Lecture 17: Angular momentum - PowerPoint PPT Presentation

1 / 7
About This Presentation
Title:

Lecture 17: Angular momentum

Description:

conservation of linear momentum. No torques, no angular momentum: ... The child catches a ball of mass 1.0 kg thrown by a friend. ... – PowerPoint PPT presentation

Number of Views:25
Avg rating:3.0/5.0
Slides: 8
Provided by: Patr520
Category:

less

Transcript and Presenter's Notes

Title: Lecture 17: Angular momentum


1
Lecture 17Angular momentum
  • Phys 2101
  • Gabriela González

2
Newtons law systems of particles
total linear momentum
total angular momentum
No forces, no linear momentum conservation of
linear momentum. No torques, no angular momentum
conservation of angular momentum.
3
Newtons law
No forces, no linear momentum conservation of
linear momentum. No torques, no angular momentum
conservation of angular momentum.
4
Conservation of angular momentum
5
Example
  • A 4.0 kg particle moves in an xy plane. At the
    instant when the particle's position and velocity
    are r (2.0 i 4.0 j ) m and v -4.0 j m/s,
    the force on the particle is F -3.0 i N. At
    this instant, determine
  • the particle's angular momentum about the origin,
  • the particle's angular momentum about the point
    x 0, y 4.0 m,
  • the torque acting on the particle about the
    origin, and
  • the torque acting on the particle about the
    point x 0, y 4.0 m.

y
F
v
r
x
6
Example
  • A 30 kg child stands on the edge of a stationary
    merry-go-round of mass 100 kg and radius 2.0 m.
    The rotational inertia of the merry-go-round
    about its axis of rotation is 150 kgm2. The
    child catches a ball of mass 1.0 kg thrown by a
    friend. Just before the ball is caught, it has a
    horizontal velocity of 12 m/s that makes an angle
    of 37 with a line tangent to the outer edge of
    the merry-go-round, as shown in the overhead view
    of the figure. What is the angular speed of the
    merry-go-round just after the ball is caught?

7
Example
  • A 1.0 g bullet is fired into a 0.50 kg block that
    is mounted on the end of a 0.60 m nonuniform rod
    of mass 0.50 kg. The blockrodbullet system then
    rotates about a fixed axis at point A. The
    rotational inertia of the rod alone about A is
    0.060 kgm2. Assume the block is small enough to
    treat as a particle on the end of the rod.
  • What is the rotational inertia of the
    blockrodbullet system about point A?
  • If the angular speed of the system about A just
    after the bullet's impact is 4.5 rad/s, what is
    the speed of the bullet just before the impact?
Write a Comment
User Comments (0)
About PowerShow.com