# Download Lecture 02: Rotational Dynamics I

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Transcript
```Physics 106: Mechanics
Lecture 02
Wenda Cao
NJIT Physics Department
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 mivi2
 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
1
K R   K i   mi ri 2w 2
i
i 2
1
1 2
2 2
K R    mi ri w  Iw
2 i
2


Where I is called the moment of inertia
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
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)

April 7, 2009
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  ...





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
I   mi ri  mb2  Ma 2  mb2  Ma 2  2Ma 2  2mb2
2
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,
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
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
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
M
I y   r dm  
x
dx
L / 2
L
1
I
ML2
12
2
L/2
2
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
I y   r dm  
2
L/2
L / 2
I

x2
M
dx
L
1
ML2
12
The moment of inertia
I y '  I CM
1
L 2 1
2
 MD  ML  M ( )  ML2
12
2
3
2
April 7, 2009
Moment of Inertia for some other
common shapes
April 7, 2009
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

t  rF
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

t  rF

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
April 7, 2009
Net Torque
The force F1 will tend to
cause a counterclockwise
 The force F2 will tend to
cause a clockwise
 St  t1  t2  F1d1 – F2d2
 If St  0, starts rotating
 Rate of rotation of an
 If St  0, rotation rate
object does not change,
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
t  rF sin 
  0° or   180 °:
torque are equal to zero
   90° or   270 °: magnitude of torque attain
to the maximum

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

t  rF sin   Fd
d = r 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°
F2
F3
2.0m
θ
F1
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 Ft
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
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 = m1r12 + m2r22 + m3r32 + …
 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
St  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
April 7, 2009
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 =
ma
Remember that a =  r
and solve the resulting
equations
April 7, 2009
```
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