Rigid Body Dynamics chapter 10 continues

1
Rigid Body Dynamicschapter 10 continues

• around and around we go

2
Rigid Body Rotation
Moment of Inertia
3
CORRESPONDENCE
4
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 Iw2)
• 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)

5
Important ConceptMoment of Inertia

• The definition of moment of inertia is
• 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

6
Moment of Inertia, cont

• 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

7
Question WHAT IS THE MOMENT OF INERTIA OF THIS
OBJECT??
WHICH AXIS ?
8
Lets Look at the possibilities
c d
9
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).
10
Remember the Various Densities

• Volumetric Mass Density gt mass per unit volume
  r m / V
• Face Mass Density gt mass per unit thickness of a
  sheet of uniform thickness, t s rt
• Linear Mass Density gt mass per unit length of a
  rod of uniform cross-sectional area l m / L
  rA

11
Moment of Inertia of a Uniform Thin Hoop

• Since this is a thin hoop, all mass elements are
  the same distance from the center

12
Moment of Inertia of a Uniform Rigid Rod

• The shaded area has a mass
• dm l dx
• Then the moment of inertia is

13
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

dVL(2prdr)
14
Moments of Inertia of Various Rigid Objects
15
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

16
Howcome??

• 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.

Not in same plane
17
From the top
Center of Mass
18
Remember the Center of Mass?
Since for our problem the sum is ABOUT the center
of mass, rCM must be zero
19
So
ZERO
Inew ICM ML2
QED
20
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

21
Moment of Inertia for a Rod Rotating Around One
End

• The moment of inertia of the rod about its center
  is
• D is ½ L
• Therefore,

22
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
23
Torque (Another Vector) t
F
24
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)

25
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 F

26
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
27
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

28
Torque vs. Force

• Forces can cause a change in linear motion
• Described by Newtons 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

29
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

30
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
• Ftrmatrm(ar)rmr2a
• tIa

31
More Associations
How about that?
32
SO?

• Worry about concepts.
• Dont worry about too many new formulas.

33
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

34
Torque and Angular Acceleration, Extended cont.

• From Newtons Second Law
• dFt (dm) at
• The torque associated with the force and using
  the angular acceleration gives
• dt r dFt atr dm ar 2 dm
• Finding the net torque
• This becomes St Ia

35
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

36
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
  Newtons Second Law
• SFy may mg - T

37
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.
38
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.
39
Torque and Angular Acceleration, Multi-body Ex., 1

• Both masses move in linear directions, so apply
  Newtons Second Law
• Both pulleys rotate, so apply the torque equation

40
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

41
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.
42
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

43
Power in Rotational Motion

• The rate at which work is being done in a time
  interval dt is
• This is analogous to P Fv in a linear system

44
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
  objects rotational kinetic energy

45
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

46
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

47
Summary of Useful Equations
48
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

49
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

50
Rolling Object, Center of Mass

• The velocity of the center of mass is
• The acceleration of the center of mass is

51
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

52
Rolling Motion Cont.

• Rolling motion can be modeled as a combination of
  pure translational motion and pure rotational
  motion

53
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

54
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

55
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
• Kf U f Ki Ui
• Kf ½ (ICM / R 2) vCM2 ½ MvCM2
• Ui Mgh
• Uf Ki 0