Download Lecture 03: Rotational Dynamics II: 2nd Law

Physics 106: Mechanics
Lecture 03
Wenda Cao
NJIT Physics Department
Rotational Equilibrium and
Rotational Dynamics II
Rotational Kinetic Energy
 Moment of Inertia
 Torque
 Newton 2nd Law for
Rotational Motion: Torque
and angular acceleration

Februaryl 3, 2011
Rotational Kinetic Energy
There is an analogy between the kinetic
energies associated with linear motion (K = ½
mv 2) and the kinetic energy associated with
rotational motion (KR= ½ Iw2)
 Rotational kinetic energy is not a new type of
energy, the form is different because it is applied
to a rotating object
 Units of rotational kinetic energy are Joules (J)

Februaryl 3, 2011
Moment of Inertia of Point Mass

For a single particle, the definition of moment
of inertia is
2
I  mr


m is the mass of the single particle
r is the rotational radius
SI units of moment of inertia are kg.m2
 Moment of inertia and mass of an object are
different quantities
 It depends on both the quantity of matter and
its distribution (through the r2 term)

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Moment of Inertia of Point Mass

For a composite particle, the definition of moment of
inertia is
2
2
2
2
2
I   mi ri  m1r1  m2 r2  m3r3  m4r4  ...



P
mi is the mass of the ith single particle
ri is the rotational radius of ith particle
SI units of moment of inertia are kg.m2
cm
m
L
P
m
cm
m
P
m
cm
m
L/2
L
L/2
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m
Moment of Inertia of Extended Objects


Divided the extended objects into many small volume
elements, each of mass Dmi
We can rewrite the expression for I in terms of Dm
I  Dmi lim0  ri 2 Dmi   r 2dm
i

With the small volume segment assumption,
I   r r 2dV

If r is constant, the integral can be evaluated with
known geometry, otherwise its variation with position
must be known
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Moment of Inertia for some other
common shapes
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Parallel-Axis Theorem



In the previous examples, the axis of
rotation coincided with the axis of
symmetry of the object
For an arbitrary axis, the parallel-axis
theorem often simplifies calculations
The theorem states
I = ICM + MD 2



I is about any axis parallel to the axis through
the center of mass of the object
ICM is about the axis through the center of
mass
D is the distance from the center of mass axis
to the arbitrary axis
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• Rotation axes perpendicular to plane of figure
• Masses on the corners of a rectangle, sides a & b
h2 = (a/2)2 + (b/2)2
 mir2,i
I
• About an axis through the CM:
ICM




b
m
• About an axis “P” through a corner:
IP  0  ma2  m b2  m a2  b2
m
h
 a2 b2 
2
2
 4 m h  4m  
  m a b
4 
 4
2
a
m

X
cm
h
m
P
 2m a2  b2

• Using the Parallel Axis Theorem directly for the same corner axis:
IP  Icm  Mtoth
2



 a2 b2 
2
2
 m a  b  4m  
  2m a  b
4 
 4
2
2
Februaryl 3, 2011

Force vs. Torque




Forces cause accelerations
What cause angular accelerations ?
A door is free to rotate about an axis through O
There are three factors that determine the
effectiveness of the force in opening the door:



The magnitude of the force
The position of the application of the force
The angle at which the force is applied
Februaryl 3, 2011
General Definition of Torque

Let F be a force acting on an object, and let r be a
position vector from a rotational center to the point of
application of the force. The magnitude of the torque is
given by
  rF sin 
  0° or   180 °:
torque are equal to zero
   90° or   270 °: torque attain to the maximum


Torque will have direction


If the turning tendency of the force is counterclockwise, the
torque will be positive
If the turning tendency is clockwise, the torque will be negative
Februaryl 3, 2011
Net Torque
The force F1 will tend to
cause a counterclockwise
rotation about O
 The force F2 will tend to
cause a clockwise
rotation about O
 S  1  2  F1d1 – F2d2
 If S  0, starts rotating
 Rate of rotation of an
 If S  0, rotation rate
object does not change,
does not change
unless the object is acted

on by a net torque
Februaryl 3, 2011
Torque on a Rotating Object





Consider a particle of mass m rotating in a circle of
radius r under the influence of tangential force F
t
The tangential force provides a tangential acceleration:
Ft = mat
Multiply both side by r, then
rFt = mrat
Since at = r, we have
rFt = mr2
So, we can rewrite it as
 = mr2
 = I
Februaryl 3, 2011
Torque on a Solid Disk
Consider a solid disk rotating about its axis.
 The disk consists of many particles at various
distance from the axis of rotation. The torque on
each one is given by
 = mr2
 The net torque on the disk is given by
S = (Smr2)
 A constant of proportionality is the moment of
inertia,
I = Smr2 = m1r12 + m2r22 + m3r32 + …
 So, we can rewrite it as
S = I

Februaryl 3, 2011
Newton’s Second Law for a
Rotating Object

When a rigid object is subject to a net torque (≠0),
it undergoes an angular acceleration
S  I
The angular acceleration is directly proportional to
the net torque
 The angular acceleration is inversely proportional to
the moment of inertia of the object
 The relationship is analogous to

 F  ma
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Februaryl 3, 2011
Example 1: second law for rotation
When she is launched from a springboard, a diver's angular speed
about her center of mass changes from zero to 6.20 rad/s in 220
ms. Her rotational inertia about her center of mass is constant at
12.0 kg·m2. During the launch, what are the magnitudes of
(a) her average angular acceleration and
(b) the average external torque on her from the board?
a) Use:
or
wf  w0  t
 ave 

b) Use:
I

w f  w0
t
Dw
Dt
6.20
 28.2 rad/s 2
0.22
 12 kg.m2
ave  I ave  12  28.2  338 N.m
Februaryl 3, 2011
Example 2:  for an unbalanced bar
 Bar is massless and originally horizontal
 Rotation axis at fulcrum point
+y
 N has zero torque
 Find angular acceleration of bar and the linear
acceleration of m1 just after you let go
Use:
net
net
 Itot    
Itot
where: Itot  I1  I2  m1L21  m2L22
net   o,i   m1gL1  m2gL2
What happened to sin() in moment arm?
m1g
N
L2
fulcrum
m2g
Constraints:
Using specific numbers:
Let m1 = m2= m
L1=20 cm, L2 = 80 cm

gL1  gL2

g(0.2 - 0.8)
L21  L22
0.2 2  0.8 2
  8.65 rad/s 2 Clockwise
net
torque

total I
about
pivot
L1
m1gL1  m2gL2
m1L21  m2L22
a1  L1   1.7 m/s 2
Accelerates UP
Februaryl 3, 2011
Newton 2nd Law in Rotation

Suppose everything is as it was in the preceding example, but the
bar is NOT horizontal. Assume both masses are equal. Which of
the following is the correct equation for the angular acceleration?
net   o,i   m1gL1 cos()  m2gL2 cos()
L1
Itot  I1  I2  m1L21  m2L22

[m1L1  m2L2 ]
m1L21

m2L22
g cos()
N
m1g
fulcrum

L2
m2g
 
(L1  L 2 )
L21

L22
gcos()
net  Itot 
Februaryl 3, 2011
Strategy to use the Newton 2nd Law
Many components in the system means several (N) unknowns….
… need an equal number of independent equations
Draw or sketch system. Adopt coordinates, name the variables, indicate
rotation axes, list the known and unknown quantities, …
• Draw free body diagrams of key parts. Show forces at their points of
application. find torques about a (common) axis
Note: can have
• May need to apply Second Law twice to each part
Fnet .eq. 0


but net .ne. 0
 Translation:
Fnet  Fi  ma
 Rotation:



net   i

 I
• Make sure there are enough (N) equations; there may be constraint
equations (extra conditions connecting unknowns)
• Simplify and solve the set of (simultaneous) equations.
• Interpret the final formulas. Do they make intuitive sense? Refer back
to the sketches and original problem
• Calculate numerical results, and sanity check anwers (e.g., right order of
magnitude?)
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Rotating Rod

A uniform rod of length L
and mass M is attached at
one end to a frictionless pivot
and is free to rotate about
the pivot in the vertical plane
as in Figure. The rod is
released from rest in the
horizontal position. What are
the initial angular
acceleration of the rod and
the initial translational
acceleration of its right end?
Februaryl 3, 2011
Rotating Rod
L
  0  Mg ( 2 )
1
I  ML 2
3



Mg ( L / 2) 3 g


2
I
ML / 3
2L
3
at  L  g
2
Februaryl 3, 2011
The Falling Object
A solid, frictionless cylindrical reel of
mass M = 2.5 kg and radius R = 0.2
m is used to draw water from a well.
A bucket of mass m = 1.2 kg is
attached to a cord that is wrapped
around the cylinder.
 (a) Find the tension T in the cord and
acceleration a of the object.
 (b) If the object starts from rest at
the top of the well and falls for 3.0 s
before hitting the water, how far does
it fall ?

Februaryl 3, 2011
Example, Newton’s Second Law
for Rotation




Draw free body diagrams
of each object
Only the cylinder is
rotating, so apply S = I 
The bucket is falling, but
not rotating, so apply SF =
ma
Remember that a =  r
and solve the resulting
equations
Februaryl 3, 2011
•
•
•
•
Cord wrapped around disk, hanging weight
Cord does not slip or stretch  constraint
Disk’s rotational inertia slows accelerations
Let m = 1.2 kg, M = 2.5 kg, r =0.2 m
For mass m:
y
T
mg
Fy  ma  mg  T
T  m (g  a)
r
Unknowns: T, a
support force
at axis “O” has
zero torque
a
FBD for disk, with axis at “o”:
N
T
  0   Tr  I

Mg
I
Tr m(g  a)r
 1
I
Mr 2
1 2
Mr
2
Unknowns: a, 
2
So far: 2 Equations, 3 unknowns Need a constraint:
Substitute and solve:
2mgr 2mr 2

2
Mr
Mr 2
mg
(1  2
m
2mg
)
M
Mr

a   r
from “no
slipping”
assumption
mg
( 24 rad/s 2 )
r(m  M/2)
Februaryl 3, 2011
•
•
•
•
Cord wrapped around disk, hanging weight
Cord does not slip or stretch  constraint
Disk’s rotational inertia slows accelerations
Let m = 1.2 kg, M = 2.5 kg, r =0.2 m
For mass m:
y
T
mg
Fy  ma  mg  T
T  m (g  a)
Unknowns: T, a

mg
( 24 rad/s 2 )
r(m  M/2)
a
mg
( 4.8 m/s 2 )
(m  M/2)
T  m ( g  a)  1.2(9.8 - 4.8)  6N
r
support force
at axis “O” has
zero torque
a
mg
1
1
x f - x f  vi t  at 2  0   4.8  32  21.6m
2
2
Februaryl 3, 2011