Two-Dimensional Rotational Dynamics 8.01 W09D2 Young and Freedman: 1.10 (Vector Product), 10.1-10.2, 10.4, 11.1-11.3;

1
Two-Dimensional Rotational Dynamics8.01W09D2
Young and Freedman 1.10 (Vector Product),
10.1-10.2, 10.4, 11.1-11.3
2
Announcements
Next Reading Assignment Young and Freedman
1.10 (Vector Product) 10.1-10.2, 10.5-10.6
11.1-11.3
3
Rigid Bodies

• Rigid body An extended object in which the
  distance between any two points in the object is
  constant in time. Examples sphere, disk
• Effect of external forces (the solid arrows
  represent forces)

4
Main Idea Fixed Axis Rotation of Rigid Body

• Torque produces angular acceleration about center
  of mass
• is the moment of inertial about the center
  of mass
• is the angular acceleration about center of
  mass

5
Recall Fixed Axis Rotation Kinematics

• Angle variable
• Angular velocity
• Angular acceleration
• Mass element
• Radius of orbit
• Moment of inertia
• Parallel Axis Theorem

6
Torque as a Vector

• Force exerted at a point P on a rigid
  body.
• Vector from a point S to the point P.

Torque about point S due to the force exerted at
point P
7
Summary Cross Product

• Magnitude equal to the area of the parallelogram
  defined by the two vectors
• Direction determined by
• the Right-Hand-Rule

8
Torque Magnitude and Direction

• Magnitude of torque about a point S due to force
  acting at point P

Direction of torque Perpendicular to the
plane formed by and .Determined
by the Right-Hand-rule.
9
Properties of Cross Products
10
Cross Product of Unit Vectors

• Unit vectors in Cartesian coordinates

11
Components of Cross Product
12
Concept Question Torque

• Consider two vectors with x gt 0 and
• with Fx gt 0 and Fz gt 0 . The cross product
• points in the
• x-direction
• -x-direction
• y-direction
• -y-direction
• z-direction
• -z-direction
• None of the above directions

13
Concept Question Torque Answer

• Answer 4. We calculate the cross product noting
  that in a right handed choice of unit vectors,
  and
• Since and , the direction of the cross
  product is in the negative y-direction.

14
Concept Question Magnitude of Torque

• In the figure, a force of magnitude F is applied
  to one end of a lever of length L. What is the
  magnitude of the torque about the point S?
• FL sin?
• FL cos?
• FL tan?
• None of the above

15
Concept Question Magnitude of Torque Answer

• Answer 2

16
Torque due to Uniform Gravitational Force

• The total torque on a rigid body due to the
  gravitational force can be determined by placing
  all the gravitational force at the center-of-mass
  of the object.

17
Recall Rotational Kinematics

• Individual element of mass
• Radius of orbit
• Tangential velocity
• Tangential acceleration
• Radial Acceleration

18
Dynamics Newtons Second Law and Torque about S

• Tangential force on mass element produces torque
• Newtons Second Law
• Torque about S
• z-component of torque about S

19
Torque, Moment of Inertia and Angular Acceleration

• Component of the total torque about an axis
  passing through S is the sum over all elements
• Recall Moment of Inertia about and axis passing
  through S
• Summary

20
Fixed Axis Rotational Dynamics
21
Worked Example Moment of Inertia Wheel

• An object of mass m is attached to a string
  which is wound around a disc of radius Rd. The
  object is released and takes a time t to fall a
  distance s. What is the moment of inertia of the
  disc?

22
Analysis Measuring Moment of Inertia

• Free body force diagrams and force equations
• Rotational equation
• Constraint
• Solve for moment of inertia
• Time to travel distance s

23
Demo Moment of Inertia Wheel

• Measuring the moment of inertia.
• Radius of disc
• Mass of disc
• Mass of weight holder
• Theoretical result

24
Concept Question Chrome Inertial Wheel

• A fixed torque is applied to the shaft of the
  chrome inertial wheel. If the four weights on the
  arms are slid out, the component of the angular
  acceleration along the shaft direction will
• increase.
• decrease.
• remain the same.

25
Concept Question Chrome Inertial Wheel Answer

• Answer 2. Torque about the central axis is
  proportional to the component of the angular
  acceleration along that axis. The proportionality
  constant is the moment of inertia about that
  axis. By pushing the weights out, the moment of
  inertia has increased. If the applied torque is
  constant then the component of the angular
  acceleration must decrease.

26
Problem Solving StrategyTwo Dimensional Rotation

• Step 1 Draw free body force diagrams for each
  object and indicate the point of application of
  each force
• Step 2 Select point to compute torque about
  (generally select center of mass)
• Step 3 Choose coordinate system. Indicate
  positive direction for increasing rotational
  angle.
• Step 4 Apply Newtons Second Law and Torque Law
  to obtain equations
• Step 5 Look for constraint condition between
  rotational acceleration and any linear
  accelerations.
• Step 6 Design algebraic strategy to find
  quantities of interest

27
Rotor Moment of Inertia
28
Table Problem Moment of Inertia Wheel

• A steel washer is mounted on a cylindrical
  rotor . The inner radius of the washer is R. A
  massless string, with an object of mass m
  attached to the other end, is wrapped around the
  side of the rotor and passes over a massless
  pulley. Assume that there is a constant
  frictional torque about the axis of the rotor.
  The object is released and falls. As the mass
  falls, the rotor undergoes an angular
  acceleration of magnitude a1. After the string
  detaches from the rotor, the rotor coasts to a
  stop with an angular acceleration of magnitude
  a2. Let g denote the gravitational constant.
• What is the moment of inertia of the rotor
  assembly (including the washer) about the
  rotation axis?

29
Torque and Static Equilibrium
30
Conditions for Static Equilibrium

• Translational equilibrium the sum of the forces
  acting on the rigid body is zero.
• (2) Rotational Equilibrium the vector sum of the
  torques about any point S in a rigid body is zero.

31
Concept Question Tipping

• A box, with its center-of-mass off-center as
  indicated by the dot, is placed on an inclined
  plane. In which of the four orientations shown,
  if any, does the box tip over?

32
Concept Question Tipping Answer

• Answer 3. In order to tip over, the box must
  pivot about its bottom left corner. Only in case
  3 does the torque about this pivot (due to
  gravity) rotate the box in such a way that it
  tips over.

33
Problem Solving Strategy Static Equilibrium

• Force
• Identify System and draw all forces and where
  they act on Free Body Force Diagram
• Write down equations for static equilibrium of
  the forces sum of forces is zero
• Torque
• Choose point to analyze the torque about.
• Choose sign convention for torque
• Calculate torque about that point for each force.
  (Note sign of torque.)
• Write down equation corresponding to condition
  for static equilibrium sum of torques is zero

34
Table Problem Standing on a Hill
A person is standing on a hill that is sloped at
an angle a with respect to the horizontal. The
persons legs are separated by a distance d, with
one foot uphill and one downhill. The center of
mass of the person is at a distance h above the
ground, perpendicular to the hillside, midway
between the persons feet. Assume that the
coefficient of static friction between the
persons feet and the hill is sufficiently large
that the person will not slip. a) What is the
magnitude of the normal force on each foot? b)
How far must the feet be apart so that the normal
force on the upper foot is just zero? (This is
the instant when the person starts to rotate and
fall over.)
35
Rotational Work

• Tangential force
• Displacement vector
• work for a small displacement

36
Rotational Work

• Newtons Second Law
• Tangential acceleration
• Work for small displacement
• Summation becomes integration for continuous body

37
Rotational Work

• Rotational work for a small displacement
• Torque about S
• Infinitesimal rotational work
• Integrate total work

38
Rotational Work-Kinetic Energy Theorem

• Infinitesimal rotational work
• Integrate rotational work
• Kinetic energy of rotation about S

39
Rotational Power

• Rotational power is the time rate of doing
  rotational work
• Product of the applied torque with the angular
  velocity

40
Table Problem Change in Rotational Energy and
Work

• Suppose that a rotor of moment of inertia Ir
  is slowing down during the interval t1, t2
  according to
• ?(t) ?(t1)-a t,
• where a ?(t1)/(t2 ). Use work energy
  techniques to find the frictional torque acting
  on the rotor.

41
Table Problem Solution Change in Rotational
Energy and Work

• Suppose that a rotor of moment of inertia Ir
  is slowing down during the interval t1, t2
  according to
• ?(t) ?(t1)-a t,
• where a ?(t1)/(t2). Use work energy
  techniques to find the frictional torque acting
  on the rotor.