Download Rigid Body Dynamics chapter 10 continues

Rigid Body Dynamics
chapter 10 continues
around and around we go …
Rigid Body Rotation
( KE ) i 
vi  ri
1
mi vi2
2


1
1
mi 2 ri 2  mi ri 2  2
2
2
1
1

( KE ) total   mi ri 2  2  I 2
2 i
2

( KE ) i 
Moment of Inertia:
Rotation Axis


I   mi ri 2 
 i

2
CORRESPONDENCE:
1 2
1 2
mv  I
2
2
v 
mI
Rotational Kinetic Energy



We there have an analogy between the kinetic
energies associated with linear motion (K = ½ mv 2)
and the kinetic energy associated with rotational
motion (KR= ½ I2)
Rotational kinetic energy is not a new type of
energy, the form is different because it is applied to
a rotating object
The units of rotational kinetic energy are also Joules
(J)
Important Concept:
Moment of Inertia

The definition of moment of inertia is
I   ri 2mi


i
The dimensions of moment of inertia are ML2
and its SI units are kg.m2
We can calculate the moment of inertia of an
object more easily by assuming it is divided
into many small volume elements, each of
mass Dmi
Moment of Inertia, cont

We can rewrite the expression for I in terms of Dm
lim
2
2
I  Dmi 
r
D
m

r
0 i
i
 dm
i


With the small volume segment assumption,
If r is constant, the integral can be evaluated with
known geometry,
its variation with position
I   rotherwise
r 2dV
must be known
Question – WHAT IS THE
MOMENT OF INERTIA OF THIS
OBJECT??
Let’s Look at the possibilities
c
d
Two balls with masses M and m are connected by a rigid rod
of length L and negligible mass as in Figure P10.22. For an
axis perpendicular to the rod, show that the system has the
minimum moment of inertia when the axis passes through
the center of mass. Show that this moment of inertia is
I = L2, where = mM/(m + M).
Remember the Various Densities



Volumetric Mass Density –> mass per unit
volume: r = m / V
Face Mass Density –> mass per unit
thickness of a sheet of uniform thickness, t :
s  rt
Linear Mass Density –> mass per unit length
of a rod of uniform cross-sectional area: l =
m / L = rA
Moment of Inertia of a Uniform
Thin Hoop

Since this is a thin
hoop, all mass
elements are the same
distance from the
center
I   r dm  R
2
I  MR 2
2
 dm
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   r dm   x
dx
L/ 2
L
1
I  ML2
12
2
L/2
2
Moment of Inertia of a Uniform
Solid Cylinder


Divide the cylinder into
concentric shells with
radius r, thickness dr
and length L
Then for I
dV=L(2rdr)
I   r 2dm 
1
I z  MR 2
2
2
r
  2r Lr dr 
Moments of Inertia of Various
Rigid Objects
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
Howcome??

ri


mi
L

Not in
same
plane
The new axis is
parallel to the old
axis of rotation.
Assume that the
object rotates about
an axis parallel to
the z axis.
The new axis is
parallel to the
original axis.
 mi
From the top:
L

NEW
ri

OLD
I new   mi (L  ri ) 2
i
note : a  a  a 2
I new   mi (L  L  2L  ri  ri 2 )
i
I new  ML2   mi r12  2L   mi ri
i
I new
i


 ML  I old  2L   mi ri 
 i

2
Remember the Center of Mass?
rCM
1

M
mr
i i
i
Since for our problem the sum is ABOUT the center of
mass, rCM must be zero
rCM  0
So:
I new


 ML  I old  2L   mi ri 
 i

2
ZERO
Inew = ICM + ML2
Parallel-Axis Theorem
Example



The axis of rotation
goes through O
The axis through the
center of mass is
shown
The moment of inertia
about the axis through
O would be IO = ICM +
MD 2
Moment of Inertia for a Rod
Rotating Around One End

The moment of inertia of
the rod about its center is
1
I CM  ML2
12

D is ½ L
Therefore,

I  I CM  MD 2
2
1
 L 1
I  ML2  M    ML2
12
2 3
Many machines employ cams for various purposes, such as
opening and closing valves. In Figure P10.29, the cam is a
circular disk rotating on a shaft that does not pass through the
center of the disk. In the manufacture of the cam, a uniform solid
cylinder of radius R is first machined. Then an off-center hole of
radius R/2 is drilled, parallel to the axis of the cylinder, and
centered at a point a distance R/2 from the center of the cylinder.
The cam, of mass M, is then slipped onto the circular shaft and
welded into place. What is the kinetic energy of the cam when it
is rotating with angular speed about the axis of the
Torque (Another Vector): t
Torque

Torque, t, is the tendency of a force to rotate
an object about some axis


Torque is a vector
t = r F sin f = Fd = rXF



F is the force
f is the angle the force makes with the horizontal
d is the moment arm (or lever arm)
More Torqueing

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 Φ
Torque


The horizontal component of F (F cos f) has
no tendency to produce a rotation
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
Right Hand Screw Rule
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
St  t1  t2  F1d1 –
F2d2
Torque vs. Force

Forces can cause a change in linear
motion


Described by Newton’s Second Law
Forces can cause a change in rotational
motion


The effectiveness of this change depends on
the force and the moment arm
The change in rotational motion depends on
the torque
Torque Units

The SI units of torque are N.m


Although torque is a force multiplied by a
distance, it is very different from work and
energy
The units for torque are reported in N.m and not
changed to Joules
Torque and Angular
Acceleration


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
 Ftr=matr=m(ar)r=mr2a

tIa
More Associations:
t  Ia
F  ma
t F
a a
I m
SO?
 Worry
about concepts.
 Don’t worry about too many new
“formulas”.
Torque and Angular
Acceleration, Extended



Consider the object consists
of an infinite number of
mass elements dm of
infinitesimal size
Each mass element rotates
in a circle about the origin,
O
Each mass element has a
tangential acceleration
Torque and Angular
Acceleration, Extended cont.

From Newton’s Second Law


The torque associated with the force and
using the angular acceleration gives


dFt = (dm) at
dt = r dFt = atr dm = ar 2 dm
Finding the net torque
2
2
t

a
r
dm

a
r



 dm

This becomes St  Ia
Torque and Angular
Acceleration, Extended final


This is the same relationship that applied to a
particle
The result also applies when the forces have
radial components


The line of action of the radial component must
pass through the axis of rotation
These components will produce zero torque about
the axis
Torque and Angular
Acceleration, Wheel Example

The wheel is rotating
and so we apply St  Ia


The tension supplies the
tangential force
The mass is moving in
a straight line, so apply
Newton’s Second Law

SFy = may = mg - T
Problem
Find the net torque on the wheel in Figure P10.31
about the axle through O if a = 10.0 cm and b = 25.0
cm.
Anudder one
An electric motor turns a flywheel through a drive belt that joins a pulley on
the motor and a pulley that is rigidly attached to the flywheel, as shown in
Figure P10.39. The flywheel is a solid disk with a mass of 80.0 kg and a
diameter of 1.25 m. It turns on a frictionless axle. Its pulley has much
smaller mass and a radius of 0.230 m. If the tension in the upper (taut)
segment of the belt is 135 N and the flywheel has a clockwise angular
acceleration of 1.67 rad/s2, find the tension in the lower (slack) segment of
the belt.
Torque and Angular
Acceleration, Multi-body Ex., 1


Both masses move in
linear directions, so
apply Newton’s Second
Law
Both pulleys rotate, so
apply the torque
equation
Torque and Angular
Acceleration, Multi-body Ex., 2


The mg and n forces on each pulley act at the
axis of rotation and so supply no torque
Apply the appropriate signs for clockwise and
counterclockwise rotations in the torque
equations
32.
The tires of a 1 500-kg car are 0.600 m in diameter
and the coefficients of friction with the road surface are s =
0.800 and k = 0.600. Assuming that the weight is evenly
distributed on the four wheels, calculate the maximum
torque that can be exerted by the engine on a driving wheel,
without spinning the wheel. If you wish, you may assume
the car is at rest.
Work in Rotational Motion


Find the work done by F on
the object as it rotates through
an infinitesimal distance ds = r
dq
dW = F . d s
= (F sin f) r dq
dW = t dq
The radial component of F
does no work because it is
perpendicular to the
displacement
Power in Rotational Motion

The rate at which work is being done in a
time interval dt is
dW
dq
Power 
t
 t
dt

dt
This is analogous to P = Fv in a linear system
Work-Kinetic Energy Theorem
in Rotational Motion

The work-kinetic energy theorem for rotational motion
states that the net work done by external forces in rotating
a symmetrical rigid object about a fixed axis equals the
change in the object’s rotational kinetic energy
d
dq
dW  tdq  Iadq  I
dq  Id
 Id
dt
dt
f
W  
i
1 2 1 2
I  d  I  f  I i
2
2
Work-Kinetic Energy Theorem,
General

The rotational form can be combined with the
linear form which indicates the net work done
by external forces on an object is the change
in its total kinetic energy, which is the sum of
the translational and rotational kinetic
energies
Energy in an Atwood Machine,
Example


The blocks undergo
changes in translational
kinetic energy and
gravitational potential
energy
The pulley undergoes a
change in rotational
kinetic energy
Summary of Useful Equations
Rolling Object

The red curve shows the path moved by a point on the rim of
the object


This path is called a cycloid
The green line shows the path of the center of mass of the
object
Pure Rolling Motion


In pure rolling motion, an object rolls without
slipping
In such a case, there is a simple relationship
between its rotational and translational
motions
Rolling Object, Center of Mass

The velocity of the
center of mass is
ds
dq
vCM 

dt
R
dt
 R
The acceleration of the
center of mass is
aCM 
dvCM
d
R
 Ra
dt
dt
Rolling Object, Other Points


A point on the rim, P,
rotates to various
positions such as Q
and P ’
At any instant, the point
on the rim located at
point P is at rest
relative to the surface
since no slipping
occurs
Rolling Motion Cont.

Rolling motion can be
modeled as a
combination of pure
translational motion
and pure rotational
motion
Total Kinetic Energy of a
Rolling Object

The total kinetic energy of a rolling object is
the sum of the translational energy of its
center of mass and the rotational kinetic
energy about its center of mass

K = ½ ICM w2 + ½ MvCM2
Total Kinetic Energy, Example

Accelerated rolling
motion is possible only
if friction is present
between the sphere
and the incline

The friction produces the
net torque required for
rotation
Total Kinetic Energy, Example
cont






Despite the friction, no loss of mechanical energy
occurs because the contact point is at rest relative
to the surface at any instant
Let U = 0 at the bottom of the plane
K f + U f = Ki + U i
Kf = ½ (ICM / R 2) vCM2 + ½ MvCM2
Ui = Mgh
U f = Ki = 0