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12.4 Zero Torque and Static Equilibrium AP PHYSICS March 21, 2019 PRACTICE LABS TESTS Unit 12 Problems (123) Balancing Lab Corrections Unit 12 Test Tuesday (3/26/19) PHYSICS ANNOUNCEMENTS Zero Torque and Static Equilibrium 12.4 I can describe, interpret, and solve problems involving static equilibrium. Conservation of Angular Momentum If no net external torque acts on an object, then its angular momentum does not change. L i =L f

12.4 Zero Torque and Static Equilibrium

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Page 1: 12.4 Zero Torque and Static Equilibrium

12.4 Zero Torque and Static Equilibrium

AP PHYSICS

March 21, 2019

PRACTICE LABS TESTSUnit 12 Problems

(1­23)

Balancing Lab

Corrections

Unit 12 Test Tuesday (3/26/19)

PHYSICS ANNOUNCEMENTS

Zero Torque and Static Equilibrium

12.4I can describe, interpret, and solve problems involving static equilibrium.

Conservation of Angular Momentum

If no net external torque acts on an object, then its angular momentum does not change.

Li = Lf

Page 2: 12.4 Zero Torque and Static Equilibrium

12.4 Zero Torque and Static Equilibrium

AP PHYSICS

March 21, 2019

The Vector Nature of Angular Velocity and Momentum

When an object rotates it is said to have an angular velocity, ω, and therefore angular momentum, L. How do we determine the direction of these two vector quantities?

The Right‐Hand RuleCurl the fingers of the right hand in the direction of rotation. The thumb will point in the direction of the angular velocity, ω, and angular momentum, L.

Gyroscopic Effect Changing M.O.I.

The center of mass of an object is the point on the object that moves in the same way that a point particle would move.

Center Of Mass Finding the Center of Mass

3.) Projectile Method

1.) Balance Method

2.) Hanging Method

GO

GO

GO

Page 3: 12.4 Zero Torque and Static Equilibrium

12.4 Zero Torque and Static Equilibrium

AP PHYSICS

March 21, 2019

Center of MassThe center of mass of a system of masses is the point where the system can be balanced in a uniform gravitational field.

Xcm =m1x1 + m2x2 m1x1 + m2x2= Mm1 + m2

Two Dimensional Center of Mass

Xcm =m1x1 + m2x2 m1x1 + m2x2= Mm1 + m2

ycm =m1y1 + m2y2 m1y1 + m2y2= Mm1 + m2

Where is the Center of Mass? Applications of Center of Mass

FosburyFlop

Conditions For Equilibrium

Net Force = 0

Net Torque = 0

1.) TranslationalEquilibrium

2.) RotationalEquilibrium

Page 4: 12.4 Zero Torque and Static Equilibrium

12.4 Zero Torque and Static Equilibrium

AP PHYSICS

March 21, 2019

Balance Method BACK Center of Mass

Center of Mass BACK Projectile Method

PULL

PULL

BACK