Transcript Rotational Motion

ROTATIONAL MOTION
AND EQUILIBRIUM
Angular Quantities of Rotational Motion
Rotational Kinematics
Torque
Center of Gravity
Moment of Inertia
Rotational Kinetic Energy
Angular Momentum
Rotational Equilibrium
ANGULAR SPEED AND ANGULAR ACCELERATION:
w = angular speed
rad/s
deg/s
rev/s
a = angular acceleration rad/s2 deg/s2
rev/s2
q
rotating
drum
axis
Dq
w=
Dt
Dw
a=
Dt
ANGULAR CONVERSIONS:
1 rev = 2p rad = 360 deg
p rad = 180 deg
Convert 246o to radians.
Convert 16.4 rev to degrees.
Convert 246o to radians.
246 = 246
o
o
p
180
o
= 1.37p rad
Convert 16.4 rev to degrees.
360o
16.4rev = 16.4rev
= 5904o
rev
ROTATIONAL KINEMATICS:
a =a
Uniform angular acceleration:
q = instantaneous angular position
w = instantaneous angular speed
q = q o  wo  a t
1
2
2
w = wo  at
w = w  2a q  q o 
2
2
o
EXAMPLE:
Through what angle does a wheel rotate if it begins from rest
and accelerates at 3 o/s2 for 15 seconds? Also calculate its final
angular speed.
ANSWER:
Through what angle does a wheel rotate if it begins from rest
and accelerates at 3 o/s2 for 15 seconds? Also calculate its final
angular speed.
q = q o  w ot  a t
1
2
q = 00
1
2
2
3  15s 
deg
s2
q = 337 deg
w = wo  a t
w = 0  3
deg
s2
w = 45deg/ s
 15s 
2
ROTATIONAL AND LINEAR QUANTITIES:
w
r
s = rq
v = rw
a = ra
Angular quantities must be expressed in
terms of radians.
EXAMPLE:
A point on the rim 5cm
from the axis of rotation moves through an
angle of 15 degrees. How far has it traveled?
axis
ANSWER:
A point on the rim 5cm from
the axis of rotation moves through an angle
of 15 degrees. How far has it traveled?
 p
s =  0.05m  15  
o
 180
o

 = 0.13m

EXAMPLE:
A wheel of radius r is rolling along the ground without slipping so that
its center moves with a constant speed v? What are the speeds of the
points at its top and at its bottom?
Since it is not slipping the speed at the center must be v = 2pr/T.
The rotational speed must be w = 2p/T.
The speed of a point on the rim, relative to the center, must be
rw = 2pr/T. This is equal to v.
At its bottom the rim is moving backwards relative to the center so that
its velocity relative to the ground if v -v = 0.
At its top the rim is moving forward relative to the center to that its
2v
velocity is v + v = 2v.
v
TORQUE:
Torque, t, is produced when a force acts on a body along a line that is
displaced from a pivot point or axis of rotation.
LINE OF ACTION: The line of action of a force is the line along
which the force acts.
LEVER ARM: The lever arm of a force is the distance from the pivot
or axis of rotation to the point where the force is applied.
Bar
t = Fr
F = F sin q 
t = F sin q  r
r
pivot
F
q
F
Force
line of action
The torque produced by this force is trying to
rotate the bar in a counter-clockwise direction.
HOW TO CALCULATE A TORQUE:
1. In a sketch, locate the pivot point or axis of rotation.
2. Locate the force producing the torque and where it is applied.
3. Determine the length of the line segment from the pivot to the
point where the force is applied. This is the lever arm.
4. Determine the angle between the force and the lever arm. The
force multiplied by the sine of this angle is the force
component perpendicular to the lever arm.
5. The magnitude of the torque is the product of the force
perpendicular component and the lever arm.
6. Determine the direction of rotation the torque is trying to
induce as being either clockwise or counter-clockwise.
EXAMPLE:
Calculate the torque produced by
the force acting on the rod as shown.
1.25m
pivot
57o
F = 16N
EXAMPLE:
Calculate the torque produced by
the force acting on the rod as shown.
1. Identify the pivot (left end of rod)
1.25m
pivot
57o
F = 16N
EXAMPLE:
Calculate the torque produced by
the force acting on the rod as shown.
1. Identify the pivot (left end of rod)
1.25m
pivot
57o
2. Identify the force ( F = 16 N)
F = 16N
EXAMPLE:
Calculate the torque produced by
the force acting on the rod as shown.
1. Identify the pivot (left end of rod)
1.25m
pivot
57o
2. Identify the force ( F = 16 N)
3. Determine the lever arm ( r = 1.25 m)
F = 16N
EXAMPLE:
Calculate the torque produced by
the force acting on the rod as shown.
1. Identify the pivot (left end of rod)
1.25m
pivot
57o
2. Identify the force ( F = 16 N)
3. Determine the lever arm ( r = 1.25 m)
4. Determine the angle between the
force and the lever arm (q = 57o)
F = 16N
EXAMPLE:
Calculate the torque produced by
the force acting on the rod as shown.
1. Identify the pivot (left end of rod)
1.25m
pivot
57o
2. Identify the force ( F = 16 N)
3. Determine the lever arm ( r = 1.25 m)
F = 16N
4. Determine the angle between the
force and the lever arm (q = 57o)
5. Calculate the torque:
t = F sin(q )r = (16N )sin(57 )(1.25m) = 16.8N  m
o
EXAMPLE:
Calculate the torque produced by
the force acting on the rod as shown.
1. Identify the pivot (left end of rod)
1.25m
pivot
2. Identify the force ( F = 16 N)
3. Determine the lever arm ( r = 1.25 m)
Rotation will be
counter-clockwise
F = 16N
4. Determine the angle between the
force and the lever arm (q = 57o)
5. Calculate the torque:
t = F sin(q )r = (16N )sin(57 )(1.25m) = 16.8N  m
o
6. Determine the orientation of the
torque. (see diagram)
57o
CENTER OF GRAVITY:
The center of gravity of an object is the point, (xcg , ycg), from which
the weight acts.
xcg
xmg

=
m g
i
i
i
i
ycg
i
ymg

=
m g
i
i
i
i
i
For most ordinary objects all gi are equal. Thus we can write:
x cg
xm

=
m
i
i
i
y cg
ym

=
m
i
i
i
When g is considered constant cg is usually referred to as center of
mass.
EXAMPLE:
A system of three masses, m1, m2 and
m3 are arranged as shown. Calculate
the coordinates of the center of mass
of this system.
y
m2 = 4 kg at (0, 3m)
m3 = 3 kg at (4m, 0)
x
m1 = 5 kg at (0, 0)
ANSWER:
y
A system of three masses, m1, m 2 and
m3 are arranged as shown. Calculate
the coordinates of the center of mass
of this system.
m2 = 4 kg at (0, 3m)
m3 = 3 kg at (4m, 0)
x
m1 = 5 kg at (0, 0)
xcg =
m1 x1  m2 x2  m3 x3
m1  m2  m3
xcg =
(5kg )(0m)  (4kg )(0m)  (3kg )(4m)
(5kg )(0m)  (4kg )(3m)  (3kg )(0m)
ycg =
5kg  4kg  3kg
5kg  4kg  3kg
xcg = 1m
m1 y1  m2 y2  m3 y3
ycg =
m1  m2  m3
ycg = 1m
MOMENT OF INERTIA:
Objects have an intrinsic resistance to changes in their rate of
rotation. This is referred to as rotational inertia and is quantified by
moment of inertia, I.
Moment of inertia depends on two factors:
1. the object’s mass, and
2. how that mass is distribute relative to the axis of rotation.
I =  mi ri
2
For many solid objects there are simple formulas for their moments
of inertia.
2
2
Any axis through center.
sphere
5
I
= MR
I disk = MR
1
2
2
Axis through center and
perpendicular to disk.
EXAMPLE:
Suppose the three masses in the diagram
are fixed together by massless rods and are
free to rotate in the xy plane about their
center of gravity. The coordinates of their
center of gravity are (1m, 1m) as calculated
in the previous example.
Calculate the moment of inertia of these
masses about this axis.
y
m2 = 4 kg at (0, 3m)
r2
r1
r3 m3 = 3 kg at (4m, 0)
x
m1 = 5 kg at (0, 0)
ANSWER:
y
Suppose the three masses in the diagram
are fixed together by massless rods and are
free to rotate in the xy plane about their
center of gravity. The coordinates of their
center of gravity are (1m, 1m) as calculated
in the previous example.
m2 = 4 kg at (0, 3m)
r2
r1
r3 m3 = 3 kg at (4m, 0)
x
Calculate the moment of inertia of these
masses about this axis.
r1 =
1m  0m   1m  0m  = 1.41m
r2 =
1m  0m   1m  3m 
r3 =
1m  4m   1m  0m 
2
2
2
m1 = 5 kg at (0, 0)
2
2
2
= 2.24m
= 3.16m
I = m1r12  m2 r22  m3r32
I = (5kg )(1.41m) 2
 (4kg )(2.24m)
2
 (3kg )(3.16m)
I = 60.00kg  m 2
2
ROTATIONAL DYNAMICS I:
If the net torque acting on an object is not zero, the object will
experience an angular acceleration, a. a is related to torque by
Newton’s second law of motion for rotation:
t = Ia
EXAMPLE:
An object has a moment of inertia of 60 kgm2 and is acted upon by a
torque of 20Nm. What is the magnitude of the resulting angular
acceleration of the object?
ANSWER:
An object has a moment of inertia of 60 kgm2 and is acted upon by a
torque of 20Nm. What is the magnitude of the resulting angular
acceleration of the object?
t = Ia
t
20 N  m
a= =
2
I 60kg  m
a = 0.33 rad s 2
ROTATIONAL DYNAMICS II:
A rotating object has kinetic energy due to its rotational motion in
addition to whatever kinetic energy it has due to the linear motion of
its center of gravity.
2
1
rot
2
The work-energy theorem applies in the following way:
KE
= Iw
DKElin  DKErot  DPE = WNC
An object’s total kinetic energy is given by:
KEtot = KElin  KErot
EXAMPLE: A solid ball rolls down a 40o incline from a height of
4m without slipping. The radius of the ball is 20cm and it begins from
rest. What is the linear speed of the ball at the bottom of the incline?
ANSWER: A solid ball rolls down a 40o incline from a height of
4m without slipping. The radius of the ball is 20cm and it begins from
rest. What is the linear speed of the ball at the bottom of the incline?
DKELin  DKERot  DPE = 0
KELin ,i = KERot ,i = 0
KELin , f = 12 mv 2
Basic Equation
Values and expressions for
initial and final quantities
KERot , f = 12 I w 2
2

 1 2
v
2
1 2
= 2  5 mr   2  = 5 mv
r 
PEi = mgh
PE f = 0
Continued on next slide
Continues from previous slide
1
2
mv  mv  mgh = 0
Substitutions made to produce
working equation
v 2 = gh
Simplify and solve for v
7
10
2
1
5
2
10 gh
v=
=
7
v = 7.48 ms

10 9.8 sm2
  4m 
7
The speed of the ball does not depend on its mass nor on its radius. It
even does not depend on the angle of the incline, just the height from
which it starts.
Do the same calculation with a cylinder and determine which will
reach the bottom of the incline first, the sphere or the cylinder if
released together.
ANGULAR MOMENTUM:
A rotating object has rotational momentum called angular momentum,
L.
L = Iw
Angular momentum and linear momentum are different physical
quantities and thus do not add.
Whenever a system’s rotational motions are governed exclusively by
internal torques, the system’s total angular momentum will be
constant. Under such circumstances angular momentum is conserved.
Li = Lf .
EXAMPLE: A merry-go-round on a playground has a radius or 1.5 m
and a mass of 225 kg. One child is sitting on its outer edge as it rotates
1 rev/s. If the mass of the child is 50 kg, what will the new rotation rate
be when the child crawls half way to the center of the merry-go-round?
ANSWER: A merry-go-round on a playground has a radius or 1.5 m
and a mass of 225 kg. One child is sitting on its outer edge as it rotates
1 rev/s. If the mass of the child is 50 kg, what will the new rotation rate
be when the child crawls half way to the center of the merry-go-round?
L i = Lf
I system ,iwi = I system , f w f

1
2
2
2
mmgr rmgr
 mc rc2,i  wi =  12 mmgr rmgr
 mc rc2, f  w f
wf =
1
2
1
2
2
mmgr rmgr
 mc rc2,i
2
mgr mgr
m r
wi
m r
2
c c, f
225kg 1.5m    50kg 1.5m 

rad
wf =
2
p

s 
2
2
1
2  225kg 1.5m    50kg  .75m 
2
1
2
w f = 8.17 rads

f = 1.3Hz or
2
rev
s
EQUILIBRIUM:
The forces acting on an object in equilibrium must satisfy both the
first and second conditions of equilibrium.
F
x
=0
F
y
=0
t = 0
All
An object in equilibrium has no linear and no angular accelerations.
That does not mean it isn’t moving, just not accelerating. When an
object is stationary it is said to be in static equilibrium. Many of the
equilibrium problems here will be static equilibrium problems.
THE SECOND CONDITION OF EQUILIBRIUM:
An object in rotational equilibrium has uniform angular velocity, w =
constant. This means its angular acceleration, a, is zero. In order for
a to be zero, the net or total torque acting on the object must be zero.
t
i
=0
t
i
=

ti =  ti
All
All
CCW

CCW
CW
ti   ti
CW
EXAMPLE:
A uniform rod of length l = 4m
and mass m = 75kg is hinged to
a wall at the left and supported
at the right by a cable. The
cable is attached to the rod l/4
from its right edge. The rod
makes a 30o angle with the
horizontal and the cable makes
a 45o angle with the rod.
Calculate the tension in the
cable and the force exerted on
the rod by the hinge.
cable
wall
rod
45o
30o
hinge
Solution on next several slides
STEP 1:
cable
wall
T
rod
45o
30o
R
hinge
W
Add Force Vectors: T = Tension, W = Weight, R = Reaction
STEP 1:
165O
cable
wall
210O
T
rod
45
pivot
o
30o
R
hinge
W
Add Force Vectors: T = Tension, W = Weight, R = Reaction
Add needed angles.
STEP 1:
165O
cable
wall
210O
T
rod
45o
3l/4= 3m
30o
R
hinge
l/2 = 2m
W
Add Force Vectors: T = Tension, W = Weight, R = Reaction
Add needed angles.
Add Lever Arms.
STEP 2: Force analysis
F cos(165o )iˆ  F sin(165 o ) ˆj
W=
0iˆ

m  gjˆ
R=
Rxiˆ

R y ˆj
F=
First Condition of Equilibrium:
F
= Rx  T cos165 = 0
F
= Ry  735 N  T sin165 = 0
x
y
o
o
On the next slide these
force equations will be
used to solve for Rx
and Ry.
STEP 3: Torque Analysis
To analyze the torques acting on the rod, construct to following table or
small spreadsheet. The column headings are:
F = force
F = force component perpendicular to lever arm
r = lever arm
t=
the torque, which is F r
dir = direction of torque (CW or CCW)
The entries in each row
correspond to values associated
with one force on the rod. One
row for each force.
The next slide continues the
example solution.
F
F
r
t
dir
g = 9.8 m/s2 was used
STEP 3: Torque Analysis
F
F
r
T
T sin(45o )
W
mg sin(60o )
R

3l
4
l
2

t
dir
3Tl sin(45o )
CCW
4
mgl sin(60o )
CW
2
0

Second Condition of Equilibrium:
t
CCW
= t CW
T  2.1213m  = 1273.06 Nm
T = 600.13N
ANSWER
From force equations on previous
slide and using T=600.13N:
Rx=579.68N
Ry=579.67N
Then using the Pythagorean
Theorem and tan-1:
R = 819.8N
q = 45o ANSWER