ANGULAR MOMENTUM Kopal yadav

1. Present To
Ghyan Rao Dhote

2. I. Rolling
- Kinetic energy
- Forces
II. Torque
III. Angular momentum
- Definition
IV. Newton’s second law in angular form
V. Angular momentum
- System of particles
- Rigid body
- Conservation

3. I. RollingI. Rolling
- Rotation + Translation combinedRotation + Translation combined.
COMvRR
dt
d
dt
ds
Rs =⋅==→⋅= ω
θ
θ
Smooth rolling motionSmooth rolling motion
Example:Example: bicycle’s wheel.bicycle’s wheel.
The motion of any round body rolling smoothly over a surface can beThe motion of any round body rolling smoothly over a surface can be
separated into purely rotational and purely translational motions.separated into purely rotational and purely translational motions.

4. - Pure rotation.Pure rotation.
Rotation axisRotation axis  through point where wheel contacts ground.through point where wheel contacts ground.
Angular speed about P (Angular speed about P (ωω)) = Angular speed (= Angular speed (ωω))about O for stationary observer.about O for stationary observer.
COMtop vRRv 2)(2)2)(( === ωω
- Kinetic energy of rolling.- Kinetic energy of rolling.
222222
2
1
2
1
2
1
2
1
2
1
COMCOMCOMp MvIMRIIK +=+== ωωωω
Instantaneous velocity vectorsInstantaneous velocity vectors = sum of translational= sum of translational
and rotational motions.and rotational motions.
2
MRII COMp +=
A rolling object has two types of kinetic energyA rolling object has two types of kinetic energy  Rotational:Rotational:
(about its COM).(about its COM).
(translation of its COM).(translation of its COM). Translational:Translational:
2
2
1
ωCOMI
2
2
1
COMMv

5. - Forces of rolling.- Forces of rolling.
(a)(a) Rolling at constant speedRolling at constant speed  no sliding at Pno sliding at P
 no friction.no friction.
(b) Rolling with acceleration(b) Rolling with acceleration  sliding at Psliding at P 
friction force opposed to sliding.friction force opposed to sliding.
Static frictionStatic friction  wheel does not slidewheel does not slide  smoothsmooth
rolling motionrolling motion  aaCOMCOM == α Rα R SlidingSliding
Increasing accelerationIncreasing acceleration
ExampleExample11:: wheels of a car moving forward while its tires are spinningwheels of a car moving forward while its tires are spinning
madly, leaving behind black stripes on the roadmadly, leaving behind black stripes on the road  rolling with slipping =rolling with slipping =
skiddingskidding Icy pavements.Icy pavements.
Anti-block braking systems are designed to ensure that tires roll withoutAnti-block braking systems are designed to ensure that tires roll without
slipping during braking.slipping during braking.

6. ExampleExample22:: ball rolling smoothly down a ramp. (No slipping).ball rolling smoothly down a ramp. (No slipping).
1.1. Frictional force causes the rotation. WithoutFrictional force causes the rotation. Without
friction the ball will not roll down the ramp,friction the ball will not roll down the ramp,
will just slide.will just slide.
Sliding
tendency
2.2. Rolling without slidingRolling without sliding  the point of contactthe point of contact
between the sphere and the surface is at restbetween the sphere and the surface is at rest
 the frictional force is the static frictional force.the frictional force is the static frictional force.
3. Work done by frictional force = 0Work done by frictional force = 0  the pointthe point
of contact is at rest (static friction).of contact is at rest (static friction).

7. Example:Example: ball rolling smoothly down a ramp.ball rolling smoothly down a ramp.
)1(sin ,, xCOMsxxnet MaMgfmaF =−→= θ
Note:Note: Do not assume fDo not assume fss = f= fs,maxs,max . The only f. The only fss requirement is that itsrequirement is that its
magnitude is just right for the body to roll smoothly down the ramp,magnitude is just right for the body to roll smoothly down the ramp,
without sliding.without sliding.
Newton’s second law in angular formNewton’s second law in angular form
 Rotation about center of massRotation about center of mass
0==
⋅=→= ⊥
NF
sf
g
s
fRFr
ττ
ττ
)2(2
,
,
R
a
If
R
a
IIfRI
xCOM
COMs
xCOM
COMCOMsnet
−=→
−
==⋅→= αατ

8. 2,
,2,2
,
,
/1
sin
sin)(sin
sin
MRI
g
a
Mga
R
I
MMaMg
R
a
If
MaMgf
com
xCOM
xCOM
COM
xCOM
xCOM
COMs
xCOMs
+
−=
=+−→+=−=
=−
θ
θθ
θ
Linear acceleration of a body rolling along aLinear acceleration of a body rolling along a
incline planeincline plane

9. - Yo-yo- Yo-yo
Potential energy (mgh)Potential energy (mgh) kinetic energy: translationalkinetic energy: translational
(0.5mv(0.5mv22
COMCOM) and rotational (0.5 I) and rotational (0.5 ICOMCOMωω22
))
Analogous to body rolling down a ramp:Analogous to body rolling down a ramp:
- Yo-yo rolls down a string at an angle- Yo-yo rolls down a string at an angle θθ =90º with=90º with
the horizontal.the horizontal.
- Yo-yo rolls on an axle of radius R- Yo-yo rolls on an axle of radius R00..
- Yo-yo is slowed by the tension on it from the- Yo-yo is slowed by the tension on it from the
string.string.
2
0
2,
/1/1
sin
MRI
g
MRI
g
a
comcom
xCOM
+
−
=
+
−
=
θ

10. - Vector quantity- Vector quantity. Fr

×=τ
Direction:Direction: right hand rule.right hand rule.
Magnitude:Magnitude: FrFrFrFr ⊥⊥ ==⋅=⋅= )sin(sin ϕϕτ
Torque is calculated with respect to (about) a point. Changing the point canTorque is calculated with respect to (about) a point. Changing the point can
change the torque’s magnitude and direction.change the torque’s magnitude and direction.
II. TorqueII. Torque

11. III. Angular momentumIII. Angular momentum
- Vector quantity.- Vector quantity. )( vrmprl

×=×=
Direction:Direction: right hand rule.right hand rule.
Magnitude:Magnitude: vmrprprprvmrvmrprl ⋅===⋅=⋅⋅=⋅⋅=⋅= ⊥⊥⊥⊥ )sin(sinsin ϕϕϕ
l positivel positive  counterclockwisecounterclockwise
l negativel negative  clockwiseclockwise
Direction of l is always perpendicular to plane formedDirection of l is always perpendicular to plane formed
by r and p.by r and p.
Units:Units: kg mkg m22
/s/s

12. IV. Newton’s second law in angular formIV. Newton’s second law in angular form
dt
pd
Fnet =

LinearLinear AngularAngular
dt
ld
net


=τ Single particleSingle particle
The vector sum of all torques acting on a particle is equal to the time rateThe vector sum of all torques acting on a particle is equal to the time rate
of change of the angular momentum of that particle.of change of the angular momentum of that particle.
Proof:Proof: ( )
( ) netnet FrFramr
dt
ld
armvvarmv
dt
rd
dt
vd
rm
dt
ld
vrml
τ







=×=×=×=
=×=×+×=





×+×=→×=
∑
)()(
V. Angular momentumV. Angular momentum
- System of particles:- System of particles: ∑=
=++++=
n
i
in lllllL
1
321 ...


13. ∑∑ ==
=→==
n
i
netinet
n
i
i
dt
Ld
dt
ld
dt
Ld
1
,
1



ττ
Includes internal torques (due to forces between particles within system)Includes internal torques (due to forces between particles within system)
and external torques (due to forces on the particles from bodies outsideand external torques (due to forces on the particles from bodies outside
system).system).
Forces inside systemForces inside system  third law force pairsthird law force pairs  torquetorqueintint sum =0sum =0  TheThe
only torques that can change the angular momentum of a system are theonly torques that can change the angular momentum of a system are the
external torques acting on a system.external torques acting on a system.
The net external torque acting on a system of particles is equal to the timeThe net external torque acting on a system of particles is equal to the time
rate of change of the system’s total angular momentum L.rate of change of the system’s total angular momentum L.

14. - Rigid body- Rigid body (rotating about a fixed axis with constant angular speed(rotating about a fixed axis with constant angular speed ωω):):
))(()90)(sin)(( iiiiii vmrprl ∆== 
MagnitudeMagnitude
Direction:Direction: llii  perpendicular to rperpendicular to rii
and pand pii
iiiiiiiiz vmrvmrll ∆=∆== ⊥))(sin(sin)( θθ
⊥⋅= rv ω
zz
i
n
i
iii
n
i
iii
n
i
i
n
i
izz
IL
rmrrmrvmlL
ω
ωω
=






∆=⋅⋅⋅∆=∆== ⊥
=
⊥⊥
=
⊥
==
∑∑∑∑ 2
1111
)(
ωIL = Rotational inertia of a rigid body about a fixed axisRotational inertia of a rigid body about a fixed axis

15. - Conservation of angular momentum:Conservation of angular momentum:
dt
Ld
net


=τNewton’s second lawNewton’s second law
If no net external torque acts on the systemIf no net external torque acts on the system 
(isolated system)(isolated system)
cteL
dt
Ld
=→=


0
Law of conservation of angular momentum:Law of conservation of angular momentum: )( systemisolatedLL fi

=
If the net external torque acting on a system is zero, the angularIf the net external torque acting on a system is zero, the angular
momentum of the system remains constant, no matter what changes takemomentum of the system remains constant, no matter what changes take
place within the system.place within the system.
Net angular momentum at time tNet angular momentum at time tii = Net angular momentum at later time t= Net angular momentum at later time tff

16. If the component of the net external torque on a system along a certainIf the component of the net external torque on a system along a certain
axis is zero, the component of the angular momentum of the systemaxis is zero, the component of the angular momentum of the system
along that axis cannot change, no matter what changes take place withinalong that axis cannot change, no matter what changes take place within
the system.the system.
This conservation law holds not only within the frame of Newton’sThis conservation law holds not only within the frame of Newton’s
mechanics but also for relativistic particles (speeds close to light) andmechanics but also for relativistic particles (speeds close to light) and
subatomic particles.subatomic particles.
ffii II ωω =
( I( Ii,fi,f,, ωωi,fi,f refer to rotational inertia and angular speed before and afterrefer to rotational inertia and angular speed before and after
the redistribution of mass about the rotational axis ).the redistribution of mass about the rotational axis ).

17. Examples:Examples:
IIff < I< Iii (mass closer to rotation axis)(mass closer to rotation axis)
Torque ext =0Torque ext =0  IIiiωωii = I= Iff ωωff
ωωff >> ωωii
Spinning volunteerSpinning volunteer

18. A solid cylinder of radiusA solid cylinder of radius 15 cm15 cm and massand mass 3.0 kg3.0 kg rollsrolls
without slipping at a constant speed ofwithout slipping at a constant speed of 1.6 m/s1.6 m/s. (a) What is. (a) What is
its angular momentum about its symmetry axis? (b) Whatits angular momentum about its symmetry axis? (b) What
is its rotational kinetic energy? (c) What is its total kineticis its rotational kinetic energy? (c) What is its total kinetic
energy? (energy? ( II cylinder= )cylinder= )
R
v
=ω
2
2
1
MRI =
2
2
1
MR
(a) The angular speed of the cylinder is , the(a) The angular speed of the cylinder is , the
rotational inertia for cylinder . The angularrotational inertia for cylinder . The angular
momentum about the symmetry axis ismomentum about the symmetry axis is
smkgsmmkg
MRv
R
v
MRIL
/36.0)/6.1)(15.0)(0.3(5.0
2
1
2
2
⋅=×
==











== ω

19. A solid cylinder of radiusA solid cylinder of radius 15 cm15 cm and massand mass 3.0 kg3.0 kg rollsrolls
without slipping at a constant speed ofwithout slipping at a constant speed of 1.6 m/s1.6 m/s. (a) What. (a) What
is its angular momentum about its symmetry axis? (b)is its angular momentum about its symmetry axis? (b)
What is its rotational kinetic energy? (c) What is its totalWhat is its rotational kinetic energy? (c) What is its total
kinetic energy? (kinetic energy? ( II cylinder= )cylinder= )
JsmkgMv
R
v
MRIKRot 9.1)/6.1)(3(
4
1
4
1
2
1
2
1
2
1 22
2
22
===











== ω
2
2
1
MR
(b)
(c)
JJJmvmvKKK Rotlintot 7.59.18.3
4
1
2
1 22
=+=+=+=

20. (a)(a) The net torque isThe net torque is
zero at the point of contact,zero at the point of contact,
so the angular momentumso the angular momentum
before and after the collisionbefore and after the collision
must be equal.must be equal.
( )2 2 21 1
2 2
iMR MR MRω ω ω
   
= + ÷  ÷    3
iω
ω =
(b)(b) ( )( ) ( ) ( )
( )
2 2
2 2 21 1 1 1 1
2 2 3 2 3 2 2
2 21 1
2 2
2
3
i iR
i
i
MR M MRE
E MR
ω ω
ω
ω
+ −∆
= = −