1. Thursday(11/05): Introduction to Chap. 10, Rotational Kinematics,

Angular Quantities, and Tangential Variables. HW: Read pages 292-307 and

solve prob. 1, 3, 5, 11, and 17 on pages 322-3.

2. Friday(11/06): Rotational Dynamics, Torques on Rigid Objects,

Equilibrium, and Center of Gravity. HW: Read pages 307-321 and solve prob.

21, 25, 31, 35, and 40 on pages 324-27.

3. Monday(11/09): Lab on the Calculation of Torque. HW: Process

lab data and write lab report (Due Thursday).

4. Tuesday(11/10): Post-Lab Discussion. HW: Write Lab Report.

5. Wednesday(11/11): No School - Veterans Day. HW: Continue working

on assigned problems.

6. Thursday(11/12): Introduction to Ch. 11, Using Cross Products to

Calculate Torques. HW: Read pages 336-345 and solve prob. 1, 3, 6, 11,

and 14 on pages 354-5.

7. Friday(11/13): Conservation of Angular Momentum. HW: Read pages

345-351 and solve prob. 21, 23, 25, and 29 on pages 356-7.

8. Monday(11/16): Review Chapters 10 & 11 - Rotational Mechanics.

HW: Complete Review Handout.

9. Tuesday(11/17): TEST on Ch.10 & 11 - Rotational Mechanics.

HW: Go to web-site for notes on Ch.12 - Simple Harmonic Motion and

Elasticity.

Very Important: If you have any questions or miss a class, see me before

school (8:00 - 8:30 AM), during Lunch, or after school.

1. When an object spins about an axis, it is said to undergo rotary motion.

The axis of rotation is the line about which the rotation occurs.

For example, the earth rotates on its axis.

2. Circular motion occurs when the entire object revolves around a single

point. We say that the earth revolves around the Sun. To find the velocity

we can use v = 2πr/T , with T being the rotational period, or time for one

complete round trip.

3. Additionally, any point on a rotating object is moving in a circular path,

demonstrating circular motion with a velocity which therefore must be

tangent to the circle.

4. It is not practical to use linear motion quantities to analyze rotary or

circular motion, so we use the rotational quantities angular displacement,

angular velocity, and angular acceleration.

5. Angular displacement is given by the Greek letter, theta (θ), and

measured in radians (rad). A radian is the measure of a central angle

which intercepts an arc equal in length to the radius of the circle.

6. For angular velocity we use omega (ω) which then would be measured in

rad/sec., and angular acceleration, alpha (α), is in rad/s² . Remember, to

convert degrees to radians, use the conversion factor that 180^o = π rad .

7. The same five linear motion equations are then transformed into the

rotational motion equations using these new quantities.

8. These now become:

Δθ = ω_avg·Δt , ω_avg= (ω_o+ω)/2 , ω = ω₀+α·Δt , ω² = ω_o²+2α·Δθ ,

and the last one is Δθ = ω_o·Δt + ½α·(Δt)² .

9. Uniform circular motion occurs when an acceleration of constant

magnitude is perpendicular to the tangential velocity and the object

maintains a constant speed but is accelerated toward the center

of the circle.

10. This introduces the concept of centripetal (center seeking)

acceleration, a_c = v²/r .

11. If a particle moves along a curved path in such a way that the

magnitude and direction of v change with time, the particle has an

acceleration vector that can be described with two component vectors.

12. The radial component vector arises from the change in direction of v ,

which is the centripetal acceleration, a_c = v²/r and the tangential

component vector a_t = Δv/Δt, is based on the change in magnitude of v .

13. The total acceleration can be found with the vector sum of these

two accelerations which occur at right angles, so we use the

Pythagorean theorem a_total = √(a_c² + a_t²) , and θ = tan^-1(a_c / a_t).

Chapter 11 - Rotational Dynamics.

1. Recall that we have equations which relate angular and linear quantities.

For example, for arc length, s = r·Δθ ; for linear velocity, we have, v = ω·r,

and acceleration, a = α·r .

motion produces torque, given by the Greek letter Tau, τ, lowercase.

Torque can be computed with the equation, τ = F·r·sin(θ). This is called

the cross-product, so τ = r x F , and we have another vector.

3. In the torque equation, r is the perpendicular distance from the line-of-

action of the force and the point of rotation. The angle θ is measured

between the force and the distance from the axis.

4. Torque can either start, stop, or change the direction of rotation and is

measured in Newton·meters, Nm. This tells us as we study Rotational

Dynamics that the cause of rotation is torque.

5. The Moment of Inertia of an object is a measure of its resistance to

changes in rotational motion. Just think of the spin of an ice skater, with

arms-out, and then arms tucked in.

6. We use the variable I to indicate this quantity which is called Rotational

Inertia or even Moment of Inertia. In general, this is given by I = Σmr² .

For the complete set of values, see p. 304.

7. For an object to be in complete equilibrium, it must be in both rotational

and translational equilibrium. For the rotational, this implies that the

Clockwise Torque (CWT) must equal the Counter Clockwise Torque (CCWT).

8. Translational equilibrium implies that the sum of the forces in the x

direction is zero, and the sum of the forces in the y direction is also zero.

9. All rotating objects must: (i) obey Newton's second law, (ii) possess

angular momentum which still operates under the conservation law, and

(iii) have rotational kinetic energy which can do work and also obey the

conservation law in the absence of any external forces.

10. Newton's 2nd Law for rotational motion states that the angular

acceleration of an object is directly proportional to the applied torque, but

varies inversely with the rotational inertia. The equation is τ = I·α .

11. To calculate the work done by a rotating object, instead of W = F·s,

we would use W = τ·θ . This also means that to compute power we would

use the equation, P = τ·θ/Δt , or P = τ·ω_avg .

12. As far as energy is concerned, Kinetic Energy can be calculated with

the equation, K = ½I·ω² .

13. Lastly, we can compute angular momentum with the equation, L = I·ω .

The applied Angular Impulse would then of course be, J = τ·Δt .

Make sure you use  these steps to solve any problem in Physics:

(i) read the problem and identify the given variables

(ii) determine what you are asked to solve for

(iii) find the correct vector formula to use

(iv) use Algebra, Trigonometry, and/or Calculus to isolate the unknown

(v) substitute-in the given information and simplify.