PHY126 Summer Session I, 2008
• Most of information is available at:
http://nngroup.physics.sunysb.edu/~chiaki/PHY126-08
including the syllabus and lecture slides.
Read syllabus and watch for important announcements.
• Homework assignment for each chapter due nominally a week later.
But at least for the first two homework assignments, you will have more
time. All the assignments will be done through MasteringPhysics, so
you need to purchase the permit to use it. Some numerical values
in some problems will be randomized.
• In addition to homework problems and quizzes, there is a reading
requirement of each chapter, which is very important.
Chapter 10: Rotational Motion
Movement of points in a rigid body
 All
points in a rigid body move in circles about the axis
of rotation
z
Axis of rotation
P
orbit of point P A rigid body has a perfectly
definite and unchanging shape
and size. Relative position of
points in the body do not change
relative to one another.
y
rigid body In this specific example on
the left, the axis of rotation
is the z-axis.
x
Movement of points in a rigid body (cont’d)
 At
any given time, the 2-d projection of any point in the
object is described by two coordinates ( r , q )
In our example, 2-d projection onto the x-y plane is
the right one.
y
P
length of the arc from the x-axis s:
s = rq
where s,r in m, and q in rad(ian)
r
q
x
A complete circle: s = 2pr
360o = 2p rad
57.3o= 1 rad
1 rev/s = 2p rad/s
1 rev/min = 1 rpm
Angular displacement, velocity and acceleration
 At
any given time, the 2-d projection of any point in the
object is described by two coordinates ( r , q )
y
Angular displacement:
Dq = q2 - q1
in a time interval
Dt = t2 – t1
Average angular velocity:
P at t2
 avg =
r
q2
P at t1
Dq q 2 - q 1
=
Dt
t 2 - t1
rad/s
Instantaneous angular velocity:
q1
x
Dq dq
 = lim
=
Dt 0 Dt
dt
rad/s
 < (>) 0 (counter) clockwise rotation
Angular displacement, velocity and acceleration
(cont’d)
 At
any given time, the 2-d projection of any point in the
object is described by two coordinates ( r , q )
y
Average angular acceleration:
D  - 
avg =
=
rad/s2
2
Dt
P at t2
t 2 - t1
Instantaneous angular acceleration:
D d
=
Dt  0 D t
dt
 = lim
r
q2
1
P at t1
q1
x
rad/s2
Correspondence between linear & angular
quantities
Linear
Displacement
Velocity
Acceleration
Dx
Angular
Dq
v = dx / dt  = dq / dt
a = dv / dt
 = d / dt
Case for constant acceleration (2-d)
 Consider
an object rotating with constant angular
acceleration 0
 = d / dt = 0
t
t
0
0
 (d / dt ) dt =   dt
0

t
t
 (dq / dt ) dt = (dq / dt )  dt
t =0
0
0
t
 0  t dt
0
 d =  t
q - q0 = 0t  (1 / 2)0t
 - 0 = 0t
q = q0  0t  (1 / 2)0t
0
0
 = 0  0t
Eq.(1)
Eq.(2)
2
2
Case for constant acceleration (2-d) (cont’d)
Eliminating t from Eqs.(1) & (2):
Eq.(1)
Eqs.(1’)&(2)
t = ( - 0 ) / 0
q = q0  0 ( - 0 ) / 0
 (1 / 2)0 ( - 0 )2 / 02
 2 = 02  2 0 (q - q0 )
Eq.(1’)
Vectors
Vectors in 3-dimension
iˆ, ˆj , kˆ
z
ĵ
k̂
: unit vector in x,y,z direction
Consider a vector:
y

a = (ax , a y , az ) = axiˆ  a y ˆj  az kˆ
x
iˆ
Also
iˆ = (1,0,0), ˆj = (0,1,0), kˆ = (0,0,1)
• Inner (dot) product
   
a  b = a b cos q = axbx  a y by  az bz
 
 
a  b = 0 if a  b

b
 
a b

a
Small change of radial vector
 Rotation by a small rotation angle
 

r2 = r cos Dq iˆ  r sin Dq ˆj
  
Dr = r2 - r1
 
r1 = r iˆ
  


ˆ
ˆ
ˆ
Dr = r2 - r1 = ( r cos Dqi  r sin Dqj ) - r i

 r Dq ˆj  cos Dq  1 and sin Dq  Dq for Dq << 1
  drˆ
Drˆ ˆ
If we define rˆ = r / r ,
= lim
q
dq Dq 0 Dq
Note:
rˆ  qˆ = 0  rˆ  qˆ
unit vector in y direction
Relation between angular & linear variables
ĵ
rˆ  qˆ = 0  rˆ  qˆ
y
qˆ = (- sin q , cos q )
unit
tangential
vector
qˆ

r
unit
radial
vector
r̂
q
A point with
Fixed radius
x
iˆ
unit vector in x direction
iˆ = (1,0), ˆj = (0,1)

r = rrˆ = r (cos q iˆ  sin q ˆj )
drˆ / dq = - sin q iˆ  cos q
dqˆ / dq = - cos q iˆ - sin q
ˆj = qˆ
ˆj = - rˆ


v = dr / dt = d ( rrˆ) / dt = r (drˆ / dt )
r is const.
= r ( drˆ / dq )( dq / dt ) = rqˆ

v = v = r
unit vector in y direction
ĵ
Relation between angular & linear variables
(cont’d)
y
qˆ
r̂

r
q
A point with
Fixed radius
x
iˆ
unit vector in x direction


a = dv / dt = d ( rqˆ) / dt
= r ( d / dt )qˆ  r ( dqˆ / dt )
= rqˆ  r ( dqˆ / dq )( dq / dt )
2
ˆ
= rq - r rˆ
= rqˆ - ( v 2 / r ) rˆ
at
tangential
component
ar
radial
component
Relation between angular & linear variables
(cont’d)
v
Example
How are the angular speeds of the
two bicycle sprockets in Fig. related
to the number of teeth on each sprocket?
The chain does not slip or stretch, so it moves at the same
tangential speed v on both sprockets:

r
v = r11 = r22  2 = 1
1 r2
The angular speed is inversely proportional to the radius. Let N1 and
N2 be the numbers of teeth. The condition that tooth spacing is the
same on both sprockets leads to:
2pr1 2pr2
r
N
=
 1 = 1
N1
N2
r2 N 2
Combining the above two equations:
 2 N1
=
1 N 2
Description of general rotation
 Why it is not always right to define a rotation by a vector
rotation about x axis
original
y
y
x
x
original
z
x
y
x
rotation about y axis
z
y
rotation about y axis
rotation about x axis
z
y
x
x
The result depends on the order of operations
y
Description of general rotation
 Up to this point all the rotations have been about the z-axis or in
x-y plane. In this case the rotations are about a unit vector n̂ where
n̂ is normal to the x-y plane. But in general, rotations are about a
general direction.
• Define a rotation about n̂ by Dq as:

n̂
Dq = Dq n̂



• In general Dq = Dq1  Dq 2  Dq 2  Dq1
but if Dq , Dq1 & Dq2 infinitesimally

small,
we can define a vector dq by





dq = dq1  dq 2 = dq 2  dq1

RH rule

dq 
d 
=  =  nˆ ;
=  =  nˆ
dt
dt
• If the axes
 of rotation

 are the
 same,

Dq = Dq1  Dq 2 = Dq 2  Dq1
Kinetic energy & rotational inertia
K = i =1 (1 / 2)Dm v
N
y
v̂ 
r
A point in a
Rigid body
x
2
i i
= i =1 (1 / 2)Dmi (ri )
N
= (1 / 2)
2

N
i =1
Dmi ri
2
2
I
rotational inertia/
moment of inertia
axis of rotation
Kinetic energy & rotational inertia (cont’d)
K = (1 / 2) I
y
2
More precise definition of I :
I = lim
v̂ 
r
N 
Dm
i=1 Dmi ri
N
2

=  r dm =  r  (r )dV
2
A point in a
Rigid body
x Compare with:
2
density volume
element
K = (1 / 2)mv2
And remember conservation of
energy:
rotation axis
K1  U1 = K 2  U 2  Wother
Kinetic energy & rotational inertia (cont’d)
 Moment of inertia of a thin ring (mass M, radius R) (I)
dm = ds ; ds = Rd q
y
linear mass density
I =  r dm =  R ds
2p
2
3
=   R ( Rd q ) = R 2p
2

R q
ds
0
= MR
x
rotation axis (z-axis)
2
2
Kinetic energy & rotational inertia (cont’d)
 Moment of inertia of a thin ring (mass M, radius R) (II)
rotation axis (y-axis)
y
dm = ds ; ds = Rd q
r = R cosq
2p
I =  ( R cos q )2 Rd q
0
r

R q
ds
x

= R
3
= R

3
2p
0
2p
0
cos2 qdq
(1 / 2)(1  cos 2q )dq
= (1 / 2)R 3[q  (1 / 2) sin 2q ] |02p
cos 2q = cos2 q - sin 2 q = 2 cos2 q - 1
= (1 / 2)R3 (2p ) = (1 / 2) MR 2
Kinetic energy & rotational inertia (cont’d)
Table of moment of inertia
Kinetic energy & rotational inertia (cont’d)
Tables of moment of inertia
Parallel axis theorem
y
x, y measured
w.r.t. COM
The axis of rotation is parallel to the z-axis
rotation axis
2
2
I
=
r
dm
=
[(
x
a
)
through P

y
P
COM
O
=  ( x 2  y 2 )dm - 2a  xdm
dm
r
y-b
- 2b ydm  (a 2  b2 )  dm
x-a
d
b
a
x

x
= I com - 2aMxcom - 2bMy com  d 2 M
= I com  Md
rotation axis
through COM
COM: center of mass
 ( y - b)2 ]dm

rcom =

i mi ri
m
i
i
=
2
total
mass

i mi ri
M
1 

r dm

M
Parallel axis theorem (cont’d)
y
rotation axis
through P
y
dm
P
COM
O
r
y-b
Knowing the moment of inertia
about an axis through COM
(center of mass) of a body,
the rotational inertia for rotation
about any parallel axis is :
I = I com  Md
x-a
d
b
a
x
rotation axis
through COM
x
2
Correspondence between linear & angular
quantities
linear
displacement
Dx
angular
Dq
 = dq / dt
acceleration
v = dx / dt
a = dv / dt
 = d / dt
mass
m
I
kinetic energy
K = (1 / 2)mv2
K = (1 / 2) I 2
velocity
Exercises
Problem 1
A meter stick with a mass of 0.160 kg is pivoted about one end so that it can
rotate without friction about a horizontal axis. The meter stick is held in a
horizontal position and released. As it swings through the vertical, calculate
(a) the change in gravitational potential energy that has occurred;
(b) the angular speed of the stick;
(c) the linear speed of the end of the stick opposite the axis.
(d) Compare the answer in (c) to the speed of a particle that has fallen 1.00m,
starting from rest.
y
cm
1
Solution
v =0
(a) U = Mgy
cm
1.00 m
 DU = Mg ( ycm 2 - ycm1 )
cm
= (0.160 kg)( 9.80 m / s )( 0.50 m - 1.00 m)
2
= -0.784 J
v2
0.500 m
x
Problem 1 (cont’d)
(b)
K1  U1 = K2  U2
K1 = 0  K2 = U1 - U 2 = -DU = (1 / 2) I22 ; I = (1 / 3) ML2
2 = 2( - DU ) / I = (6  0.784 J ) /[( 0.160 kg)(1.00 m)2 ] = 5.42 rad / s
(c)
(d)
v = r = (1.00 m)( 5.42 rad / s ) = 5.42 m / s
v 2 = v02y  2a y ( y - y0 ); v0 y = 0, y - y0 = -1.00 m; a y = -9.80 m / s2
v = - 2a y ( y - y0 ) = -4.43 m / s
Problem 2
The pulley in the figure has radius R and a moment of inertia I. The rope
does not slip over the pulley, and the pulley spins on a frictionless axle.
The coefficient of kinetic friction between block A and the tabletop is k .
The system is released from rest, and block B descends. Block A has mass
mA and block B has mass mB. Use energy methods to calculate the speed of
Block B as a function of distance d that it has descended.
Solution
Energy conservation:
v1 = 0  v2  0
1 = 0  2  0
A
I
E = K1  U1 = K2  U2  W
K1 = 0,U1 = mB gd
K2 = (1 / 2)mAv22  (1 / 2)mB v22  (1 / 2) I2
2
y
U 2 = 0, W = k mA gd , 2 = v2 / R
B
mB gd = (1 / 2)( m A  mB  I / R 2 )v22  k m A gd
v2 =
2 gd (mB - k mA )
mA  mB  I / R 2
x
d
Problem 3
You hang a thin hoop with radius R over a nail at the rim of the hoop. You
displace it to the side through an angle b from its equilibrium position and let
it go. What is its angular speed when it returns to its equilibrium position?
y
Solution
K1  U1 = K 2  U 2 ;
pivot point
R
K1 = 0, K 2 = (1 / 2) I22
I = I cm  Md 2 = MR 2  MR 2 = 2 MR 2
U1 = Mgy cm ,1 = MgR (1 - cos b ), U 2 = 0
ycm ,1
b
R
x
MgR(1 - cos b ) = MR 222
2 = g (1 - cos b ) / R
the origin =
the original location of
the center of the hoop