(Click on even # problem to check answer. Odds in back of book.)

1. Thursday(10/28): Introduction to Chap. 8, Rotational Kinematics,

Angular Quantities, and Tangential Variables. HW: Read pages 207-218 and

solve prob. 5, 9, 13, 20, and 29 on pages 225-6.

2. Friday(10/29): Centripetal and Tangential Acceleration, and

Rolling Motion. HW: Read pages 218-224 and solve prob. 39, 41, 45, 46,

and 48 on page 227.

3. Monday(11/01): Introduction to Ch. 9, Rotational Dynamics, Torques

on Rigid Objects, Equilibrium, and Center of Gravity. HW: Read pages

232-241 and solve prob. 4, 8, 12, 16, and 21 on page 258-9.

4. Wednesday(11/03): Newton's 2nd Law for Rotational Motion,

Rotational Work and Energy, and Angular Momentum. HW: Read pages

242-253 and solve prob. 28, 34, 39, 45, and 52 on pages 260-262.

5. Thursday(11/04): Lab on the Calculation of Torque. HW: Process

lab data and write lab report (Due Monday).

6. Friday(11/05): Lab Discussion and Review Chapter 8&9- Rotational

Mechanics. HW: Complete Review Handout.

7. Monday(11/08): TEST on Ch.8&9 - Rotational Mechanics. HW: Go

to web-site for notes on Ch.10 - Simple Harmonic Motion and Elasticity.

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 9 - Rotational Dynamics.

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

For example, for arc length, s = r·θ ; 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. Our textbook has a table of values for the Moment of Inertia of various

objects on page 244. 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 .

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 .

And still, we need 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.

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