040sin)4)(355(40sin)3.6)(875(50sin8
040sin40sin50sin
0
1
1
11
r L=4.
0m
39
SHELF PROBLEM: A safe of mass 430kg hangs by a rope from the end of a boom that is 3.0 m long. The boom consists of a hinged beam and a horizontal cable that connects the beam to a wall. The beam is uniform and has a mass of 85 kg; the mass of the cable and rope are negligible. What is the tension in the cable?
37o
FHx
FHy rB=1.5m
r=3.0m
WS
WBT1
T1
T2
////
//Pivotr=0
40
SHELF PROBLEM: What is the tension in the cable? Find the vertical and horizontal hinge forces
37o
FHx
FHy rB=1.5m
r=3.0m
WS
WBT1
T1
T2
////
//
Pivotr=0
NTT
TggTrTrWr TTBB
TTWB
61460805.135.100969.997
037sin)3(53sin)430)(3(53sin)85(5.1037sin53sin53sin
00
2
2
2
2211
21
Which of these objects is in static equilibrium? D
A. B.
C. D.
A square piece of plywood on a horizontal tabletop is subjected to the two horizontal forces shown at right. Where should a third force of magnitude 5 N be applied to put the piece of plywood into equilibrium? A
Problem: A cat walks along a uniform plank that is 4.00 m long and has a mass of 7.00 kg. The plank is supported by two sawhorses, one 0.440 m from the left end of the board and the other 1.50 m from its right end. When the cat reaches the right end, the plank just begins to tip. What is the mass of the cat?
Rolling MotionCombination of rotation and translation
vT = wR
vT
vCM
vCM
vCM
+
pure ROTATION
pure TRANSLATION
When an object rolls without slipping, there is a relationship between the rotational and translational motion.
When there is slidding or slipping, there is NO relationship between the rotational and translational motions.
vCM
R
Rolling Motion(without slipping)
rvCM
)/( sradin
Linear speed
Rotational speed
raCM
)/( 2sradin
Linear acceleratio
n
Rotational acceleration
DxCM = s = rqhttp://tube.geogebra.org/student/m23809
Rolling MotionCombination of rotation and translation
Rolling and Slipping
CM Travels more than 1 circumference for every full rotation
No relationship between the translational motion (vCM) and rotational motion (Rw) dCM > Rq vCM > Rw aCM > Ra
Rolling without Slipping
CM Travels 1 circumference for every full rotation
There is a relationship between the translational motion (vCM) and rotational motion (Rw) dCM = Rq vCM = Rw aCM = Ra
ROLLING WITHOUT SLIPPINGVelocity at any point on a rolling object
Velocity at point in
pure ROTATION
(in CM ref frame)
Rw=vT
vT
vCM= vT
vCM
vCM = -vT
2vCM
vCM
0
+ =
Velocity at pointin
pure TRANSLATION
(relative to ground)
Velocity at pointin
COMBINED ROLLING MOTION
(relative to ground)
A wheel rolls without slipping. Which is the correct velocity vector for point P on the wheel?
vCMP
A. B. C. D. E.
C
vCM
vT(In CMframe)
v=2vCM
vT(In CM frame)
v
vCM
49
Example An Accelerating Car: An automobile starts from rest and for 20.0 s has a constant linear acceleration of 0.800 m/s2 to the right. During this period, the tires do not slip. The radius of the tires is 0.330 m. At the end of the 20.0-s interval, what is the angle through which each wheel has rotated?
st
sradra
20?
/42.233.0/8.0/
0
2
0
rad
tt
484)20)(42.2(0 2
21
221
0
Rotational Dynamics and
Newtons 2nd Law
What is the RELATIONSHIP between TORQUE and ANGULAR ACCELERATION?
r
w
F FT
Fr
22
mrr
vmmaFF T
ccr
FT causes tangential acceleration; it causes a change in the speed or magnitude of the velocity.
Fr changes only the direction of velocity. It does not affect the speed; it does no work. It produces no torque.
tT maF
NONuniform Circular Motion and Newtons 2nd Law
m
r
w
F FT
Fr
2
90sinmr
rmarF
maF
TT
TT
NONuniform Circular Motion and Newtons 2nd Law
This is the relationship between torque and angular acceleration for a single particle moving in circular motion. To look at rotational dynamics, must expand idea to an extended rigid object.
(Only if angle unit is
radians)
w
2
2222
211
321
.....
...
ii
nn
n
rmrmrmrm
Rotational Motion and Newtons 2nd Law
m1
r1
m2r2
m3
r3
F2
F1
F3
t1 = r1FT1 = r1m1aT1 = m1r12a
t2 = r2FT2 = r2m2aT2 = m2r22a
t3 = r3FT3 = r3m3aT3 = m3r32a
If we did this for ALL particles in the object, the net torque on the entire object would be
ALL pts on rigid body rotate at same
a
Rotational Inertia, I
w 2
iirm
Rotational Motion and Newtons 2nd Law
m1
r1
m2r2
m3
r3
F2
F1
F3
ALL pts on rigid body rotate at same
a
Rotational Inertia, I
I (angle unit must be in radians)
Newtons 2nd Law for Rotation of a Rigid Body
Rotational equivalent of
FORCE
Rotational equivalent of MASS
2iirmI
Rotational Inertia (Moment of Inertia)
Moment of inertia is the rotational equivalent of mass.Rotational Inertia of an object depends on the mass AND on how the mass is distributed about an axis of rotation. It is constant for a particular rigid body and a particular axis.
ri is the perpendicular distance between the mass and the rotation axis.
(Units are kg m2)
Both involve same mass.Left is easier to rotate since the mass is distributed closer to the rotation axis.
2iirmI
Rotational Inertia (Moment of Inertia)ri is the perpendicular distance between the mass and the rotation axis.
(Units are kg m2)
I depends on where the axis of rotations
isBoth involve same mass.Top is easier to rotate since the mass is distributed closer to the rotation axis.
Which rod has a greater Moment of Inertia (is harder to rotate)?
Larger I (harder to rotate)
Smaller I (easier to rotate)
Problem: Two 3kg masses are at the ends of a 4m bar with negligible mass. What is the moment of inertia of the object when
2
22
222
211
2
24)2(3)2(3
kgm
rmrmrmI ii
a) when the masses rotate around the center of the rod
b) when the masses rotate around one end of the rod
2
2
222
211
2
48)4(3)0(3
kgm
rmrmrmI ii
3kg
w
4m
3kg3kg
w
2m3kg
2m
22111 36kgmrmI
Rotational Inertia (Moment of Inertia)Three small spheres that rotate about a vertical axis are shown.
The perpendicular distance between the axis of rotation and center of mass of each sphere is given. Rank the 3 spheres from greatest to least rotational inertia.
1 m
2 m
3 m
wm1=36kg
m2=9kg
m3=4kg
2
22222
36)2)(9(
kgmrmI
2
22333
36)3)(4(
kgmrmI
2iirmI
Rotational Inertia (Moment of Inertia)
For a solid object
ri is the perpendicular distance between the mass and the rotation axis.
(Units are kg m2)
dmrI 2
Moments of Inertia of Common Shapes
Linear Dynamics Rotational Dynamics
Net Force (N):
Fnet
Net Torque (N m):
tnet
Mass (kg): mMoment of Inertia (kg m2):
I
Acceleration (m/s2): aAngularAcceleration (rad/s): a
Newtons 2nd Law
Newtons 2nd Law
m
Fa net
Inet
Acceleration is caused by forces.
The larger the mass, the smaller the acceleration.
Angular acceleration is caused by torques.
The larger the moment of inertia, the smaller the angular acceleration.
Using Newton’s 2nd Law for RotationIn the Scottish game of caber toss, contestants toss a heavy uniform pole, landing it on its end. A 5.9m pole with a mass of 79kg has just landed on its end. It is tipped 25o from the vertical and is starting to rotate about the end that touches the ground. A) Find the angular acceleration. B) What is the tangential acceleration of the left end of the log?
FgFN
Pivotr=02
231
21
231
/1.12
25sin325sin)(
sradL
gMLMgL
ML
I
W
231MLI
25o
gg
LraT 63.02
25sin3
Ropes and PulleysLinear Motion Angular
MotionFrictionless, massless pulley
Pulley does NOT rotate
Rope slides over frictionless pulley
Rope tension does NOT change when it passes over a frictionless, massless pulley
Pulley has mass and friction
Pulley rotates
If pulley TURNS WITHOUT THE ROPE SLIPPING ON IT, then vrope=vT pulley=Rw arope = aTpulley=Ra
Rope tension CHANGES when it passes over a rotating pulley with friction.
T1/T1 / T3
T2
w
R
a Pulley RotationAt rim, vT = Rw aT = Ra
Rope Translationvrope = Rwarope = Ra
Ropes and Pulleys
Motion constraints for an object connected to a pulley of radius R by a nonslipping rope
T3T2
w
R
a
vobject = Rw
aobject = Ra
Note that these are magnitudes only
maFmg T
A light string is wrapped around a uniform pulley (disk) of mass M=2.5kg and radius R=20cm. Suspended from the end of the string is a block of mass of m=1.2kg. Find the acceleration of the falling block, the angular acceleration of the pulley and the tension in the string. The string does not slip and there is no friction in the axle. If the block starts at rest, how many revolutions does the pulley in 2s?
MR
m
mg
FT
FT
Mg
F1
//
////
MaF
R
aMRRF
MR
I
T
TT
FT
21
221
221
Block Pulley
2
21
21
/8.42
2
)(
smMm
mga
mgMma
maMamg
a+
a+t
A light string is wrapped around a uniform pulley (disk) of mass M=2.5kg and radius R=20cm. Suspended from the end of the string is a block of mass of m=1.2kg. Find the acceleration of the falling block, the angular acceleration of the pulley and the tension in the string. The string does not slip and there is no friction in the axle. If the block starts at rest, how many revolutions does the pulley in 2s?
MR
m
mg
FT
FT
Mg
F1
//
//// Pulley
2/24
2.0
8.4
)2(
2
srad
MmR
mgRa
NMm
mMgMm
mgMMaFT
22.12
2
2
2
121
A light string is wrapped around a uniform pulley (disk) of mass M=2.5kg and radius R=20cm. Suspended from the end of the string is a block of mass of m=1.2kg. Find the acceleration of the falling block, the angular acceleration of the pulley and the tension in the string. The string does not slip and there is no friction in the axle. If the block starts at rest, how many revolutions does the pulley in 2s?
MR
m
mg
FT
FT
Mg
F1
//
//// Pulley
2/242.0
8.4
)2(
2
sradMmR
mgRaT
NMm
mMgMm
mgMMaFT
22.12
2
2
2
121
Rotational Work and Energy
WorkWhen a force acts on an object over a distance, it is said that work was done upon the object. Work tells us how much a force transfers energy to a system.
sF
FsW
//
cos
Work done by a constant force that points in the same direction as the displacement
FrWFsW
R
Constant force pulls rope out a distance, s
Wheel rotates through the angle q = s/r
Rotational Work
Unit: J must be in rad
r
FT
sr
FT
q
s
Work done by a force
Work done by a torque
w
2
21
2221
222122
222122
1121
321
.....
...
IK
rmKrmrmrm
KKKKK
R
iiR
nn
nRRRRR
Rotational Kinetic Energy
m1
r1
m2r2
m3
r3
K1 = ½m1v1T2 = ½m1r1
2w2
K2 = ½m2v2T2 = ½m2r2
2w2
K3 = ½m3v3T2 = ½m3r3
2w2
If we did this for ALL particles in the object, the total rotational kinetic energy would be
ALL pts on rigid body rotate at same
w
Rotational Inertia, I
vT2
vT1
vT3
Kinetic Energy
(angle unit must be in radians)Rotational
Kinetic Energy
Rotational equivalent of MASS
221 IKR
221 mvKT
Translational Kinetic Energy
Rotational Kinetic Energy
Tnet KW RnetR KW
PowerTranslational
Rotational
vFt
sF
t
W
t
EP //
//
tt
W
t
EP RR
A solid disk and a hoop with mass M and radius R start from rest and roll without slipping down an incline. What is the velocity at the bottom in terms of h?
fNCi EWE
s
q
w Disk
Hoop
Sphere
SphericalShell
221 mrI
2mrI
232 mrI
252 mrI
h
ghv
MvMvMgh
MRMvMgh
IMvMgh
UKKUKKEE
B
BB
Rv
B
BB
gBRBTBgTRTTT
BT
B
34
2412
21
2221
212
21
2212
21
))((
0
Disk
Disk Hoop ghvB
Roll without slipvCM=Rw
System is isolated, WNC=0 since FN and fS do no work on object
Find the acceleration of an object rolling down an inclined plane. The object rolls without slipping. It has a mass of M and radius R.
MafMgMaF
s
x
sin
FN
Fg
Fg sinq
Fgcosq
fS
//
//
q
q
2R
Iaf
IIRfI
s
Ra
s
+ ,a t Torque around Rotation axis at CG
22
2
2
1
sinsin
sin)(
sin
MRI
RI
RI
RIa
g
M
Mga
MgMa
MaMg
Disk
Hoop
Sphere
SphericalShell
221 mrI
2mrI
232 mrI
252 mrI
+a
Which wins the RACE? 4 different spherical objects with the same mass and radius are released from rest and roll down an inclined plane. The objects roll without slipping. Rank the objects according to the order that they reach the finish line
AB
CD
Spherical Shell
DiskHoop
Sphere
C < A < D < B
Disk
Hoop
Sphere
SphericalShell
221 mrI
2mrI
232 mrI
252 mrI
http://tube.geogebra.org/student/m837651
NONuniform Circular MotionA single particle moving in circular motion has rotational kinetic energy and angular momentum and can experience a torque if a force acts tangentially.
(angle unit is radians)
r
w
F
FT
Fr
vT
Moment of inertia of single particle moving in a circle of radius r is mr2
(all mass is traveling at r)
Imr
rmarFrF TT
2
sin
22122
212
21 ImrmvK TR
Angular Momentum
Momentum
(angle unit must be in radians)
IL mvp Linear momentum
Angular Momentum
t
pFnet
t
Lnet
If there are no external forces on a system of particles, momentum of the system is conserved
If there are no external torques on a system of particles, angular momentum of the system is conserved
Conservation of Angular Momentum
Conservation of Angular Momentum
Problem: Suppose an empty “carousel" having a radius, R of 1.8 m and a mass, M of 50 kg (I = ½mr2) is initially rotating on a frictionless axis counterclockwise at 15 rev/min.
Simultaneously, four students walk up and place four 5-kg boxes (I = mr2) symmetrically along its outer edge so that each box's center of mass is located 1.50 m from the axis of the carousel. What will become the carousel's new angular velocity at the instant the boxes are in place?
before after
rpmrpm
IIILL
BCCC
64.9'))25.11(481(')15)(81(
'4''
22
2221
25.1181
mkgmrImkgMRI
B
C
Problem: Suppose an empty “carousel" having a radius, R of 1.8 m and a mass, M of 50 kg (I = ½mr2) is initially rotating on a frictionless axis counterclockwise at 15 rev/min.
Suppose instead that they had placed the boxes so that each box's center of mass was located 50 cm from the axis of the merry-go-round, what would have been its new angular velocity?
before after
rpmrpm
IIILL
BCCC
1.14'))25.1(481(')15)(81(
'4''
22
2221
25.181
mkgmrImkgMRI
B
C
Linear and angular momentum are SEPARATELY CONSERVED
One form of momentum does NOT transform into the other form
w
p=mv
L = 0 p = 0L ≠ 0
Linear momentum of the system is NOT conserved since there is an external force on the system (friction between turntable and surface). Linear momentum gets transferred to the EarthAngular momentum of the system IS conserved since there are no external torques on the system (the axis provides no torque). The angular momentum of the bullet transfers to the turntable.
r
v
Angular Momentum of a Point ParticleIn general, a single particle moving in any way has angular momentum
relative to an axis point, O
q
r
O 2
mrIL
sinrpmvrL
Radius is closest distance to rotation
axis, r
Ballistic Pendulum
Linear momentum of the system is NOT conserved since the pendulum is fixed to the earth at the axis of rotation and this provides and external force. Linear momentum gets transferred to the Earth
Angular momentum of the system IS conserved since there are no external torques on the system (the axis provides no torque). The angular momentum of the ball gets transferred to the pendulum.
w’
Ballistic Pendulum
ddM
dm, v
w’
dMm
mv
Mdmdmvd
IIIImvd
LLLL
LL
RbRb
RbRb
)('
)('
)('''
''
'
31
2312
Angular Momentum in Astronomy
L=mvR
L=mvR
The Angular momentum of the orbiting object is the same at every point on the orbit since the force of gravity produces no external torque. Therefore the orbiting object travels faster when it is closer to the rotation axis and slower when it is further away. This is Kepler’s 3rd Law
Linear Dynamics
Rotational Dynamics
Displacement s, d qVelocity v wAcceleration a aCause of acceleration
Fnet, net force tnet, net torque
Inertia m, Mass I, Moment of Inertia
Newton’s 2nd Law F = ma = t Ia
Work W = F//s WR = tq
Kinetic Energy K = ½mv2 K R= ½Iw2
Momentum p = mv L = Iw
A baseball catcher sits on a rotating stool. He reaches out 85.0cm tocatch a 40.0m/s fastball. After catching the ball he spins at a rate of60.0rpm. His mass is 80.0kg and the mass of the ball is 150g. Find therotational inertia of the catcher and the stool.
85 cm
0.81 kg m2
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