Thursday, February 13, 2014
Simple Machines Resource
Sunday, February 2, 2014
Work and Power Resource
Thursday, January 30, 2014
Unit 4 Blog Reflection
Things I learned this Unit:
Rotational Motion
Tangential speed- the distance an object covers in a given time. Depends on radial distance.
Rotational speed- the amount of rotations or portion of a rotation an object performs in a given time
Gears: Same tangential
speed, the smaller gear will have to have a faster rotational velocity
to account for the equivalent tangential speed.
Example- 8 teeth vs 4 teeth.
Smaller gear will have twice the rotational velocity as the bigger one
because it has to rotate 2 times to go the same distance it takes the
bigger gear to rotate once.
DIFFERENT RADIUS FROM AXIS OF ROTATION= DIFFERENT TANGENTIAL VELOCITY
Rotational Inertia + Conservation of Angular Momentum
Rotational inertia- property
of an object to resist changes in the spin. Distribution of mass affects
RI. Easier to spin= less rotational inertia, harder to spin= more
rotational inertia.
The greater the distance a mass is from its axis of rotation, the more rotational inertia it will have.
Angular momentum before= Angular momentum after
RI x RV before= RI x RV after
A hoop will have more
rotational inertia than a steel ball because its mass is further from
its axis of rotation, it will therefore lose a downhill race. (same goes
for frozen water bottle vs water bottle)
Torque and Center of Mass
Torque- something that causes a rotation, = Force x Lever Arm
Lever arm- the distance from an object's axis of rotation.
When an object is balanced its counterclockwise torque is equal to its clockwise torque
Center of mass- average position of mass
An object will fall when its center of gravity is outside of its base of support
Examples of torque + center of mass problems:
Why are football players told to keep their legs bent and shoulder width?
To widen their base of
support and get their center of gravity closer to the base of support,
they will then have to be pushed harder to get their center of gravity
outside of their base of support and fall.
Large wrench= more torque
10 N ball rests on one end of
a meter stick with a 1m lever arm. How much must a ball on the other
side weigh if it is 5m away from its axis of rotation?
10 x 1= 10
10/5= 2N
Centripetal Force
Centripetal force is a center seeking force
Centrifugal force does not exist
Why are racetracks banked?
So that they can have more centripetal force from the x component of the support force coming from the track.
In this unit I had difficulty with the concept of the racetrack being banked, but after being shown the force vectors I felt completely fine with the subject. I overcame this difficulty by learning to think of force vectors whenever provided such a picture.
My Problem Solving Skills, Effort, and Learning
Throughout the unit my problem solving skills improved by learning to think of things in different ways such as in force vectors or as in centripetal forces interacting with objects in the world. My effort in the class has been fairly consistent with a few dips in effort due to the fact that I grasped the material sooner than classmates and waited for them to see the reasoning behind a problem.
My goal for next unit is to have my effort back to where it was before.
My previous goal of not procrastinating has improved greatly and I often feel better prepared for class.
This unit helps me understand why in football I have to keep my legs spread out and my knees bent as to not be hit over so easily.
My group's unit podcast
Monday, January 20, 2014
Meter Stick Mass Blog
The above image includes the pictures for step 1 parts A, B, and C.
A. The equations that we used for our initial plan are as follows:
Torque= Force x Lever Arm
Torque Clockwise = Torque Counter-Clockwise (when balanced)
W=mg
Our initial plan was to multiply the mass of the added weight by 9.8 to give us the total weight added to the ruler. We would then find the torque on the side of the weight (we'll use counter-clockwise torque to describe this) by multiplying the added weight by the distance from the pivot point (Lever arm, 30cm).
Since the center of mass of the meter stick itself wasn't changed by adding weight, its center of mass remained at 50 cm. We then found the distance to the pivot point from 50cm and found that it was 20cm for the clockwise lever arm. We did not know the weight of the meter stick, so we plugged in x for that.
The Clockwise torque = the Counter-Clockwise torque when balanced, so we set the two equations of (0.98N)(30cm)= (X)(20cm).
We would then solve for the weight and convert the weight to mass by dividing it by 9.8.
The above image includes all the work that I did in step 3 describing our procedure.
B. Our initial plan worked.
Balancing point was at 70cm, the lever arm of the weighted side was 30cm, the added weight was 0.1 kg, which weighs .98 N. The center of mass of the meter stick was at 50cm. the lever arm from the center of mass of the stick to the pivot point was 20cm. Our final mass of the meter stick was 150g.
C.
For the measurements we took see part B.
Our initial plan worked to solve the problem. When things are balanced the Counter-Clockwise torque will equal the Clockwise torque. The center of mass of the meter stick was unchanged throughout the whole lab. We remembered that when we found the weight of the meter stick we needed to convert it to mass by dividing it by 9.8. The actual mass of the meter stick was only 0.8 off of what we had calculated.
D.
A. The equations that we used for our initial plan are as follows:
Torque= Force x Lever Arm
Torque Clockwise = Torque Counter-Clockwise (when balanced)
W=mg
Our initial plan was to multiply the mass of the added weight by 9.8 to give us the total weight added to the ruler. We would then find the torque on the side of the weight (we'll use counter-clockwise torque to describe this) by multiplying the added weight by the distance from the pivot point (Lever arm, 30cm).
Since the center of mass of the meter stick itself wasn't changed by adding weight, its center of mass remained at 50 cm. We then found the distance to the pivot point from 50cm and found that it was 20cm for the clockwise lever arm. We did not know the weight of the meter stick, so we plugged in x for that.
The Clockwise torque = the Counter-Clockwise torque when balanced, so we set the two equations of (0.98N)(30cm)= (X)(20cm).
We would then solve for the weight and convert the weight to mass by dividing it by 9.8.
The above image includes all the work that I did in step 3 describing our procedure.
B. Our initial plan worked.
Balancing point was at 70cm, the lever arm of the weighted side was 30cm, the added weight was 0.1 kg, which weighs .98 N. The center of mass of the meter stick was at 50cm. the lever arm from the center of mass of the stick to the pivot point was 20cm. Our final mass of the meter stick was 150g.
C.
For the measurements we took see part B.
Our initial plan worked to solve the problem. When things are balanced the Counter-Clockwise torque will equal the Clockwise torque. The center of mass of the meter stick was unchanged throughout the whole lab. We remembered that when we found the weight of the meter stick we needed to convert it to mass by dividing it by 9.8. The actual mass of the meter stick was only 0.8 off of what we had calculated.
D.
Thursday, January 16, 2014
Center of mass/ Center of gravity resource
Sunday, January 12, 2014
Angular Momentum Resource
Monday, December 9, 2013
Unit 3 Blog Reflection
Things I learned this unit:
-Action/Reaction pairs/ Newton's 3rd Law:
In this section of the unit we studied Newton's 3rd law (every action has an equal and opposite reaction) and how it works in accordance with objects in the world. For something to be classified as an action/reaction pair they have to use The Same verb and The Same Two objects acting in opposite directions. For example: apple pushes down on table, table pushes up on apple. Not: earth pulls down on apple, table pushes apple up. We also did a discovery lab for this section that showed us that when a mack truck hits a smaller car, the force experienced by both of them is The same.
-Tug of war/ Horse and buggy:
This section of the unit came directly off of Newton's 3rd law in how we applied it. Take for example a tug of war match. In this match merely pulling on the rope with your arms will be useless as the rope/ other team will pull back on you with equal and opposite force, yielding no net gain of ground. To win a tug of war match you must be able to push harder on the ground as seen when we had the boys put on socks and girls have shoes on. The way to answer a horse and buggy/ tug of war problem is by first addressing the action/ reaction pairs seen in the problem (ie forces on rope, forces on ground w/ winning team pushing harder on the ground therefore the ground pushes the winning team further and or stronger). Next you have to explain how the object or team that pushes harder on the ground will be pushed harder by the ground and will therefore have more force. This is one of the big problems of the section and will undoubtedly appear on tests and exams in the future.
-Adding Forces/Vectors at angles:
In this section we addressed the fact that when vectors are at different angles from each other we cannot simply add the components, but rather we must find the resulting force of different components, and then use the Pythagorean theorem to find the real net force. This in part explains why a box will slide down an inclined surface. As gravity acts only downward, there must be a support force. However this support force must be perpendicular to the surface that the box is on. When we use our box method for finding a resulting force we find that the fnet points down the slope. Another problem that we encounter in this section is one involving the tension of multiple ropes that a ball or object hangs from. To do this we use the force of gravity to find a support force upward. Then we have a line connecting the tip of the arrow and intersecting one of the ropes. Note that this line must be parallel to the other rope! Then we shade in the resulting areas that have been cut off by these lines. The rope with a longer ftension will be more likely to break.
-Gravity and Tides:
In this section is the concept of universal gravitation. This means that every object with mass in the universe attracts every other object with mass in the universe. The universal gravitational force equation is as follows: F=G(m1)(m2)/d^2 where G= 7.0x10^-11. It is important to note that the Distance is Squared. To solve one of these problems we plug in given information, then separate the x10^ etc... from the other numbers. We solve both of the separate information sets by adding exponents when multiplying, subtracting exponents when dividing, and multiply exponents when raising an exponent to another exponent. We also addressed tides in this section. It is important to note that the tides are caused by The difference in forces acting on the Earth by the MOON not the sun. There is a bulge around the earth since the tides are caused by the difference in forces. The kinds of tides we have are Spring and neap tides, spring tides occur when the moon and earth and sun are all in a line, this causes higher high tides and lower low tides. A neap tide occurs when the moon is perpendicular to the earth in relation to the sun which causes higher lows and lower highs.
-Momentum/ Impulse + their relationship:
Momentum is defined as an object's mass times its velocity. In symbols it is shown as p=mv. An object's change in momentum is defined by the equation deltap= pfinal - pinitial. Impulse = force times the change in time. Written as J=f(delta t). Change in momentum = Impulse. Therefore pfinal-pinitial=force(delta t). This is important so that we can solve for a force or momentum given a problem with the right information. A big problem of this unit is a question about airbags and why they keep us safe. We have to explain that no matter how a person is stopped they go from moving to not moving, therefore their change in momentum will be the same. Then we explain that since their change in momentum is the same, so will be their impulse. With an airbag the person decelerates over a longer period of time which means that there is less force on the individual= less injury.
-Conservation of Momentum:
In this section we learned that momentum is conserved in any collision. This means that momentum before a collision will equal the momentum after a collision. The ways that we can solve problems for if the object stick or don't stick are as follows:
Don't stick- MaVa+MbVb=MaVa+MbVb.
Stick- MaVa+MbVb=Ma+b(Vab).
It is important to know that an individual object's momentum can change without violating the law of conservation of momentum as the momentum of the system is not changed.
What I have found difficult about what we have studied is when we learned about tides and how the tides were in a bulge shape around the Earth. I had a hard time seeing why the bulge on the other side of the earth from the moon was formed.
I overcame this difficulty after Ms. Lawrence explained that it is because of the Difference in forces on the earth that caused the bulges and comparing that to the center of the earth means that one side will have a negative force, meaning it is going in the opposite direction.
My problem solving skills, effort, and learning:
My effort towards class was held constant and high at all times during this unit towards homework and classwork as well as projects and blogs. I tried my best to find understanding when I was confused and have noted the mistakes I have made along the way.
I had a little trouble having patience with problems as some of them were repetitive and I had to resist the urge to say didn't we already do this? I still enjoyed the unit no matter when that question popped up. In my communication with partners I often try to offer what knowledge I may have because I can easily grasp the concepts provided in class, sometimes faster than other students. I enjoy helping others reach understanding of a certain topic and seeing them do well on it in the future.
My goal for the next unit is to not procrastinate and do work at the last minute.
The previous goal I had of asking questions in class was reached in this unit I felt pretty adequately and I will continue on with this goal throughout the year.
I can connect This unit to every day life by seeing car crashes or seeing examples of newton's 3rd law (jumping off a boat) or conservation of momentum (catching a ball and rolling backwards in a rolly chair).
-Action/Reaction pairs/ Newton's 3rd Law:
In this section of the unit we studied Newton's 3rd law (every action has an equal and opposite reaction) and how it works in accordance with objects in the world. For something to be classified as an action/reaction pair they have to use The Same verb and The Same Two objects acting in opposite directions. For example: apple pushes down on table, table pushes up on apple. Not: earth pulls down on apple, table pushes apple up. We also did a discovery lab for this section that showed us that when a mack truck hits a smaller car, the force experienced by both of them is The same.
-Tug of war/ Horse and buggy:
This section of the unit came directly off of Newton's 3rd law in how we applied it. Take for example a tug of war match. In this match merely pulling on the rope with your arms will be useless as the rope/ other team will pull back on you with equal and opposite force, yielding no net gain of ground. To win a tug of war match you must be able to push harder on the ground as seen when we had the boys put on socks and girls have shoes on. The way to answer a horse and buggy/ tug of war problem is by first addressing the action/ reaction pairs seen in the problem (ie forces on rope, forces on ground w/ winning team pushing harder on the ground therefore the ground pushes the winning team further and or stronger). Next you have to explain how the object or team that pushes harder on the ground will be pushed harder by the ground and will therefore have more force. This is one of the big problems of the section and will undoubtedly appear on tests and exams in the future.
-Adding Forces/Vectors at angles:
In this section we addressed the fact that when vectors are at different angles from each other we cannot simply add the components, but rather we must find the resulting force of different components, and then use the Pythagorean theorem to find the real net force. This in part explains why a box will slide down an inclined surface. As gravity acts only downward, there must be a support force. However this support force must be perpendicular to the surface that the box is on. When we use our box method for finding a resulting force we find that the fnet points down the slope. Another problem that we encounter in this section is one involving the tension of multiple ropes that a ball or object hangs from. To do this we use the force of gravity to find a support force upward. Then we have a line connecting the tip of the arrow and intersecting one of the ropes. Note that this line must be parallel to the other rope! Then we shade in the resulting areas that have been cut off by these lines. The rope with a longer ftension will be more likely to break.
-Gravity and Tides:
In this section is the concept of universal gravitation. This means that every object with mass in the universe attracts every other object with mass in the universe. The universal gravitational force equation is as follows: F=G(m1)(m2)/d^2 where G= 7.0x10^-11. It is important to note that the Distance is Squared. To solve one of these problems we plug in given information, then separate the x10^ etc... from the other numbers. We solve both of the separate information sets by adding exponents when multiplying, subtracting exponents when dividing, and multiply exponents when raising an exponent to another exponent. We also addressed tides in this section. It is important to note that the tides are caused by The difference in forces acting on the Earth by the MOON not the sun. There is a bulge around the earth since the tides are caused by the difference in forces. The kinds of tides we have are Spring and neap tides, spring tides occur when the moon and earth and sun are all in a line, this causes higher high tides and lower low tides. A neap tide occurs when the moon is perpendicular to the earth in relation to the sun which causes higher lows and lower highs.
-Momentum/ Impulse + their relationship:
Momentum is defined as an object's mass times its velocity. In symbols it is shown as p=mv. An object's change in momentum is defined by the equation deltap= pfinal - pinitial. Impulse = force times the change in time. Written as J=f(delta t). Change in momentum = Impulse. Therefore pfinal-pinitial=force(delta t). This is important so that we can solve for a force or momentum given a problem with the right information. A big problem of this unit is a question about airbags and why they keep us safe. We have to explain that no matter how a person is stopped they go from moving to not moving, therefore their change in momentum will be the same. Then we explain that since their change in momentum is the same, so will be their impulse. With an airbag the person decelerates over a longer period of time which means that there is less force on the individual= less injury.
-Conservation of Momentum:
In this section we learned that momentum is conserved in any collision. This means that momentum before a collision will equal the momentum after a collision. The ways that we can solve problems for if the object stick or don't stick are as follows:
Don't stick- MaVa+MbVb=MaVa+MbVb.
Stick- MaVa+MbVb=Ma+b(Vab).
It is important to know that an individual object's momentum can change without violating the law of conservation of momentum as the momentum of the system is not changed.
What I have found difficult about what we have studied is when we learned about tides and how the tides were in a bulge shape around the Earth. I had a hard time seeing why the bulge on the other side of the earth from the moon was formed.
I overcame this difficulty after Ms. Lawrence explained that it is because of the Difference in forces on the earth that caused the bulges and comparing that to the center of the earth means that one side will have a negative force, meaning it is going in the opposite direction.
My problem solving skills, effort, and learning:
My effort towards class was held constant and high at all times during this unit towards homework and classwork as well as projects and blogs. I tried my best to find understanding when I was confused and have noted the mistakes I have made along the way.
I had a little trouble having patience with problems as some of them were repetitive and I had to resist the urge to say didn't we already do this? I still enjoyed the unit no matter when that question popped up. In my communication with partners I often try to offer what knowledge I may have because I can easily grasp the concepts provided in class, sometimes faster than other students. I enjoy helping others reach understanding of a certain topic and seeing them do well on it in the future.
My goal for the next unit is to not procrastinate and do work at the last minute.
The previous goal I had of asking questions in class was reached in this unit I felt pretty adequately and I will continue on with this goal throughout the year.
I can connect This unit to every day life by seeing car crashes or seeing examples of newton's 3rd law (jumping off a boat) or conservation of momentum (catching a ball and rolling backwards in a rolly chair).
My group's Podcast for this Unit:
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