## November 11, 2008

### Problem for Wednesday

Hello everyone-

I'll start lecture on Wednesday by discussing problem 7, which is very similar to the ruler problem we did in class on Monday.

jay-

## November 6, 2008

### Homework 9 key; problem for tomorrow

Hello everyone-

The key for HW 9 is posted; I'll post HW 10 problems first thing tomorrow.

Tomorrow, I am going to start the lecture by discussing the following problem:

You fire a bullet, of mass 9.00g, at 860m/s into a wooden door of mass 18kg and width 0.85m. The bullet hits the door at a right angle, 0.75cm away from the hinges (the axis of rotation), and becomes lodged in the wood. The hinges are frictionless, of course. What is the resulting angular velocity of the door?

## October 24, 2008

### A problem for Monday

Hello everyone-

Here's the schedule for the next few days:

Monday 27 October Chapter 10.1-4 (Rotational Kinematics)
Wednesday 29 Oct Chapter 10.5-10 (Moment of Inertia) (Homework 8 due, Key released)
Thursday 30 Oct Review of chapters 5-9
Friday 31 October Exam II, covering chapters 5-9.

I appear to be about a day behind where I'd planned on being.

For Monday, a good problem to get you on track will be 10.6.

jay-

## October 17, 2008

### Monday's problem

We will discuss Problem 9.3 in class as well.

Another good practice problem from 8 is 8.53.

### Arrrrrrrgh.....

Okay, I found the sign error. We were looking at #8.59. I noted that there is a change in the mechanical energy of the system, due to energy being lost to friction. The mechanical energy is initially all kinetic (K1) and at the end state is all spring potential (U2).

So we know that

Wf=Emec2-Emec1=U2-K1

And the problem asks for the initial velocity of the mass, which I can figure out given the Initial kinetic energy:

K1=U2-Wf

We determined Wf to be -0.46J (friction does negative work). Where I got flummoxed was that I had dropped in -0.9J for the spring potential energy, whereas that is actually the work done by the spring- the spring potential HAS to be positive: +0.9J. Drat.

Anyway, the initial KE then is

K1=U2-Wf=(+0.9J) - (-0.46J) = 1.35J

From which I can calculate the initial velocity: 1.0m/s.

## October 15, 2008

### problem for tomorrow

We'll start tomorrow with a discussion of problem #22 in chapter 8.

jay-

## October 12, 2008

### Problem for Monday

I'll start the lecture on Monday with a discussion of problem 65 from Chapter 7.

## October 2, 2008

### Problem for tomorrow

I'll start tomorrow's lecture with a discussion of problem 23 from chapter 6.

If somebody lost a calculator in class today (Thursday) let me know.

## October 1, 2008

### Practice problem for tomorrow

Tomorrow, I will start the lecture by going through problem 32 in chapter 5. You're not required to do it but you will find the discussion of the problem much more useful if you do!

jay-

## September 26, 2008

### Net Monday

Hello all-

I've started grading the exams- hopefully I'll be able to finish this weekend and get grades posted by tuesday at the latest. I will hand them back on Wednesday, most likely. I'll try to post a key as well.

Another professor will be filling in for me on Monday. I am going to be down at UW-Madison giving a lecture.

Have a great weekend!

jay-

## September 22, 2008

### Tonight's problem

Hello everyone-

I forgot to mention the problem we'll talk about in class tomorrow- we'll start tomorrow's lecture with a discussion of #64 in chapter 4.

Ive posted the equation sheet for exam 1. You are not allowed to bring a print-out into the exam; however, you will get the EXACT same sheet handed out with the exam. This is what you'll have available to you.

## September 19, 2008

### Problem solving

Again, what I'd like to do is pick one book problem a day and give you a chance to solve it at home. I'll go over it first thing in the next day's lecture.

The problem I'll present today (Friday, 19 September) will be chapter 4, #44.