Ah, Friday the ninth of December! A very very cold day. A day that really made me miss my home in North Carolina, where a couple inches of snow was the most I ever saw each winter. I remember I never liked having much snow. But now that I'm here, in unpredictable Ohio, I prefer the couple inches over a couple feet. Anyways, on to Physics now. Well, really, I'm sure we could find the final velocity of falling snowflakes by applying our formulas, maybe having to call up the weather man to help us out with the initial velocity. We already know the acceleration due to gravity is 9.8 m/s squared. If the snowflake wouldnt melt before we weighed it, we could find its mass on our own...I should stop now before I give Coats-Haan any ideas. But really, I am moving on to Physics class now.
Well, I really hope this is not a surprise to any of you, and I am rather scared for you if it is (unless, of course, if you are Chris Roseblossom, my lab partner, who is just really too smart for his own good), we took a test over Newton's Law Application. I studied using my Newton's Law Application Test Review, which we turned in on Friday. The answers were on the physics website, and Coats-Haan even showed her work for us, like usual. The test reviews are always helpful for tests and also exams, so I save every one of them. Psst! You should save them too! They make filling out the quarterly exams a breeze!
QOD: What do you know about astronomy? About the only thing I know about astronomy is that it involves the study of the stars. However, I'm sure there is some in-depth physics reasoning behind astronomy that we are about to delve into on Monday.
Sunday, December 11, 2011
Friday, December 9, 2011
Thursday, December 8, 2011
12/8 Scheitlin
If you missed today you really didn't miss much. First off we turned in our pair check from yesterday if we hadn't already and then checked our homework from last night (centripetal force worksheet) and asked questions. Then we had the rest of the class to work on our Newton's Law Application test review which is due Friday (our only homework besides studying and the lab report due next week) and ask Coats-Haan any remaining questions we have. We also found out today that Coats-Haan has an inferiority complex to Kreider and (her words not mine) she has never beaten him at anything except that her shoes match her shirt.
Question of the Day: We only learned one new equation this entire unit. What is it? Why was there nothing else that was new?
Well I am really confused because I thought the fun equation and the center of mass definition equation were new but I am assuming that one of them is just a rearrangement of a previous one. However, I cannot figure out which one we have had before.
Question of the Day: We only learned one new equation this entire unit. What is it? Why was there nothing else that was new?
Well I am really confused because I thought the fun equation and the center of mass definition equation were new but I am assuming that one of them is just a rearrangement of a previous one. However, I cannot figure out which one we have had before.
12/8 qod
We only learned one new equation this entire unit. What is it? Why was there nothing else that was new?
--ch
--ch
Wednesday, December 7, 2011
12/7 Tamayo
Today in Physics we received quite a bit of items in our folders. We got back our Newton’s Laws test, our Lab Report which was turned in on Tuesday, Our Center of Mass Pair Check, and our Center of Mass Reading Question sheet. We did not turn in anything in class today. Today we compared our homework which was questions out of the book with the key while Coats-Haan checked them for completion. After the homework we took notes on Uniform Circular Motion, which dealt with centripetal forces. During these notes Coats-Haan pulled out the Japanese anime doll on a string to illustrate the concepts of today’s lesson. This is the second appearance that the doll has made in Physics this year. We revisited the equation for centripetal acceleration which is a=v2/r. r is the radius and v is the velocity, and a is acceleration. We learned to apply Newton’s Second Law ∑f=ma=mv2/r as centripetal force f is directed radially inward. We then proceeded to complete problems 10 and 11 in the example packet. We then completed a pair check. It was helpful to complete the pair check problems out of order because the second to last problem proved to be the most difficult. Our homework was to complete the centripetal force worksheet, which consists of an explanation of the problems and five problems to be completed.
Work turned in. None (checked Homework)
Work assigned. Centripetal Force Worksheet Due Thursday, Newton’s Laws Applications Test Review due Friday.
QOD. What are some forces that can act as centripetal forces? What is the direction of a centripetal force?
Three forces that can act as centripetal forces are gravity – a satellite orbiting earth, Tension – twirling an object on a string, and Friction – a car travelling on a circular ramp.
The direction of the centripetal force is directed radially inward.
12/7 qod
What are some forces that can act as centripetal forces?
What is the direction of a centripetal force?
--ch
What is the direction of a centripetal force?
--ch
Tuesday, December 6, 2011
Monday, December 5, 2011
12/5 Tuazon
My birthday is now 22 days away. I should just take everyone's blogs until break so I can continue this countdown.
I am going to be honest--I didn't remember that I had to write the blog again until the very end of class, but rest assured I will make sure you are well-informed about what took place today.
While Coats-Haan passed out a packet to do tonight's homework on p. 97 of our lab manuel, we checked our homework from over the weekend, which was the Forces and the Laws of Motion: Overcoming Friction worksheet. No one seemed to have any questions about the homework, so we went straight on to our next order of business, which was finishing #6 on our examples sheet.
Before we began, Coats-Haan told us a little fun fact. She said it made her "conceptually happy" to draw triangles with the slope going down to the right (see figure 1) when something is going downhill, as opposed to drawing the slope going up to the right (see figure 2) when something is going uphill . By doing this, the motion of the object will always go in the positive x direction. Coats-Haan says that if you're better at spatial manipulation than she is, however, you can always draw the triangle going a single direction. Either way you choose to go, she won't take off points from your test for the way you draw the triangle.
(I apologize that one triangle is labeled and the other is not. I'm also sorry for the big gap after this figure. I don't know how to fix it.)
Now for #6 on our examples sheet. Coats-Haan began drawing the triangle in the same direction as figure 1. The three forces acting on the skier are a normal force, a gravitational force, and a frictional force going up the slope.
First we start with the three steps:
1. Sigma Fy = 0; N = mgcostheta
2. f = uN = umgcostheta
3. Sigma Fx = ma (Because the skier is not at rest or going at a constant velocity, we cannot set this equal to zero)
Note that we neglect air resistance. In #7 on our pair check, we have to include a 50.0N air resistance in our calculations, yet doing so is not much different than simply adding another force in the x direction.
The sum of the forces in the x direction (or ma) is the x component of mg minus friction, which ends up being mgsintheta – umgcostheta = ma. All the "m's" cancel out.
From this we get a = g(sintheta – ucostheta) = 2.76 m/s. Using the fourth kinematics equation, we solve for Vf to get 27 m/s.
To find time, Coats-Haan says we can use either the third kinematics equation or the first kinematics equation. Again, either way you choose to go, she will not take points off on the test. When you calculate t, you get 8.4 s.
---
While Coats-Haan was showing us how to do the above example, she dropped her SMART Board eraser and had to chase it around behind her desk while mumbling incoherent things to herself.
Soon afterwards, she dusted herself off and went to the back of the room to explain the Coefficient of Friction lab, for which we have to complete a lab report, due next Tuesday. (Coats-Haan says that there is only one more lab report to do after this one for the rest of the school year. We'll hold her accountable for that.) In a bunch of bins at the back counter there were a variety of wooden blocks, a protractor, a weight, a ramp, etc. We were to use these things to measure the angle at which various items begin to slide down the ramp.
The remainder of class was spent finishing up our Two Dimension Friction Pair Checks, which we turned in. I was pretty thrilled by working through it. I tried explaining things on the pair check that I myself was trying to understand, but somehow, explaining what I kind-of knew to them helped me to fully understand it now.
It was a pretty typical day in second period physics.
THE QUICKIE DETAILS:
Activities:
Checked our homework (Forces and the Laws of Motion: Overcoming Friction, #1-7)
Completed #6 on our example sheet
Finished #5-8 on the Two Dimensional Friction Pair Check (which we turned in)
Started working on the Coefficient of Friction lab, which we will finish up tomorrow
Returned to us:
Nothing (Except for me. I got my blog grade sheet for last Friday's blog returned.)
Checked Homework:
Forces and the Laws of Motion: Overcoming Friction, #1-7 (a worksheet)
Homework:
Pendulum Lab Report (due tomorrow)
Hewitt Center of Mass questions (p. 97 of lab manuel) based on a packet that Coats-Haan passed out
Coefficient of Friction Lab Report due next Tuesday, 12/13
Question of the Day:
What are the two situations where the sum of the forces are zero in the x direction for inclined plane friction problems?
Answer:
Before I answer, I'd just like to say that I wrote this down in the margin of my example sheet with a gut feeling that I would have to know it later.
It is later.
Two situations where the sum of the forces are zero in the x direction for inclined plan friction problems are when the object is:
1. at rest.
2. moving at a constant velocity.
In fact, we had one of these constant velocity situations on our Two Dimensional Friction Pair Check in #6-7. Onur and I got the correct answers (without looking at the board), which end up being 675.58 N and 69.12 N, respectively.
I am going to be honest--I didn't remember that I had to write the blog again until the very end of class, but rest assured I will make sure you are well-informed about what took place today.
While Coats-Haan passed out a packet to do tonight's homework on p. 97 of our lab manuel, we checked our homework from over the weekend, which was the Forces and the Laws of Motion: Overcoming Friction worksheet. No one seemed to have any questions about the homework, so we went straight on to our next order of business, which was finishing #6 on our examples sheet.
FIGURE 1
FIGURE 2
(I apologize that one triangle is labeled and the other is not. I'm also sorry for the big gap after this figure. I don't know how to fix it.)
Now for #6 on our examples sheet. Coats-Haan began drawing the triangle in the same direction as figure 1. The three forces acting on the skier are a normal force, a gravitational force, and a frictional force going up the slope.
First we start with the three steps:
1. Sigma Fy = 0; N = mgcostheta
2. f = uN = umgcostheta
3. Sigma Fx = ma (Because the skier is not at rest or going at a constant velocity, we cannot set this equal to zero)
Note that we neglect air resistance. In #7 on our pair check, we have to include a 50.0N air resistance in our calculations, yet doing so is not much different than simply adding another force in the x direction.
The sum of the forces in the x direction (or ma) is the x component of mg minus friction, which ends up being mgsintheta – umgcostheta = ma. All the "m's" cancel out.
From this we get a = g(sintheta – ucostheta) = 2.76 m/s. Using the fourth kinematics equation, we solve for Vf to get 27 m/s.
To find time, Coats-Haan says we can use either the third kinematics equation or the first kinematics equation. Again, either way you choose to go, she will not take points off on the test. When you calculate t, you get 8.4 s.
---
While Coats-Haan was showing us how to do the above example, she dropped her SMART Board eraser and had to chase it around behind her desk while mumbling incoherent things to herself.
Soon afterwards, she dusted herself off and went to the back of the room to explain the Coefficient of Friction lab, for which we have to complete a lab report, due next Tuesday. (Coats-Haan says that there is only one more lab report to do after this one for the rest of the school year. We'll hold her accountable for that.) In a bunch of bins at the back counter there were a variety of wooden blocks, a protractor, a weight, a ramp, etc. We were to use these things to measure the angle at which various items begin to slide down the ramp.
The remainder of class was spent finishing up our Two Dimension Friction Pair Checks, which we turned in. I was pretty thrilled by working through it. I tried explaining things on the pair check that I myself was trying to understand, but somehow, explaining what I kind-of knew to them helped me to fully understand it now.
It was a pretty typical day in second period physics.
THE QUICKIE DETAILS:
Activities:
Checked our homework (Forces and the Laws of Motion: Overcoming Friction, #1-7)
Completed #6 on our example sheet
Finished #5-8 on the Two Dimensional Friction Pair Check (which we turned in)
Started working on the Coefficient of Friction lab, which we will finish up tomorrow
Returned to us:
Nothing (Except for me. I got my blog grade sheet for last Friday's blog returned.)
Checked Homework:
Forces and the Laws of Motion: Overcoming Friction, #1-7 (a worksheet)
Homework:
Pendulum Lab Report (due tomorrow)
Hewitt Center of Mass questions (p. 97 of lab manuel) based on a packet that Coats-Haan passed out
Coefficient of Friction Lab Report due next Tuesday, 12/13
Question of the Day:
What are the two situations where the sum of the forces are zero in the x direction for inclined plane friction problems?
Answer:
Before I answer, I'd just like to say that I wrote this down in the margin of my example sheet with a gut feeling that I would have to know it later.
It is later.
Two situations where the sum of the forces are zero in the x direction for inclined plan friction problems are when the object is:
1. at rest.
2. moving at a constant velocity.
In fact, we had one of these constant velocity situations on our Two Dimensional Friction Pair Check in #6-7. Onur and I got the correct answers (without looking at the board), which end up being 675.58 N and 69.12 N, respectively.
12/5 qod
What are the two situations where the sum of the forces are zero in the x direction for inclined plane friction problems/
--ch
--ch
Friday, December 2, 2011
12/2 Tuazon
My birthday is 25 days away––just thought I'd let you know.
But as always, physics calls.
While we were going over homework, Coats-Haan was speaking her mind like she usually does (you understand), and for some reason, she was staring at Jeff and said, "You know why I think fourth period is stupider? It's because I don't have any gingers in it." I suppose she has a point––fourth period did have a class average of 73.10% on the Newton's Laws Test compared to everyone else's 80% and up averages.
As for the homework, someone asked to go over #43, which if you recall is about to a 25 kg block sliding on a frictionless surface with a 4 kg block on its side, and you had to find the push force necessary to put on the 25 kg block to prevent the 4 kg block from falling down. In this problem, friction equals gravity, and you have to use the f=uN equation to help solve this problem. The answer ends up being a 40 N push. If I were you, I would ask Coats-Haan for the details, because there are several forces working in this problem and they are easy to mix up on a blog.
We put away our homework as Coats-Haan passed out a worksheet, which is our homework for the weekend. It's seven problems about overcoming friction, and it doesn't look too thrilling. But as Coats-Haan said today, "I'm sure you had f=uN on last night's homework."
Jeff shook his head. Kelly smiled. Onur stared at her like usual.
After that, Coats-Haan told us to get out our example sheet (which I really insist on calling a packet) to do #4, which asks what force is needed to push a 100 N cartoon [of gumm[y] worms] up a 30 degree incline if the coefficient of static friction is 0.400 and the coefficient of kinetic friction is 0.350.
Now I'm going to warn you, I don't feel qualified to explain this over a blog because I don't have the skill to put detailed visual diagrams in your head. Plus, Coats-Haan even told us in class that grasping the concept may be a little challenging.
Basically, because the problem is set on an incline, we have to turn our x and y axes so that x is parallel to the ground surface and y is perpendicular to the ground surface. By doing this, the normal force, push force, and kinetic friction force are along the axis, and the only force which we need to find the components of is mg.
Next, we made a triangle using mg as the hypotenuse and it's x and y components as its other sides. Using geometry, we calculated that the angle opposite from the x component side is always 90 degree – theta, which is equal to the incline given to us in the problem (30 degrees). The x component = mgsintheta, and the y component = mgcostheta. Now you may be wondering why x is proportional to sin rather than cos like it usually is, but I have made a more detailed expression of why this is true in the QotD below.
Furthermore, there are three steps to solve these incline problems:
1. Sigma Fy = 0; N = mgcostheta (This and step 2 are not really steps for Honors Physics because all of our incline problems will always be set up with these conditions. AP Physics may be different, however.)
2. f = uN = umgcostheta
3. Sigma Fx = ma; (which in this problem is P = umgcostheta + mgsintheta = (0.35)(100N)cos(30) + (100N)sin(30) = 80.3 N)
On the examples packet, #5 is similar to #4 because it is on an incline, only you need to use step 3 to solve for u. Also, friction moves up the incline instead of downwards because as you walk, you push back on the ground, and the ground pushes you forward.
For the pair check we got today, Coats-Haan only wanted us to do #1-4 for Monday. She told us that #3, in which we are wearing glass slippers on top of a glass hill, is more like #5 on the examples packet than #4. Onur and I breezed through these questions, but Jeff refused to believe in the possibility of him wearing glass slippers.
THE QUICKIE DETAILS:
Activities:
Checked our homework (see below)
Learned about two-dimensional physics, specifically on an incline.
Completed #4-5 on our examples packet
Worked on Two-Dimensional Friction Pair Check
Returned to us:
1. Pair Check on One-Dimensional Friction
2. Hewitt Reading on Friction (a worksheet)
Checked Homework:
Textbook homework problems, p.128 #34, 35, 37, 39-44 (Turned nothing in)
Homework for the weekend:
Forces and the Laws of Motion: Overcoming Friction, #1-7 (a worksheet)
Pair Check Problems #1-4 (The rest of the pair check will be finished Monday)
Pendulum Lab Report (due Tuesday)
Question of the Day:
Why is the x component not always proportional to cos theta?
Answer:
Cos theta is defined as (adjacent/hypotenuse), which is y/mg if theta is opposite of x like in problem #4 on our examples packet. In this case, y is proportional to cos theta. The x component is instead proportional to sin theta, which is (opposite/hypotenuse) or x/mg.
Therefore in these problems:
sin theta=x/mg, or x=mgsintheta
cos theta=y/mg, or y=mgcostheta (Coats-Haan asked Onur if he was fine with this being true. Onur's fine with it.)
But as always, physics calls.
While we were going over homework, Coats-Haan was speaking her mind like she usually does (you understand), and for some reason, she was staring at Jeff and said, "You know why I think fourth period is stupider? It's because I don't have any gingers in it." I suppose she has a point––fourth period did have a class average of 73.10% on the Newton's Laws Test compared to everyone else's 80% and up averages.
As for the homework, someone asked to go over #43, which if you recall is about to a 25 kg block sliding on a frictionless surface with a 4 kg block on its side, and you had to find the push force necessary to put on the 25 kg block to prevent the 4 kg block from falling down. In this problem, friction equals gravity, and you have to use the f=uN equation to help solve this problem. The answer ends up being a 40 N push. If I were you, I would ask Coats-Haan for the details, because there are several forces working in this problem and they are easy to mix up on a blog.
We put away our homework as Coats-Haan passed out a worksheet, which is our homework for the weekend. It's seven problems about overcoming friction, and it doesn't look too thrilling. But as Coats-Haan said today, "I'm sure you had f=uN on last night's homework."
Jeff shook his head. Kelly smiled. Onur stared at her like usual.
After that, Coats-Haan told us to get out our example sheet (which I really insist on calling a packet) to do #4, which asks what force is needed to push a 100 N cartoon [of gumm[y] worms] up a 30 degree incline if the coefficient of static friction is 0.400 and the coefficient of kinetic friction is 0.350.
Now I'm going to warn you, I don't feel qualified to explain this over a blog because I don't have the skill to put detailed visual diagrams in your head. Plus, Coats-Haan even told us in class that grasping the concept may be a little challenging.
Basically, because the problem is set on an incline, we have to turn our x and y axes so that x is parallel to the ground surface and y is perpendicular to the ground surface. By doing this, the normal force, push force, and kinetic friction force are along the axis, and the only force which we need to find the components of is mg.
Next, we made a triangle using mg as the hypotenuse and it's x and y components as its other sides. Using geometry, we calculated that the angle opposite from the x component side is always 90 degree – theta, which is equal to the incline given to us in the problem (30 degrees). The x component = mgsintheta, and the y component = mgcostheta. Now you may be wondering why x is proportional to sin rather than cos like it usually is, but I have made a more detailed expression of why this is true in the QotD below.
Furthermore, there are three steps to solve these incline problems:
1. Sigma Fy = 0
2. f = uN
3. Sigma Fx = ma
The below info is more detailed for this problem.
1. Sigma Fy = 0; N = mgcostheta (This and step 2 are not really steps for Honors Physics because all of our incline problems will always be set up with these conditions. AP Physics may be different, however.)
2. f = uN = umgcostheta
3. Sigma Fx = ma; (which in this problem is P = umgcostheta + mgsintheta = (0.35)(100N)cos(30) + (100N)sin(30) = 80.3 N)
On the examples packet, #5 is similar to #4 because it is on an incline, only you need to use step 3 to solve for u. Also, friction moves up the incline instead of downwards because as you walk, you push back on the ground, and the ground pushes you forward.
For the pair check we got today, Coats-Haan only wanted us to do #1-4 for Monday. She told us that #3, in which we are wearing glass slippers on top of a glass hill, is more like #5 on the examples packet than #4. Onur and I breezed through these questions, but Jeff refused to believe in the possibility of him wearing glass slippers.
THE QUICKIE DETAILS:
Activities:
Checked our homework (see below)
Learned about two-dimensional physics, specifically on an incline.
Completed #4-5 on our examples packet
Worked on Two-Dimensional Friction Pair Check
Returned to us:
1. Pair Check on One-Dimensional Friction
2. Hewitt Reading on Friction (a worksheet)
Checked Homework:
Textbook homework problems, p.128 #34, 35, 37, 39-44 (Turned nothing in)
Homework for the weekend:
Forces and the Laws of Motion: Overcoming Friction, #1-7 (a worksheet)
Pair Check Problems #1-4 (The rest of the pair check will be finished Monday)
Pendulum Lab Report (due Tuesday)
Question of the Day:
Why is the x component not always proportional to cos theta?
Answer:
Cos theta is defined as (adjacent/hypotenuse), which is y/mg if theta is opposite of x like in problem #4 on our examples packet. In this case, y is proportional to cos theta. The x component is instead proportional to sin theta, which is (opposite/hypotenuse) or x/mg.
Therefore in these problems:
sin theta=x/mg, or x=mgsintheta
cos theta=y/mg, or y=mgcostheta (Coats-Haan asked Onur if he was fine with this being true. Onur's fine with it.)
Thursday, December 1, 2011
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