## Thursday, March 16, 2017

### Activity 98: Electrostatics

Read sections 18.1, 18.2 and 18.3 in the textbook and answer the following questions at the end of the chapter.

#10, 11, 12, 14, 16, 21, 24

When done check your answers here.

## Monday, March 13, 2017

### Activity 92:Sound

Video on calculation of the speed of sound using resonance

Do problems

1,3,8,40,41,45,47 at the end of chapter 17

## Monday, March 6, 2017

## Friday, March 3, 2017

### Activity 90 Harmonic Motion

Link to notes on Harmonic Motion

Read section 16.1, 16.2 & 16.3 in text.

Homework:

Complete problems 1, 3, 7, 10, 13, 14, 18. on page 590 in text.

When done check your answers here.

## Wednesday, March 1, 2017

## Monday, February 27, 2017

### Activity 88:Conservation of Angular Momentum

Linear Momentum = mv

Angular (Rotational) Momentum = (I)(Angular Velocity)

Conservation of Momentum holds true for both.

mv before = mv after

Angular Momentum Before = Angular Momentum After

Seminar Skater Problem

Do Problems 38 through 41 at the end of chapter 10

Check your answers here

Seminar Skater Problem

Do Problems 38 through 41 at the end of chapter 10

Check your answers here

## Thursday, February 16, 2017

### Activity 87: Rotational Kinetic Energy

Derivation of Rotational Energy Equation

net W = (net F )Δs

net W = (r net F )Δs/r

net W = (net τ)θ

net τ = Iα

net W = Iαθ

Consider a person who spins a large grindstone by placing her hand on its edge and exerting a force through part of a revolution . In this example, we verify that the work done by the torque she exerts equals the change in rotational energy.

(a) How much work is done if she exerts a force of 200 N through a rotation of 1.00 rad(57.3º) ? The force is kept perpendicular to the grindstone’s 0.320-m radius at the point of application, and the effects of friction are negligible.

(b) What is the final angular velocity if the grindstone has a mass of 85.0 kg?

(c) What is the final rotational kinetic energy? (It should equal the work.)

(a) How much work is done if she exerts a force of 200 N through a rotation of 1.00 rad(57.3º) ? The force is kept perpendicular to the grindstone’s 0.320-m radius at the point of application, and the effects of friction are negligible.

(b) What is the final angular velocity if the grindstone has a mass of 85.0 kg?

(c) What is the final rotational kinetic energy? (It should equal the work.)

**Assignment:**

Do Problems 21, 22,23,24,25, 27 & 28 on page 357 in the textbook.

When you are done check your answers here.

When done check your answer here.

## Tuesday, February 14, 2017

### Activity 86: Introduction to Moment of Inertia

Moment of Inertia:

mr2 is called the moment of inertia - the resistance of a mass a distance r to a change in its angular acceleration.

This is analogous to the way that inertial mass resists translational acceleration.

Inertial Mass (m) and Moment of Inertia (I) resist translational and angular acceleration.

Consider the father pushing a playground merry-go-round.

He exerts a force of 250 N at the edge of the 50.0-kg merry-go-round, which has a 1.50 m radius. (Consider the merry-go-round itself to be a uniform disk with negligible retarding friction.)

Calculate the angular acceleration produced:

(a) when no one is on the merry-go-round

(b) when an 18.0-kg child sits 1.25 m away from the center. Consider the merry-go-round itself to be a uniform disk with negligible retarding friction.

Activity 82.10

Activity 82.10

## Friday, February 10, 2017

### Activity 85: Rotational Kinematics

Rotational (Angular) Kinematics is similar to Linear (Translational) Kinematics

Our previous knowledge of translational kinematics and motion can be extended to describe rotational kinematic motion.

A radian 𝚹 in rotational kinematics is analogous to displacement in linear (translational) kinematics

When linear distance, (d,s,x,etc…), was measured, we used meters or similar metric unit. Now we must use radians to measure angular or rotational distance, θ.

A radian “𝚹” is analogous to a meter

Thus an angle will be analogous to distance .

Relationship

s = rRotational (Angular) Kinematics is similar to Linear (Translational) Kinematics

Our previous knowledge of translational kinematics and motion can be extended to describe rotational kinematic motion.

A radian 𝚹 in rotational kinematics is analogous to displacement in linear (translational) kinematics

When linear distance, (d,s,x,etc…), was measured, we used meters or similar metric unit. Now we must use radians to measure angular or rotational distance, θ.

A radian “𝚹” is analogous to a meter

Thus an angle will be analogous to distance .

Relationship

s = r

**𝚹****𝚹**= s/r

Read Section 10.1 in the textbook.

Pick up a copy of this Seminar Problem 1 and complete.

Activity 85.20

Pick up a copy of this Angular Motion & Acceleration Worksheet and complete. When done check your answers here.

The equations and concepts you learned in Translational Kinematics have an analogous relationship to Rotational Kinematics.

Activity 85.30

Activity 85.40

## Tuesday, February 7, 2017

## Sunday, February 5, 2017

### Activity 83: Slipping and Tipping and Stability

A

With a lab partner read and execute this Slipping and Tipping Activity. It is on pages 72 - 73 in the lab book. This is not a formal lab write-up. Simply do the exercises and answer the "light-bulb" questions on paper ( per group) and hand it in.

With a lab partner read and execute this Slipping and Tipping Activity. It is on pages 72 - 73 in the lab book. This is not a formal lab write-up. Simply do the exercises and answer the "light-bulb" questions on paper ( per group) and hand it in.

## Thursday, February 2, 2017

### Activity 82: Center of Mass Lab Activity

Go to this Google Classroom Activity and read and complete the Center of Mass Assignment

The important part of this assignment is a well documented procedure. Be sure to put all team names on the submission.

### Activity 81: Solving Force & Torque Equilibrium Problems

Activity 81.10

Watch this video on solving torque and force problems.

Watch this video on solving torque and force problems.

Activity 81.20

Pick up a paper copy of this worksheet and solve. When you are done please check your answers here.

## Wednesday, February 1, 2017

### Activity 80: Center of Gravity

Activity 77A.10

Please watch this video on torque and the center of gravity:

From this video it should become clear that when doing torque problems you can "pretend" the entire weight of an object is located at it's center of mass.

Pick up a paper copy of torque problem set number #3. When you are done check your answers here.

Activity 77A.20

Pick up a paper copy of Catwalk and solve.

## Monday, January 30, 2017

### Activity 79: Introduction to Torque

Activity 75.10

Read Chapter Sections 9.1, 9.2, and 9.3 in the textbook.

Activity 75.20 -

Watch this video on torque

- Torque is the result of applying a force on an object somewhere other than it's center of gravity.
- Torque causes an object to rotate.
- At its most basic level Torque is calculated by multiplying the Force applied times the Distance the force is from the object's pivot point.
- Torque = Force X Distance.
- Torque = (Force) (sin(angle)) (Distance)
- The angle referenced above is the angle between the applied force and the axis of the rotation of the object.

**Example**

In the picture above the two people are sitting on a see saw.

**Question:**How far from the center does the person on the right need to sit to balance the see saw?

The person on the left is generating a force of (30 kg X 9.8 m/s/s) = 294 N . This force is creating a torque of (294N ) (2.0m) = 588Nm of counter clockwise torque of the left side of the see saw.

To put the see saw in balance (

**called rotational equilibrium**) the counterclockwise torque must equal the clockwise torque. This is often written**Tc = Tcc**.
In this case the person sitting on the right side must generate 588Nm of clockwise torque to put the see saw in rotational equilibrium. Since they weigh (50kg x 9.8m/s/s) or 490N they must sit (588Nm/490N) or 1.2m from the center.

Complete this Introduction to Torque Google Classroom activity collecting the data fro the class demonstration.

Activity 79.22

Pick up a paper copy of these seminar problems. Look at the first problem and come up with an approach to solve it. Check your answer against the solution in the answer key. When you are done with the rest of the problems please check your answers here.

Activity 79.25

Pick up a paper copy of this document on concurrent forces. Make sure you understand the difference between concurrent and nonconcurrent forces.

Activity 79.30

Pick up a paper copy of torque problem set #1. When you are done check your answers here.

Activity 79.40

Pick up a paper copy of torque problem set #2. When you are done check your answers here.

## Thursday, January 26, 2017

### Activity 77:Banked Track & Newtons' Universal Law of Gravitation

Newton's Universal Law of Gravity

Calculate g

Constants for Homework

Mass of the Moon: 7.35E22 kg

Distance from the Earth to the Moon: 3.84E8 m

Radius of the Moon: 1.74E6m

Homework Problems 23, 24, 29, 31, 33, 34, 35

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