Transcript Chapter 10

Chapter 10
Rotation
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
  2  1
Angular velocity
• Average angular velocity
avg
 2  1



t 2  t1
t
• Instantaneous angular velocity – the rate of change
in angular position
 d
  lim

t 0 t
dt
Angular acceleration
• Average angular acceleration
 avg
2  1



t 2  t1
t
• Instantaneous angular acceleration – the rate of
change in angular velocity
 d
  lim

t 0 t
dt
Rotation with constant angular
acceleration
• Similarly to Chapter 2 (case of 1D motion with a
constant acceleration) we can derive a set of
formulas (Table 10-1)
Chapter 10
Problem 6
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
ds d (r )
d

r
s  r v 
 r
dt
dt
dt
• ω is measured in rad/s
• Period (recall Ch. 4)
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 (Ch. 4)
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

dV

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 9
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
• More information at:
http://scienceworld.wolfram.com/physics/MomentofInertia.html
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

dV


2
volume
r
dm

2
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

rcom 




 r 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
 I com  Mh 2
I  I com  Mh
2
Chapter 10
Problem 39
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  (mr 2 )  I
  I
• Similar derivation for rigid body
Chapter 10
Problem 51
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
(Table 10-3)
Answers to the even-numbered problems
Chapter 10:
Problem 2
14 rev
Answers to the even-numbered problems
Chapter 10:
Problem 10
(a) 30 s;
(b) 1.8 × 103 rad
Answers to the even-numbered problems
Chapter 10:
Problem 22
(a) 3.0 rad/s;
(b) 30 m/s;
(c) 6.0 m/s2;
(d) 90 m/s2
Answers to the even-numbered problems
Chapter 10:
Problem 36
(a) 7.1%;
(b) 64%
Answers to the even-numbered problems
Chapter 10:
Problem 46
(a) 8.4 N · m;
(b) 17 N · m;
(c) 0
Answers to the even-numbered problems
Chapter 10:
Problem 50
1.28 kg · m2
Answers to the even-numbered problems
Chapter 10:
Problem 64
(a) 0.15 kg · m2;
(b) 11 rad/s
Answers to the even-numbered problems
Chapter 10:
Problem 78
(a)1.57m/s2;
(b) 4.55 N;
(c) 4.94 N