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Goals: Chapter 12 Define center of mass Analyze rolling motion Use Work Energy relationships Introduce torque Equilibrium of objects in response to forces & torques – PowerPoint PPT presentation

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Title: Goals:


1
Lecture 17
  • Goals
  • Chapter 12
  • Define center of mass
  • Analyze rolling motion
  • Use Work Energy relationships
  • Introduce torque
  • Equilibrium of objects in response to forces
    torques
  • Assignment
  • HW7 due tomorrow
  • Wednesday, Exam Review

2
Combining translation and rotation
  • Objects can have translational energy
  • Objects can have rotational energy
  • Objects can have both K ½ m v2 ½ I w2

3
1st A special point for rotationCenter of Mass
(CM)
  • A free object will rotate about its center of
    mass.
  • Center of mass Where the system is balanced !
  • A mobile exploits this centers of mass.

mobile
4
System of Particles Center of Mass
  • How do we describe the position of a system
    made up of many parts ?
  • Define the Center of Mass (average position)
  • For a collection of N individual point like
    particles whose masses and positions we know

RCM
m2
m1
r2
r1
y
x
(In this case, N 2)
5
Sample calculation
m at ( 0, 0) 2m at (12,12) m at (24, 0)
  • Consider the following mass distribution

XCM (m x 0 2m x 12 m x 24 )/4m meters YCM
(m x 0 2m x 12 m x 0 )/4m meters XCM 12
meters YCM 6 meters
6
System of Particles Center of Mass
  • For a continuous solid, one can convert sums to
    an integral.

dm
r
y
x
where dm is an infinitesimal mass element but
there is no new physics.
7
Connection with motion...
  • An unconstrained rigid object with rotation and
    translation rotates about its center of mass!
  • Any point p rotating

8
Work Kinetic Energy
  • Work Kinetic-Energy Theorem ?K WNET
  • Applies to both rotational as well as linear
    motion.
  • What if there is rolling without slipping ?

9
Same Example Rolling, without slipping, Motion
  • A solid disk is about to roll down an inclined
    plane.
  • What is its speed at the bottom of the plane ?

10
Rolling without slipping motion
  • Again consider a cylinder rolling at a constant
    speed.

2VCM
CM
VCM
11
Motion
  • Again consider a cylinder rolling at a constant
    speed.

Both with VTang VCM
Rotation only VTang wR
Sliding only
2VCM
VCM
CM
CM
CM
VCM
If acceleration acenter of mass - aR
12
Example Rolling Motion
  • A solid cylinder is about to roll down an
    inclined plane. What is its speed at the bottom
    of the plane ?
  • Use Work-Energy theorem

Mgh ½ Mv2 ½ ICM w2 and v wR Mgh ½
Mv2 ½ (½ M R2 )(v/R)2 ¾ Mv2 v 2(gh/3)½
13
How do we reconcile force, angular velocity and
angular acceleration?
14
From force to spin (i.e., w) ?
A force applied at a distance from the rotation
axis gives a torque
a
FTangential
F
q
Fradial
r
r
FTangential
Fradial
FTang sin q
  • If a force points at the axis of rotation the
    wheel wont turn
  • Thus, only the tangential component of the force
    matters
  • With torque the position angle of the force
    matters
  • ?NET r FTang r F sin q

15
Rotational Dynamics What makes it spin?
  • ?NET r FTang r F sin q
  • Torque is the rotational equivalent of force
  • Torque has units of kg m2/s2 (kg m/s2) m N m

?NET r FTang r m aTang
r m r a (m r2)
a For every little part of the wheel
16
Torque
  • The further a mass is away from this axis the
    greater the inertia (resistance) to rotation


?NET I a
  • This is the rotational version of FNET ma
  • Moment of inertia, I Si mi ri2 , is the
    rotational equivalent of mass.
  • If I is big, more torque is required to achieve
    a given angular acceleration.

17
Rotational Dynamics
a
FTangential
F
Fradial
  • ?NET I a r F sin q

r
  • A constant torque gives constant angular
    acceleration if and only if the mass distribution
    and the axis of rotation remain constant.

18
Torque, like w, has pos./neg. values
  • Magnitude is given by (1) r F sin q
  • (2) Ftangential r
  • (3) F rperpendicular to line of
    action
  • Direction is parallel to the axis of rotation
    with respect to the right hand rule
  • And for a rigid object ? I a

r sin q
line of action
F cos(90-q) FTang.
r
a
90-q
q
F
F
F
Fradial
r
r
r
19
Statics
Equilibrium is established when
In 3D this implies SIX expressions (x, y z)
20
Example
  • Two children (30 kg 60 kg) sit on a horizontal
    teeter-totter. The larger child is at the end of
    the bar and 1.0 m from the pivot point. The
    smaller child is trying to figure out where to
    sit so that the teeter-totter remains horizontal
    and motionless. The teeter-totter is a uniform
    bar of length 3.0 m and mass 30 kg.
  • Assuming you can treat both children as point
    like particles, what is the initial angular
    acceleration of the teeter-totter when the large
    child lifts up their legs off the ground (the
    smaller child cant reach)?
  • The moment of inertia of the bar about the pivot
    is 30 kg m2.
  • For the static case

21
Example Soln.
30 kg
  • Draw a Free Body diagram (assume g 10 m/s2)
  • 0 300 d 300 x 0.5 N x 0 600 x 1.0
  • 0 2d 1 4
  • d 1.5 m from pivot point

22
Recap
  • Assignment
  • HW7 due tomorrow
  • Wednesday review session
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