R. Field 10/22/2013 University of Florida

PHY 2053 Page 1

Angular Momentum• The vector angular momentum of the point

mass m about the point P is given by:

prL

perprprpL sin

x-axis

r

P

m

p

y-axis pperp

The position vector of the mass m relative to the point P is:

(units = kg∙m2/s)

yyxxr ˆˆ

The momentum vector of the mass m is:

ymvxmvypxpp yxyx ˆˆˆˆ

The magnitude of the angular momentum is:

xyzyx ypxpLLL 00The components of the angular momentum are:

Distance from the Point P to the mass m times the perpendicular component of the momentum.

R. Field 10/22/2013 University of Florida

PHY 2053 Page 2

Torque• The torque vector about the point P due to

the force F acting at r is given by:

Fr

perprFrF sin

x-axis

r

P

F

y-axis Fperp

The position vector of the mass m relative to the point P is:

(units = N∙m)

yyxxr ˆˆ

The force acting on the mass m is:

yFxFF yx ˆˆ

The magnitude of the torque is:

xyzyx yFxF 00The components of the torque are:

Distance from the Point P to the mass m times the perpendicular

component of the force.

R. Field 10/22/2013 University of Florida

PHY 2053 Page 3

Rotation: Angular Variables

The arc length s is related to the angle (in radians = rad) as follows:

• Arc Length:

rs r

s

if • Angular Displacement and Angular Velocity:

rdt

dsvv tt

dt

d

tt

0

lim

• Tangential Velocity and Angular Velocity:

(radians/s2)

• Angular Acceleration:

(360o = 2 rad)

Tangential Velocity

rvt

dt

d

tt

0

lim

(radians/second)

R. Field 10/22/2013 University of Florida

PHY 2053 Page 4

Rolling Without Slipping: Rotation & Translation

R

v

s x-axis

x

• If a cylinder of radius R rolls without slipping along the x-axis then:

rsx

R

dt

dR

dt

dxv

Translational Speed

Rotational Speed

R. Field 10/22/2013 University of Florida

PHY 2053 Page 5

Translation vs Rotation• Translation:

Mass: m

• Rotation:Moment of Inertia: I

Force: Torque: F

Position: x Angular Position:

Velocity: vx Angular Velocity:

Acceleration: ax Angular Acceleration:

dt

pdamF

dt

LdI

vmp

IL

221 mvKEtrans

221 IKErot

0 F

0If p

then L

If constant

then constant

Momentum Conservation! Angular Momentum Conservation!

R. Field 10/22/2013 University of Florida

PHY 2053 Page 6

Exam 2 Fall 2010: Problem 11

• A non-uniform cylinder with mass M and radius R rolls without sliding along the floor. If its translational kinetic energy is three times greater than its rotational kinetic energy about the rotation axis through its center of mass (i.e. the central axis of the cylinder), what is its moment of inertia about the central axis?Answer: MR2/3 % Right: 44%

221 MvKE ntranslatio

Ra

Rv

Rs

221 IKErotation

2

22

212

21

2

3)(33

R

IvIKEMvKE rotationalntranslatio

231 MRI

R. Field 10/22/2013 University of Florida

PHY 2053 Page 7

Example: Rolling without Slipping R

h

• If a cylinder with moment of inertia I and radius R starts from rest at a height h above the ground and rolls without slipping down an incline. What is its translational speed when it reaches the ground?

)( RhMgEi

MgRMRIMvMgRIMvE f ))/(1( 22212

212

21

Rv

MgRMRIMvRhMg ))/(1()( 2221

fi EE

))/(1/(2 2MRIghv • Example: I = MR2/2 (solid cylinder), h =

9.8 m then

smsmghv /3.11)/8.9( 34

34

R. Field 10/22/2013 University of Florida

PHY 2053 Page 8

Example: Rolling without Slipping R

h

d

• If a cylinder with moment of inertia I and radius R starts from rest at a height h above the ground and rolls without slipping down an incline. If the cylinder starts from rest at t = 0, when does it reach the ground?

))/(1/(2 2MRIghv adv 22 d

va

2

2

d

hsin

))/(1(

sin

))/(1( 22 MRI

g

MRId

gha

221 atd

a

dt

2

sin

hd

2

2

sin

))/(1(2

g

MRIht

• Example: I = MR2/2 (solid cylinder), h = 9.8 m, = 45o then

sg

h

g

ht 45.2

6

sin

32

R. Field 10/22/2013 University of Florida

PHY 2053 Page 9

Exam 2 Fall 2010: Problem 14

x

P

Fwall

Mg

L

r

• A 100-N uniform plank leans against a frictionless wall as shown. What is the magnitude of the torque (about the point P) applied to the plank by the wall?

Answer: 150 N∙m % Right: 42%

3 m

4 m

P 0 gravitywall

mNmNWxL

xWL

MgL

gravitywall

150)3)(100(2

sin2

21

21

yyxxr ˆˆ21

21

yMgF ˆ

xWxMgyFxF xyz 21

21

R. Field 10/22/2013 University of Florida

PHY 2053 Page 10

Exam 2 Fall 2011: Problem 13

hinge • A thin stick with mass M, length L, and moment

of inertia ML2/3 is hinged at its lower end and allowed to fall freely as shown in the figure. If its length L = 2 m and it starts from rest at an angle = 20o, what is the speed (in m/s) of the free end of the stick when it hits the table?Answer: 7.43 m/s % Right: 14% cos

2

LMgMghEi 2

2

212

21

L

vIIE f

ff

smmsm

gLmL

mgL

I

mgLv f

/43.7)20cos()2)(/8.9(3

cos3coscos

2

231

33

fi EE

cos2

Lh