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Department of Physics and Applied Physics 95.141, Fall 2013, Lecture 21 Lecture 21 Chapter 12 Static Equilibrium 11.27.2013 Physics I Course website: http://faculty.uml.edu/Andriy_Danylov/Teaching/PhysicsI Lecture Capture: http://echo360.uml.edu/danylov2013/physics1fall.html

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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

Department of Physics and Applied Physics95.141, Fall 2013, Lecture 21

Thank youSee you on Monday