physics Mechanics.ppt

Physics 106: Mechanics
Lecture 02
Wenda Cao
NJIT Physics Department

April 7, 2009
Rotational Equilibrium and
Rotational Dynamics
 Rotational Kinetic Energy
 Moment of Inertia
 Torque
 Angular acceleration
 Newton 2nd Law for
Rotational Motion: Torque
and angular acceleration

April 7, 2009
Rotational Kinetic Energy
 An object rotating about z axis with
an angular speed, ω, has rotational
kinetic energy
 Each particle has a kinetic energy of
 Ki = ½ mivi
2
 Since the tangential velocity
depends on the distance, r, from the
axis of rotation, we can substitute
vi = wri

April 7, 2009
Rotational Kinetic Energy, cont
 The total rotational kinetic energy of the rigid
object is the sum of the energies of all its
particles
 Where I is called the moment of inertia
2 2
2 2 2
1
2
1 1
2 2
R i i i
i i
R i i
i
K K m r
K m r I
w
w w
 
 
 
 
 
 


April 7, 2009
Rotational Kinetic Energy, final
 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)

April 7, 2009
Moment of Inertia of Point Mass
 For a single particle, the definition of moment
of inertia is
 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)
2
mr
I 

April 7, 2009
Moment of Inertia of Point Mass
 For a composite particle, the definition of moment of
inertia is
 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
 Consider an unusual baton made up of four sphere
fastened to the ends of very light rods
 Find I about an axis perpendicular to the page and
passing through the point O where the rods cross
...
2
4
4
2
3
3
2
2
2
2
1
1
2






 r
m
r
m
r
m
r
m
r
m
I i
i
2
2
2
2
2
2
2
2
2 mb
Ma
Ma
mb
Ma
mb
r
m
I i
i 








April 7, 2009
The Baton Twirler
 Consider an unusual baton made
up of four sphere fastened to the
ends of very light rods. Each rod is
1.0m long (a = b = 1.0 m). M =
0.3 kg and m = 0.2 kg.
 (a) Find I about an axis
perpendicular to the page and
passing through the point where
the rods cross. Find KR if angular
speed is w
 (b) The majorette tries spinning
her strange baton about the axis y,
calculate I of the baton about this
axis and KR if angular speed is w

April 7, 2009
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
 With the small volume segment assumption,
 If r is constant, the integral can be evaluated with
known geometry, otherwise its variation with position
must be known
lim 2 2
0
i
m i i
i
I r m r dm
D 
 D 
 
2
I r dV
r
 

April 7, 2009
Moment of Inertia
of a Uniform Rigid Rod
 The shaded area has
a mass
 dm = l dx
 Then the moment of
inertia is
/ 2
2 2
/ 2
2
1
12
L
y L
M
I r dm x dx
L
I ML

 

 

April 7, 2009
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

April 7, 2009
Moment of Inertia
of a Uniform Rigid Rod
 The moment of inertia
about y is
 The moment of inertia
about y’ is
/ 2
2 2
/ 2
2
1
12
L
y L
M
I r dm x dx
L
I ML

 

 
2
2
2
2
'
3
1
)
2
(
12
1
ML
L
M
ML
MD
I
I CM
y 





April 7, 2009
Moment of Inertia for some other
common shapes

April 7, 2009
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

April 7, 2009
Torque Definition
 Torque, t, is the tendency of a force to rotate
an object about some axis
 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, with F
perpendicular to r. The magnitude of the
torque is given by
rF

t

April 7, 2009
Torque Units and Direction
 The SI units of torque are N.m
 Torque is a vector quantity
 Torque magnitude is given by
 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
rF

t

April 7, 2009
Net Torque
 The force will tend to
cause a counterclockwise
rotation about O
 The force will tend to
cause a clockwise
rotation about O
 St  t1  t2  F1d1 – F2d2
 If St  0, starts rotating
 If St  0, rotation rate
does not change
1
F
2
F
 Rate of rotation of an
object does not change,
unless the object is acted
on by a net torque

April 7, 2009
General Definition of Torque
 The applied force is not always perpendicular to the
position vector
 The component of the force perpendicular to the
object will cause it to rotate
 When the force is parallel to the position vector, no
rotation occurs
 When the force is at some angle, the perpendicular
component causes the rotation

April 7, 2009
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
   0° or   180 °:
torque are equal to zero
   90° or   270 °: magnitude of torque attain
to the maximum

t sin
rF


April 7, 2009
Understand sinθ
 The component of the force
(F cos  ) has no tendency
to produce a rotation
 The moment arm, d, is the
perpendicular distance from
the axis of rotation to a line
drawn along the direction of
the force
d = r sin
Fd
rF 
 
t sin

April 7, 2009
The Swinging Door
 Two forces are applied to the door, as shown in
figure. Suppose a wedge is placed 1.5 m from
the hinges on the other side of the door. What
minimum force must the wedge exert so that
the force applied won’t open the door? Assume
F1 = 150 N, F2 = 300 N, F3 = 300 N, θ = 30°
F1
2.0m
F2
θ
F3

April 7, 2009
Torque on a Rotating Object
 Consider a particle of mass m rotating in a circle of
radius r under the influence of tangential force
 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
t = mr2
t = I
t
F

April 7, 2009
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
t = mr2
 The net torque on the disk is given by
St = (Smr2)
 A constant of proportionality is the moment of
inertia,
I = Smr2 = m1r1
2 + m2r2
2 + m3r3
2 + …
 So, we can rewrite it as
St = I

April 7, 2009
Newton’s Second Law for a
Rotating Object
 When a rigid object is subject to a net torque (≠0),
it undergoes an angular acceleration
 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
I
t 
S 
  ma
F

April 7, 2009
The Falling Object
 A solid, frictionless cylindrical reel of
mass M = 3.0 kg and radius R =
0.4m is used to draw water from a
well. A bucket of mass m = 2.0 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 ?

April 7, 2009
Example, Newton’s Second Law
for Rotation
 Draw free body diagrams
of each object
 Only the cylinder is
rotating, so apply St = I 
 The bucket is falling, but
not rotating, so apply SF =
m a
 Remember that a =  r
and solve the resulting
equations

physics Mechanics.ppt

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physics Mechanics.ppt