Transcript Chapter 10

Chapter 10:Rotation of a rigid object
about a fixed axis
Reading assignment:
Chapter 10.1 to10.4, 10.5 (know concept of moment of
inertia, don’t worry about integral calculation), 10.6 to 10.9
Homework 10.1 (due Tuesday, Oct. 23):
CQ1, CQ2, AE1, 2, 3, 6, 7, 12
Homework 10.2 (due Wednesday, Oct. 24): CQ8, QQ3, QQ4, OQ3, OQ4, OQ6,
OQ8, AE3, 13, 15, 19, 26, 29
Homework 10.3 (due Friday, Oct. 26):
CQ13, 35, 36, 38, 43, 49, 55, 56, 59
• Rotational motion
• Angular displacement, angular velocity, angular acceleration
• Rotational energy
• Moment of Inertia
• Torque
Rotational motion
Look at one point P:
Arc length s:
s  r 
Thus the angle (angular
position) is:
s

r
Planar, rigid object rotating about origin O.
 is measured in degrees or radians (SI unit: radian)
Full circle has an angle of 2p radians.
Thus, one radian is 360°/2p  57.3
Radian
2p
p
p/2
1
degrees
360°
180°
90°
57.3°
Define quantities for circular motion
(note analogies to linear motion!!)
Angular displacement:
   f   i
 f i


Average angular speed:  
t f  ti
t
 d
Instantaneous angular speed:   lim

t 0 t
dt
Average angular acceleration:

Instantaneous angular acceleration:
 f  i
t f  ti


t
 d
  lim

t 0 t
dt
Angular velocity
is a vector
Right-hand rule for
determining the direction
of this vector.
Every particle (of a rigid object):
• rotates through the same angle,
• has the same angular velocity,
• has the same angular acceleration.
, ,  characterize rotational
motion of entire object
Linear motion with constant
linear acceleration, a.
Rotational motion with constant
rotational acceleration, .
Exactly the same equations, just different symbols!!
v xf  v xi  a x t
 f   i  t
x f  xi  12 (vxi  vxf )t
 f   i  12 (i   f )t
1 2
x f  xi  v xi t  a x t
2
1 2
 f   i   i t  t
2
vxf  vxi  2ax ( x f  xi )
 f  i  2 ( f   i )
2
2
2
2
Black board example 10.1
A wheel starts from rest and
rotates with constant angular
acceleration and reaches an
angular speed of 12.0 rad/s in
3.00 s.
1. What is the magnitude
2. Through what angle does 3. Through what angle does
of the angular
the wheel rotate in these
the wheel rotate between
acceleration of the wheel
3 sec (in rad)?
2 and 3 sec (in rad)?
(in rad/s2)?
A. 18
A. 5
A. 0
B. 24
B. 10
B. 1
C. 30
C. 15
C. 2
D. 36
D. 20
D. 3
E. 48
E. 25
E. 4
Relation between angular
and linear quantities
Arc length s:
s  r 
Tangential speed of a
point P:
vt  r  
Tangential acceleration of a
point P:
at  r  
Note: This is not the centripetal acceleration ar
This is the tangential acceleration at
Black board example 10.2
vt
A fly is sitting at the end of a ceiling fan blade. The length of the blade is 0.50 m and
it spins with 40.0 rev/min.
a)
Calculate the (tangential) speed of the fly.
b) What are the tangential and angular speeds of another fly sitting half way in?
c)
Starting from rest it takes the motor 20 seconds to reach this speed. What is the
angular acceleration?
d) At the final speed, with what force does the fly (m = 0.01 kg, r = 0.50 m) need
to hold on, so that it won’t fall off?
(Note difference between angular and centripetal acceleration).
Demo:
Both sticks have the same weight.
Why is it so much more difficult to
rotate the blue stick?
Rotational
kinetic energy
A rotating object (collection
of i points with mass mi) has
a rotational kinetic energy of
1
2
K R  I 
2
Where:
I   mi  ri
i
2
Moment of inertia or
rotational inertia
Black board example 10.3
2
i-clicker
Four small spheres are mounted on
the corners of a weightless frame as
shown.
M = 5 kg;
m = 2 kg;
a = 1.5 m;
b=1m
a)
1
3
4
What is the rotational energy of the system if it is rotated about the z-axis (out
of page) with an angular velocity of 5 rad/s
b) What is the rotational energy if the system is rotated about the y-axis?
i-clicker for question b):
A) 281 J
B) 291 J
C) 331 J
D) 491 J
E) 582 J
Moment of inertia (rotational inertia) of an
object depends on:
- the axis about which the object is rotated.
- the mass of the object.
- the distance between the mass(es) and the axis
of rotation.
I   mi  ri
i
2
Calculation of Moments of inertia for
continuous extended objects
I
lim  ri
mi 0 i
2
 mi   r dm   r dV
2
2
Refer to Table10.2
Note that the moments of inertia are different for different axes
of rotation (even for the same object)
1
I  ML
3
I
1
ML
12
1
I  MR
2
Moment of inertia for some objects
Page 287
Black board example 10.4
Rotational energy earth.
The earth has a mass M = 6.0×1024 kg and a radius of R = 6.4×106 m. Its distance
from the sun is d = 1.5×1011 m What is the rotational kinetic energy of
a)
its motion around the sun?
b) its rotation about its own axis?
Parallel axis
theorem
 Rotational inertia for a rotation about an axis that is
parallel to an axis through the center of mass
I CM
I  I CM  Mh 2
h
Blackboard example 10.5
What is the rotational energy of a sphere (mass m = 1 kg, radius R = 1m) that is
rotating about an axis 0.5 away from the center with  = 2 rad/sec?
Conservation of energy (including rotational energy):
Again:
If there are no non-conservative forces energy is conserved.
Rotational kinetic energy must be included in energy
considerations!
Ei  E f
U i  Klinear,initial  K rotational,initial  U f  Klinear, final  K rotational, final
Black board example 10.6
Connected cylinders.
Two masses m1 (5.0 kg) and m2 (10 kg)
are hanging from a pulley of mass M
(3.0 kg) and radius R (0.10 m), as
shown. There is no slip between the
rope and the pulleys.
(a) What will happen when the masses
are released?
(b) Find the velocity of the masses after they have fallen a
distance of 0.5 m.
(c) What is the angular velocity of the pulley at that moment?
Torque
F  sin f
r
F
f
F  cos f
A force F is acting at an angle f on a lever that is rotating around
a pivot point. r is the distance between the pivot point and F.
This force-lever pair results in a torque t on the lever
t  r  F  sin f
Black board example 10.7
i-clicker
Two mechanics are trying to open a rusty
screw on a ship with a big ol’ wrench.
One pulls at the end of the wrench (r = 1 m)
with a force F = 500 N at an angle F1 = 80°;
the other pulls at the middle of wrench with
the same force and at an angle F2 = 90°.
What is the net torque the two mechanics are applying to the screw?
A. 742 Nm
B. 750 Nm
C. 900 Nm
D. 1040 Nm
E. 1051 Nm
Torque t and
angular acceleration .
Particle of mass m rotating in a
circle with radius r.
Radial force Fr to keep particle on
circular path.
Tangential force Ft accelerates
particle along tangent.
Ft  mat
Torque acting on particle is proportional to
angular acceleration :
t  I
 
dW  F  ds
 
W  F s
Definition of work:
Work in linear motion:
 
dW  F  ds
 
W  F  s  F  s  cos 
Component of force F along
displacement s. Angle 
between F and s.
Work in rotational motion:
 
dW  F  ds
Torque t and angular
dW  t  d
W  t 
displacement .
Linear motion with constant
linear acceleration, a.
Rotational motion with constant
rotational acceleration, .
v xf  v xi  a x t
 f   i  t
x f  xi  12 (vxi  vxf )t
 f   i  12 (i   f )t
1 2
x f  xi  v xi t  a x t
2
1 2
 f   i   i t  t
2
vxf  vxi  2ax ( x f  xi )
 f  i  2 ( f   i )
2
2
2
2
Summary: Angular and linear quantities
Linear motion
1
2
K

m

v
Kinetic Energy:
2
Force:
F  ma
Momentum:
p  mv
Work:
 
W  F s
Rotational motion
1
K R  I  2
2
Kinetic Energy:
Torque:
t  I
Angular Momentum:
Work:
L  I
W  t 
Rolling motion
Superposition principle:
Rolling motion
=
Kinetic energy
of rolling motion:
Pure translation +
Pure rotation
1
1
2
2
K  MvCM  I CM  
2
2
Black board example 10.8
Demo
A ring, a disk and a sphere (equal mass and diameter) are rolling
down an incline.
All three start at the same position; which one will be the fastest
at the end of the incline?
A. All the same
B. The disk
C. The ring
D. The sphere