Transcript Chapters 10&11 - Texas Christian University

Chapters 10, 11
Rotation and angular momentum
Rotation of a rigid body
• We consider rotational motion of a rigid body about
a fixed axis
• Rigid body rotates with all its parts locked together
and without any change in its shape
• Fixed axis: it does not move during the rotation
• This axis is called axis of rotation
• Reference line is introduced
Angular position
• Reference line is fixed in the body, is perpendicular
to the rotation axis, intersects the rotation axis, and
rotates with the body
• Angular position – the angle (in radians or degrees)
of the reference line relative to a fixed direction (zero
angular position)
Angular displacement
• Angular displacement – the change in angular
position.
• Angular displacement is considered positive in the
CCW direction and holds for the rigid body as a
whole and every part within that body
   f  i
Angular velocity
• Average angular velocity
avg
 f  i



t f  ti
t
• Instantaneous angular velocity – the rate of change
in angular position
 d
  lim

t  0  t
dt
Angular acceleration
• Average angular acceleration
 avg
 f  i



t f  ti
t
• Instantaneous angular acceleration – the rate of
change in angular velocity
 d
  lim

t 0 t
dt
Rotation with constant angular
acceleration
• Similarly to the case of 1D motion with a constant
acceleration we can derive a set of formulas:
Relating the linear and angular
variables: position
• For a point on a reference line at a distance r from
the rotation axis:
s  r
• θ is measured in radians
Relating the linear and angular
variables: speed
s  r
ds d (r )
d
v

r
 r
dt
dt
dt
• ω is measured in rad/s
• Period
2r 2
T

v

Relating the linear and angular
variables: acceleration
dv d (r )
d
at 

r
 r
dt
dt
dt
• α is measured in rad/s2
• Centripetal acceleration
v
(r )
2
 r
ac 

r
r
2
2
Rotational kinetic energy
• We consider a system of particles participating in
rotational motion
• Kinetic energy of this system is
2
i i
mv
K 
2
i
• Then
mv
mi (i ri )
K 

2
2
i
i
2
i i
2


2
2
 m (r )
i
i
i
2
Moment of inertia
• From the previous slide
K

2
2
 m (r )
i
i
i
• Defining moment of inertia (rotational inertia) as
I   mi (ri )
2
i
• We obtain for rotational kinetic energy
I
K
2
2
2
Moment of inertia: rigid body
• For a rigid body with volume V and density ρ(V) we
generalize the definition of a rotational inertia:
I
r
2
dV   r dm
2
volume
• This integral can be calculated for different shapes
and density distributions
• For a constant density and the rotation axis going
through the center of mass the rotational inertia for 8
common body shapes is given in Table 10-2 (next
slide)
Moment of inertia: rigid body
Moment of inertia: rigid body
• The rotational inertia of a rigid body depends on the
position and orientation of the axis of rotation relative
to the body
Chapter 10
Problem 25
Four equal masses m are located at the corners of a square of side L,
connected by essentially massless rods. Find the rotational inertia of this
system about an axis (a) that coincides with one side and (b) that bisects two
opposite sides.
Parallel-axis theorem
• Rotational inertia of a rigid body
with the rotation axis, which is
perpendicular to the xy plane and
going through point P:
I
r
volume
2
dV 
r
2
dm
volume
• Let us choose a reference
frame, in which the center of
mass coincides with the origin
Parallel-axis theorem
I   r dm   [( x  a )  ( y  b) ]dm
2
2
2
  ( x 2  y 2 )dm   (a 2  b 2 )dm
 2a  xdm  2b  ydm

rCM 

 (iˆx  ˆjy)dm / M

 iˆ  xdm  ˆj  ydm / M  0
Parallel-axis theorem
I   r dm   [( x  a )  ( y  b) ]dm
2
2
2
  ( x 2  y 2 )dm   (a 2  b 2 )dm
  ( R 2 )dm
R
  ( h 2 ) dm
 ICM  Mh2
I  I CM  Mh
2
Parallel-axis theorem
I  I CM  Mh
2
Chapter 10
Problem 51
A uniform rectangular flat plate has mass M and dimensions a by b. Use the
parallel-axis theorem in conjunction with Table 10.2 to show that its rotational
inertia about the side of length b is Ma2/3.
Torque
• We apply a force at point P to a rigid body that is
free to rotate about an axis passing through O
• Only the tangential component Ft = F sin φ of the
force will be able to cause rotation
Torque
• The ability to rotate will also depend on how far from
the rotation axis the force is applied
• Torque (turning action of a force):
  ( Ft )(r )  ( F sin  )(r )
• SI unit: N*m (don’t confuse with J)
Torque
• Torque:
  ( Ft )(r )  ( F sin  )(r )  ( F )(r sin  )
• Moment arm: r┴=
r sinφ
• Torque can be redefined as:
force times moment arm
τ = F r┴
Newton’s Second Law for rotation
• Consider a particle rotating under the influence of a
force
• For tangential components
  Ft r  mat r  m(r )r  (mr2 )  I
  I
• Similar derivation for rigid body
Newton’s Second Law for rotation
  I
   i
i
Chapter 10
Problem 57
A 2.4-kg block rests on a slope and is attached by a string of negligible mass to
a solid drum of mass 0.85 kg and radius 5.0 cm, as shown in the figure. When
released, the block accelerates down the slope at 1.6 m/s2. Find the coefficient
of friction between block and slope.
Rotational work
• Work
dW  Ft ds  Ft rd  d
f
W   d
i
• Power
dW d
P

 
dt
dt
• Work – kinetic energy theorem
I
I
K 

W
2
2
2
f
2
i
Corresponding relations for
translational and rotational motion
Smooth rolling
• Smooth rolling – object is rolling without slipping or
bouncing on the surface
• Center of mass is moving at speed vCM
• Point of momentary contact between the two
surfaces is moving at speed vCM
s = θR
ds/dt = d(θR)/dt = R dθ/dt
vCM = ds/dt = ωR
Rolling: translation and rotation
combined
• Rotation – all points on the wheel move with the
same angular speed ω
• Translation – all point on the wheel move with the
same linear speed vCM
Rolling: translation and rotation
combined
I CM 
Mv
K

2
2
2
2
CM
Rolling: pure rotation
• Rolling can be viewed as a pure rotation around the
axis P moving with the linear speed vcom
• The speed of the top of the rolling wheel will be
vtop = (ω)(2R)
= 2(ωR) = 2vCM
Friction and rolling
• Smooth rolling is an idealized mathematical
description of a complicated process
• In a uniform smooth rolling, P is at rest, so there’s
no tendency to slide and hence no friction force
• In case of an accelerated smooth rolling
aCM = α R
fs opposes tendency to slide
Rolling down a ramp
Mghi 
ghi 
Mv
2
f
2
v
2
f
2


I
2
f
2
Iv
2
f
2MR
2
2
2 ghi MR
vf 
MR2  I
Chapter 10
Problem 39
What fraction of a solid disk’s kinetic energy is rotational if it’s rolling without
slipping?
Vector product of two vectors
• The result of the vector (cross) multiplication of two
vectors is a vector
  
a b  c
• The magnitude of this vector is
c  absin 

• Angle
 φ is the smaller of the two angles between b
and a
Vector product of two vectors

• Vector c is perpendicular
to the plane that contains

vectors a and b and its direction is determined by
the right-hand rule
• Because of the right-hand rule, the order of
multiplication is important (commutative law does not
apply)
 
 
b  a  (a  b )
• For unit vectors
ˆi  iˆ  0  ˆj  ˆj  kˆ  kˆ
iˆ  ˆj  kˆ ˆj  kˆ  iˆ kˆ  iˆ  ˆj
Vector product in unit vector notation
 
a  b  (axiˆ  a y ˆj  az kˆ)  (bxiˆ  by ˆj  bz kˆ)
axiˆ  bxiˆ  axbx (iˆ  iˆ)  0
a xiˆ  by ˆj  axby (iˆ  ˆj )  axby kˆ
 
a  b  (a y bz  by az )iˆ 
 (az bx  bz ax ) ˆj  (axby  bx a y )kˆ
Torque revisited
• Using vector product, we can redefine torque
(vector) as:
     
  r  F  r  F  r  F

  rF sin   r sin F
Angular momentum
• Angular momentum of a particle of mass m and

velocity v with respect to the origin O is defined as
  
 
L  r  p  m(r  v )
• SI unit: kg*m2/s
Newton’s Second Law in angular form
  
 
L  r  p  m(r  v )



   
dL
  dv dr  
 m r    v   mr  a  v  v 
dt
dt dt


 
 

  
 mr  a   r  ma  r  Fnet   r  Fi
i


  i   net

i
dL 
  net
dt
Angular momentum of a system of
particles


L   Ln
 n



dLn
dL
  net,n   net

dt
dt
n
n

dL 
  net
dt
Angular momentum of a rigid body
• A rigid body (a collection of elementary masses
Δmi) rotates about a fixed axis with constant angular
speed ω
• For sufficiently symmetric objects:


L  I
Conservation of angular momentum
• From the Newton’s Second Law

dL 
  net
dt
• If the net torque acting on a system is zero, then

dL
0
dt

L  const
• If no net external torque acts on a system of
particles, the total angular momentum of the system
is conserved (constant)
• This rule applies independently to all components
 net , x  0 Lx  const
Conservation of angular momentum
L  I  const
I ii  I f  f
Conservation of angular momentum

L  const
More corresponding relations for
translational and rotational motion
Chapter 11
Problem 28
A skater has rotational inertia 4.2 kg·m2 with his fists held to his chest and 5.7
kg·m2 with his arms outstretched. The skater is spinning at 3.0 rev/s while
holding a 2.5-kg weight in each outstretched hand; the weights are 76 cm from
his rotation axis. If he pulls his hands in to his chest, so they’re essentially on
his rotation axis, how fast will he be spinning?
Questions?
Answers to the even-numbered problems
Chapter 10
Problem 24
0.072 N⋅m
Answers to the even-numbered problems
Chapter 10
Problem 30
2.58 × 1019 N⋅m
Answers to the even-numbered problems
Chapter 10
Problem 40
hollow
Answers to the even-numbered problems
Chapter 11
Problem 16
69 rad/s; 19° west of north
Answers to the even-numbered problems
Chapter 11
Problem 18
(a) 8.1 N⋅m kˆ
(b) 15 N⋅m kˆ
Answers to the even-numbered problems
Chapter 11
Problem 24
1.7 × 10-2 J⋅s
Answers to the even-numbered problems
Chapter 11
Problem 26
(a) 1.09 rad/s
(b) 386 J
Answers to the even-numbered problems
Chapter 11
Problem 30
along the x-axis
or 120° clockwise from the x-axis
Answers to the even-numbered problems
Chapter 11
Problem 42
26.6°