static equilibrium - faculty server...
Post on 21-Aug-2018
219 Views
Preview:
TRANSCRIPT
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 21
Lecture 21
Chapter 12
Static Equilibrium
11.27.2013Physics I
Course website:http://faculty.uml.edu/Andriy_Danylov/Teaching/PhysicsI
Lecture Capture: http://echo360.uml.edu/danylov2013/physics1fall.html
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 21
Chapter 12
The Conditions for Equilibrium Solving Statics Problems Stability and Balance
Outline
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 21
Exam III Info
Exam III Wed Dec. 4, 9:00-9:50am, OH 150.Exam III covers Chapters 9-11
Same format as Exam IIPrior Examples of Exam III posted
Ch. 9: Linear Momentum (no section 10)Ch. 10: Rotational Motion (no section 10)
Ch. 11: Angular Momentum; General Rotation (no sections 7-9)
Exam Review Session TBA, Ball 210
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 21
Conservation of Angular Momentum. Kepler’s second law
Kepler’s second law states that each planet moves so that a line from the Sun to the planet sweeps out equal areas in equal times. Use
conservation of angular momentum to show this.
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 21
Kepler’s second law
Kepler’s 2nd law is a consequence of conservation of ang. momentum!So if L is constant, dA/dt is a constant
)sin()(21 vdtrdA
dAdt
L2m
r
pv,
morigin
Let’s find ang. moment. of the Earth(treat the Earth as a point):
L r psinrpL
sinmrvNext step. Let’s find area swept by r in dt:
vdtdl
sin21 rvdt
mmrv
dtdA
2sin
)ˆ( rFrFr gext Let’s show that L is conserved:
gF
,0sin netce netdtLd
0constLthen
)sin( vdt
ConcepTest 1 Figure Skater
A) the same
B) larger because she’s rotating faster
C) smaller because her rotational inertia is smaller
A figure skater spins with her arms extended. When she pulls in her arms, she reduces her rotational inertiaand spins faster so that her angular momentum is conserved. Comparedto her initial rotational kinetic energy, her rotational kinetic energy after she pulls in her arms must be:
Because L is conserved, larger means larger Krot. The “extra” energy comes from the work she does on her arms.
2
2IKrot
ILuse 2)( I
2LKrot
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 21
Bicycle wheel precession.A bicycle wheel is spun up to high speed and is suspended from the ceiling by a wire attached to one end of its axle.
Now mg try to tilt the axle downward.
You expect the wheel to go down (and it would if it weren’t rotating), but it unexpectedly swerves to the left and starts rotating!Let’s explain that.
gm
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 21
Bicycle wheel precession.
r
gm
dtLd
Apply Rotational N.2nd law:
dtLd
mg produces torque in +y direction
)ˆ( jrmggmr
Ld
inL
finL
LdLL infin
z
x
y
dtLin
Torgue changes direction of angular momentum
Since L I
then angular velocity changes its direction
It is called precession
wir
e
TF
http://www.youtube.com/watch?v=lF9Hf5vZZKISuitcase with a gyroscope
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 21
There are three branches of Mechanics:
Kinematics Motion Forces Dynamics Motion Forces Statics Motion Forces
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 21
An object with forces acting on it, but with zero net force, is said to be in equilibrium.
The 1st Condition for Equilibriumprevents translational motion
0F
0 xF 0 yF 0 zF
amF
N. 2nd law describes translational motion
He doesn’t want to have any sliding of a ladder, i.e. 0a
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 21
The 2nd Condition for Equilibriumprevents rotational motion
0 ext
Iext
Rotational N. 2nd law describes rotational motion:v
He doesn’t want to have any rotation of a ladder, i.e. 0
There must be no net external torque around any axis (the choice of axis is arbitrary).
0 x 0 y 0 z
ConcepTest 2 Static equilibriumConsider a light rod subject to the two forces of equal magnitude as shown in figure. Choose the correct statement with regard to this situation:(A) The object is in force equilibrium but not torque equilibrium.(B) The object is in torque equilibrium but not force equilibrium(C) The object is in both force equilibrium and torque equilibrium(D) The object is in neither force equilibrium nor torque
equilibrium
0 FFF
mext
force equilibrium
torque equilibrium X
Here, the 1st condition is satisfied but the 2nd isn’t, so there will be rotation.So, to have static situation, both conditions must be satisfied.
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 21
Reduce # of Equilibrium EquationsFor simplicity, we will restrict the applications to situations in
which all the forces lie in the xy plane.
1st condition:
2nd condition:
0 xF 0 yF 0 zF
0 x 0 y 0 z
0)1 xF 0)2 yF 0)3 z
There are three resulting equations, which we will use
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 21
Axis of rotation for the 3rd equation
Does it matter which axis you choose for calculating torques?NO. The choice of an axis is arbitrary
0 z
0F
Any axis of rotation works
If an object is in translational equilibrium and the net torque is zero about one axis, then the net torque must be zero
about any other axis
We should be smart to choose a rotation axis to simplify problems
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 21
Any axis of rotation works for the 3rd
equation (proof)The choice of an axis is arbitrary
0 z
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 21
Concurrent/Nonconcurrent forces
1F 2F
3F
Concurrent forces:when the lines of action of the forces intersect at a common point, there will be no rotation. So
1F 2F
3F
0 xF 0 yF0 z 0 xF 0 yF0 z
Nonconcurrent forces:when the lines of action of the forces do not intersect at a common point, there will be rotation. So
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 21
Example:traffic light
Find the tension in the two wires supporting the traffic light
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 21
Torque due to gravityHere, we will often have objects in which there is a torque exerted by gravity. There is a simple rule how to get gravitational torque:
gMRCM
gMW
CMR
CM
For extended objects, gravitational torque acts as if all mass were concentrated at the center of mass
sinMgRCMR
gM
As if the whole mass were here
Department of Physics and Applied Physics95.141, Fall 2013, Lecture 21
1. Choose one object at a time, and make a free-body diagram by showing all the forces on it and where they act.
2. Choose a coordinate system and resolve forces into components.
3. Write equilibrium equations for the forces.
4. Choose any axis perpendicular to the plane of the forces and write the torque equilibrium equation. A clever choice here can simplify the problem enormously.
5. Solve.
Solving Statics Problems
top related