Experiment 5: Rotational Dynamics and Angular Momentum 8.01 W10D1 Young and Freedman: 10.5-10.6 - PowerPoint PPT Presentation

About This Presentation
Title:

Experiment 5: Rotational Dynamics and Angular Momentum 8.01 W10D1 Young and Freedman: 10.5-10.6

Description:

Experiment 5: Rotational Dynamics and Angular Momentum 8.01 W10D1 Young and Freedman: 10.5-10.6 – PowerPoint PPT presentation

Number of Views:163
Avg rating:3.0/5.0
Slides: 34
Provided by: padour
Learn more at: http://web.mit.edu
Category:

less

Transcript and Presenter's Notes

Title: Experiment 5: Rotational Dynamics and Angular Momentum 8.01 W10D1 Young and Freedman: 10.5-10.6


1
Experiment 5 Rotational DynamicsandAngular
Momentum8.01W10D1Young and Freedman
10.5-10.6
2
Announcements
Vote Tomorrow Math Review Night Tuesday from
9-11 pm Pset 9 Due Nov 8 at 9 pm No Class
Friday Nov 11 Exam 3 Tuesday Nov 22 730-930
pm W010D2 Reading Assignment Young and Freedman
10.5-10.6
3
Rotor Moment of Inertia
4
Review 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?

5
Review Solution Moment of Inertia of Rotor
  • Force and rotational equations while weight is
    descending
  • Constraint
  • Rotational equation while slowing down
  • Solve for moment of inertia

Speeding up
Slowing down
6
Review Worked Example Change in Rotational
Energy and Work
  • While the rotor is slowing down, use work-energy
    techniques to find frictional torque on the
    rotor.

7
Experiment 5 Rotational Dynamics
8
Review Cross Product
  • Magnitude equal to the area of the parallelogram
    defined by the two vectors
  • Direction determined by
  • the Right-Hand-Rule

9
Angular Momentum of a Point Particle
  • Point particle of mass m moving with a velocity
  • Momentum
  • Fix a point S
  • Vector from the point S to the location
    of the object
  • Angular momentum about the point S
  • SI Unit

10
Cross Product Angular Momentum of a Point
Particle
  • Magnitude
  • moment arm
  • perpendicular momentum

11
Angular Momentum of a Point Particle Direction
  • Direction Right Hand Rule

12
Concept Question
  • In the above situation where a particle is moving
    in the x-y plane
  • with a constant velocity, the magnitude of the
    angular momentum
  • about the point S (the origin)
  • decreases then increases
  • increases then decreases
  • is constant
  • is zero because this is not circular motion

13
Concept Question
Solution 3. As the particle moves in the positive
x-direction, the perpendicular distance from the
origin to the line of motion does not change and
so the magnitude of the angular momentum about
the origin is constant.
14
Table Problem Angular Momentum and Cross Product
A particle of mass m 2 kg moves with a uniform
velocity At time t, the position vector of the
particle with respect ot the point S is Find
the direction and the magnitude of the angular
momentum about the origin, (the point S) at time
t.
15
Angular Momentum and Circular Motion of a Point
Particle
  • Fixed axis of rotation z-axis
  • Angular velocity
  • Velocity
  • Angular momentum about the point S

16
Concept Question
  • A particle of mass m moves in a circle of radius
    R at an angular speed ? about the z axis in a
    plane parallel to but above the x-y plane.
    Relative to the origin
  • 1. is constant.
  • 2. is constant but is
    not.
  • 3. is constant but is not.
  • 4. has no z-component. .

17
Concept Question

Answer 2. The vector represents
a unit vector in the direction of . The
angular momentum about the origin is shown in the
figure to the right The magnitude is constant.
As the particle moves in a circle, the angular
momentum sweeps out a cone (shown in the figure
below) and so the direction of is
changing and hence is not a constant unit
vector.
18
Worked Example Changing Direction of Angular
Momentum
  • A particle of mass m moves in a circle of radius
    R at an angular speed ? about the z axis in a
    plane parallel to but a distance h above the x-y
    plane.
  • Find the magnitude and the direction of the
    angular momentum relative to the origin.
  • Also find the z component of .

19
Solution Changing Direction of Angular Momentum
20
Solution Changing Direction of Angular Momentum
21
Table Problem Angular Momentum of a Two
Particles About Different Points
Two point like particles 1 and 2, each of mass m,
are rotating at a constant angular speed about
point A. How does the angular momentum about the
point B compare to the angular momentum about
point A? What about at a later time when the
particles have rotated by 90 degrees?
22
Table Problem Angular Momentum of Two Particles
  • Two identical particles of mass m move in a
    circle of radius R, 180º out of phase at an
    angular speed ? about the z axis in a plane
    parallel to but a distance h above the x-y plane.
  • a) Find the magnitude and the direction of the
    angular momentum relative to the origin.
  • b) Is this angular momentum relative to the
    origin constant? If yes, why? If no, why is it
    not constant?

23
(No Transcript)
24
Table Problem Angular Momentum of a Ring
  • A circular ring of radius R and mass dm rotates
    at an angular speed ? about the z-axis in a plane
    parallel to but a distance h above the x-y plane.
  • a) Find the magnitude and the direction of the
    angular momentum relative to the origin.
  • b) Is this angular momentum relative to the
    origin constant? If yes, why? If no, why is it
    not constant?

25
(No Transcript)
26
Concept Question
  • A non-symmetric body rotates with constant
    angular speed ? about the z axis. Relative to the
    origin
  • 1. is constant.
  • 2. is constant but is
    not.
  • 3. is constant but is not.
  • 4. has no z-component.

27
Concept Question
  • Answer 2. For this non-symmetric rigid body,
    the angular momentum about the origin has time
    varying components in the x-y plane,
  • (where is a radial unit vector pointing
    outward from the origin) The magnitude of is
    constant because both and are
    constant. The direction of in the x-y
    plane depends on the instantaneous orientation of
    the body and so the direction of is
    changing.

28
Concept Question
  • A rigid body with rotational symmetry body
    rotates at a constant angular speed ? about it
    symmetry (z) axis. In this case
  • 1. is constant.
  • 2. is constant but is
    not.
  • 3. is constant but is not.
  • 4. has no z-component.
  • 5. Two of the above are true.

29
Concept Question
  • Answer 1. For a symmetric body, all the non-z
    components of the angular momentum about any
    point along the z-axis cancel in pairs leaving
    only a constant non-zero z-component of the
    angular momentum so is constant.

30
Angular Momentum of Cylindrically Symmetric Body
  • A cylindrically symmetric rigid body rotating
    about its symmetry axis at a constant angular
    velocity with
    has angular momentum about any point on its
    axis

31
Angular Momentumfor Fixed Axis Rotation
  • Angular Momentum about the point S
  • Tangential component of momentum
  • z-component of angular momentum
  • about S

32
Concept Question Angular Momentum of Disk
A disk with mass M and radius R is spinning with
angular speed ? about an axis that passes through
the rim of the disk perpendicular to its plane.
The magnitude of its angular momentum is
33
Concept Question Angular Momentum of Disk
Answer 6. The moment of inertia of the disk
about an axis that passes through the rim of the
disk perpendicular to its plane is I Icm MR2
(3/2)MR2 . So the magnitude of its angular
momentum is L (3/2)MR2 ?.
Write a Comment
User Comments (0)
About PowerShow.com