Unit 4 Reflection

Rotational and Tangential Velocity
Rotational Velocity:  

• number of complete rotations, revolutions, cycles, or turns per time unit
• same= with merry-go-rounds, train wheels, etc
• know something has the same rotational velocity when connected by a shaft

Tangential velocity:

• same= with gears, belts, etc
• different=  tapered train wheels
• know something has same tangential velocity when covers same amount of distance in the same amount of time.

The further away from the axis of rotation, the more tangential velocity something has.

Train Problem

We know that the wheels have the same rotational velocity because they are connected by a shaft, making them rotate with the same velocity. We know that the larger-inner parts of the wheels have a greater tangential velocity than the smaller-outer parts because they cover more distance in the same amount of time. This makes the wheels self correct because if the wheels start to get off track, the larger-inner parts will go faster, making the trains "correct" again.

Inertia
 Why does a Frisbee roll faster than a hula-hoop down a hill?

The Frisbee will go faster than the hoop because the distribution of mass is closer to its axis of rotation, causing the Frisbee to have less rotational inertia and the hoop.

Mass close to the axis of rotation= less rotational inertia
Mass away from the axis of rotation=more rotational inertia

Conservation of Angular Momentum
Rotational Inertia X Rotational Velocity = Angular/Rotational Momentum

Takes distribution of mass into account

Angular P before= Angular P after
(Rotational Inertia X Rotational Velocity) BEFORE = (Angular/Rotational Momentum) AFTER

The distribution of mass changing causes the Rotational inertia to change, which causes the Rotational velocity to change.

           

Torque
3 Ways to Cause a Larger Torque:

1. Larger lever arm
2. Large Force
3. Both

DISCLAIMER!!!!!!------- Torque is NOT a FORCE

The lower the center of gravity, and the wider the base of support, the more stable something will be

Center of Mass/Gravity

STEP 1:

This picture depicts an unbalanced meter-stick with a torque. It is unbalanced because the center of gravity is not supported. The force comes out of the center of gravity, causing it to tilt counter-clockwise. The lever arm is the distance from the Center of Gravity to the base of support. A Torque=(Force)(Lever Arm)

Step 2: 

This picture depicts a balanced meter-stick. It is balanced because the Center of Gravity is supported. The force comes out of the Center of Gravity, and since it is within the support, the meter-stick will not tilt or topple over. There is no lever arm because the meter-stick is balancing on its Center of Gravity.

Step 3:

A 100g weight was added to the meter-stick, which changed the system's center of gravity, but not the Center of Gravity of the meter-stick itself. On lever arm is from the edge to the base of support, and the other is from the base of support to the meter-stick's center of gravity. Since the Lever arm from the weight to the base of support is longer than the other one, to make the two Torques equal, which has to happen for something to be balanced, there needs to be a greater force on the side with the shorter lever arm. That makes the equation (Force(a))X(Lever Arm(a))=(Force(b))X(Lever Arm(b)) true.

Part 3:

Torque=Force X Lever Arm

We found the Center of Gravity of the meter-stick by itself, which was 50.25 cm. When we added the 100g weight, the lever arm with the weight is 29.5 cm, and the other lever arm is 20.75 cm. w=mg, which means w=0.1Kg*9.8-- Weight=0.98N Since I know the lever arm and the force of one side of the meter-stick, I can find the Torque. Torque=(0.1X9.8)(29.5)--- Torque=28.9 N cm. This is the Torque for both sides because for something to be ballanced, the clockwise Torque= the counter-clockwise Torque. This means that 28.91=20.75(F)--- 1.39=Force. w=mg---- 1.39=m(9.8)---1.39/9.8=0.1418 which, when converted from Kg to g, you get 141.8g. When we weighed our meter-stick, the weight was 142.9g.

Centripetal/Centrifugal Force
Why, when rounding a curve, you lean up against the door:

I am going straight, and want to keep going straight because of Newton's first law. Because of this, I have inertia and keep wanting to go straight. The car door pushes me inward, and I push the car door outward.

Centripetal Force: Center-seeking force
Centrifugal Force: Fictitious Force-- Center-fleeing force

Finding the Mass of a Meter Stick Without Using a Scale Lab

STEP 1:

This picture depicts an unbalanced meter-stick with a torque. It is unbalanced because the center of gravity is not supported. The force comes out of the center of gravity, causing it to tilt counter-clockwise. The lever arm is the distance from the Center of Gravity to the base of support. A Torque=(Force)(Lever Arm)

Step 2: 

This picture depicts a balanced meter-stick. It is balanced because the Center of Gravity is supported. The force comes out of the Center of Gravity, and since it is within the support, the meter-stick will not tilt or topple over. There is no lever arm because the meter-stick is balancing on its Center of Gravity.

Step 3:

A 100g weight was added to the meter-stick, which changed the system's center of gravity, but not the Center of Gravity of the meter-stick itself. On lever arm is from the edge to the base of support, and the other is from the base of support to the meter-stick's center of gravity. Since the Lever arm from the weight to the base of support is longer than the other one, to make the two Torques equal, which has to happen for something to be balanced, there needs to be a greater force on the side with the shorter lever arm. That makes the equation (Force(a))X(Lever Arm(a))=(Force(b))X(Lever Arm(b)) true.

Part 3:

Torque=Force X Lever Arm

We found the Center of Gravity of the meter-stick by itself, which was 50.25 cm. When we added the 100g weight, the lever arm with the weight is 29.5 cm, and the other lever arm is 20.75 cm. w=mg, which means w=0.1Kg*9.8-- Weight=0.98N Since I know the lever arm and the force of one side of the meter-stick, I can find the Torque. Torque=(0.1X9.8)(29.5)--- Torque=28.9 N cm. This is the Torque for both sides because for something to be ballanced, the clockwise Torque= the counter-clockwise Torque. This means that 28.91=20.75(F)--- 1.39=Force. w=mg---- 1.39=m(9.8)---1.39/9.8=0.1418 which, when converted from Kg to g, you get 141.8g. When we weighed our meter-stick, the weight was 142.9g.

Torque Resource

In this video, Mr. Hewitt explains Torque. By showing that an L-shaped piece of wood will either stand straight or topple over, depending on how it is positioned. Torque is equal to the force multiplied by the lever arm. Torque IS NOT equivalent to force! Lever arm is the distance from the axis to the center of gravity. With the L-shaped wood example, Mr. Hewitt shows that if the object's center of gravity is supported, it will stay standing up. However, if there is no support for the center of gravity, the object will fall down. This video is really helpful!

Angular/ Rotational Momentum

Angular/Rotational Momentum is the rotational inertia multiplied by the rotational velocity. Rotational inertia is an object's resistance to spinning around an axis. Rotational velocity is the number of rotations per unit of time. If the guy who flew off of the merry-go-round, had been closer to the axis of rotation (the center of the merry-go-round), it would have been harder for him to fly off of it, due to the tangential speed. Tangential speed increases the further away one is from the axis of rotation, therefore, if the guy had been closer to the center, he would not have spun as quickly, possible saving himself from flying off. The Distribution of mass is important because it affects the rotational inertia. The closer to the axis of rotation the mass is, the quicker the object will spin. The further away the mass is from the axis of rotation, the slower the object will spin. The Distribution of mass changing will cause the rotational inertia to change, which causes the rotational velocity to change.

Unit 3 Reflection

This unit, I learned about…

A.    Newton’s 3^rd Law and  Action/Reaction Pairs

• Newton’s 3^rd Law of Motion: “every action has an equal and opposite reaction.”
• Action Reaction Pair: Hammer pushes nail down, Nail pushes hammer up

B.        Tug of war/horse and buggy

• The Horse exerts a greater force on the ground than the buggy exerts on the ground, and therefore, the system is able to move.

C.        Forces in perpendicular directions

1. Draw Fgravity/Fweight straight down
2. Draw Fsupport Perpendicular to the GROUND
3. Draw Fnet

If done correctly, Fnet will be parallel to the ground and pointing downhill.

"Therefore, the box accelerates downhill."

D.        Gravity and Tides

Spring: Higher Highs
             Lower Lows
Neap:   Lower Highs
             Higher Lows

Gravity:

F=(Gm₁m₂)/d²
F=1/d²
x2 distance   = 1/4 original force
x3 distance   = 1/9 original force
x4 distance   = 1/16 original force
x1/2 distance= x2 original force
x1/3 distance= x9 original force
x1/4 distance= x16 original force

E.        Momentum – and Impulse momentum relationship

P=mv

Change in Momentum= P final- P initial

Change in Momentum= mv final-mv initial

Change in Momentum= Impulse

Impulse=(Force)(change in time)

J=Newtons
P=Kgm/s

Correct way to solve egg toss problem:

P=mv
Change in Momentum= P final- P initial

Regardless of how the egg is stopped, it will go from moving to not moving, therefore the change in momentum is the same, regardless of how it is stopped.

Change in Momentum= Impulse

Since the change in momentum is the same regardless of how the egg is stopped, the impulse is also the same.

J=Ft

The Winners increased the time it took to stop the egg, thus because the impulse is constant we can predict that the force will be less on the egg. A small force is less likely to break.

F.         Conservation of Momentum

****Individual objects can change their momentum, but systems cannot*****

Ptotal before=Ptotal after

Together in the Beginning:

m_av_a+m_bv_b = m_av_a+m_bv_b

Together in the Ending:

m_av_a+m_bv_b=m_a+bv_ab

Cart Problems:

****REMEMBER THAT IF THE CARTS ARE MOVING IN OPPOSITE DIRECTIONS, TO MAKE ONE OF THE VELOCITIES NEGATIVE***********************************************

Extra stuff:

Lab equation Steps:

1. Write equation of a line (y=mx+b)
2. Fill in y and x units (Ptotal before=slope(Vab))
3. Slope=___
4. Compare the theoretical to the actual slope
5. if close, yo confirm the law

Bouncing:

Bouncing creates more force. x2 impulse because 1 for stopping and 1 for pushing off. This is why police vests absorb bullets instead of bouncing them off.

Reflection:

What I have found difficult is remembering the simple things.

I have (hopefully) overcome these difficulties by writing myself disclaimers and making things I find tricky very noticeable.

I don't feel like me creativity, persistence, use of creativity, self-confidence in physics, skills in solving difficult problems, communicating both by my spoken and written words, collaborating with my group members, has changed since the first unit. I still feel confident, though the quizzes after the break took their toll on me a bit, but I think I am  back. It was just hard remembering all the things we had learned before the break, and then having no review time for what I, and the majority of the class, had forgotten. I think that I need to find better ways to not forget things. 

Tides Video Source

I found this video by Hewitt to be very helpful. Since he is the one who wrote the textbook we use, this information can be seen as accurate. Hewitt explains that the elongation of the earth is dependent of the lunar pulls on it. He also explains that the sun plays an affect of the tides. When the sun, moon, and the earth are aligned, there are high tides, called spring tides (although they have no correlation with the spring season). Spring tides will occur at the time of a full or a new moon. When the moon is positioned at 90 degrees between the sun and the earth, neap tides will occur.

Unit 2 Reflection

Topics Covered This Unit

Free Falling (straight down)         Free Falling (throwing things up at an angle)

Free Falling (at an angle)                    Free Falling (throwing things straight up)

 Newton’s Second Law                        Falling with Air Resistance (Skydiving)

Important Relationships

“Acceleration is directly related to force and is inversely proportional to mass.”

a=F/m

Important Equations

d=(1/2)gt²        a²+b²=c²

v=gt                     v=d/t

What Equations go with What

	Vertical	Horizontal
How Far	d=(1/2)gt2	v=d/t
How Fast	v=gt	v=d/t

Newton’s First Law

a=F/m              w=mg              a=1/m              a~F

“Acceleration is directly related to force and is inversely proportional to mass.”

To increase acceleration, increase the force, or decrease the mass. To decrease the acceleration, decrease the force, or increase the mass

Translating it to graph:

y=mx+b à acceleration=Fnet (1/m)            slope= fnet

Falling with Ari Resistance (Skydiving)

	Increasers of Air Resistance
1.)	Increase of Surface Area
2.)	Increase of Speed

When a person falls through the air, their acceleration decreases, their velocity increases and their Fnet decreases

acceleration=(Fweight-Fair)/(mass)

When a person opens their parachute, their acceleration changes direction (because Fair is larger than Fweight), their velocity stays same direction but slows down (because person is still falling), and their Fnet changes direction (upwards) because Fair is larger than Fweight.

Why does a lead ball hit the ground before a ping pong ball when dropped from a building and not when falling from a table?

This is because from the building, there would be enough time for the two balls to reach terminal velocity. The steel ball will go faster because it has a greater weight than the ping pong ball. This makes the lead ball have to compensate by going faster, which increases it’s Fair.

How do the velocities,  acceleration, and net-forces compare when a skydiver is skydiving without the parachute open, and after the parachute is open (both times in terminal velocity)?

The only thing that is different between the two is that the velocity is slower. Netforce is the same, and acceleration is the same because the weight of the diver does not change, meaning their F-weight does not change, meaning that if the diver is to reach Terminal Velocity, it must retain the same F-air, meaning the net-force is the same.

During Terminal Velocity…

Acceleration is 0m/s²                                   Velocity is constant

Netforce is 0N        Diver is at their fastest point possible

Free Falling (Straight Down)

THE ONLY FORCE IS GRAVITY

Weight does not matter

“When an object falls due to the effect of gravity ONLY”

How Far	d=(1/2)gt2
How Fast	v=gt

acceleration=gravity

Free Falling (at an angle)

The only thing that determines time in the air is vertical height

	Vertical	Horizontal
How Far	d=(1/2)gt2	v=d/t
How Fast	v=gt	v=d/t

Falls in a parabolic curve

Free Falling (throwing things up at an angle)

a²+b²=c² will help you find actual velocity

These are special triangles that will appear on a test. The Hypotenuse is the actual velocity (c)

	Vertical	Horizontal
How Far	d=(1/2)gt2	v=d/t
How Fast	v=gt	v=d/t

Acceleration at the top of an object’s path will be 10m/s² because of gravity

Velocity at the top of an object’s path will be the horizontal velocity because vertical velocity=0

Free Falling (throwing things straight up)

Always measure things from rest, meaning measure time and distance from the top of an object’s path, down.

Velocity at the top of an object’s path will be 0m/s

Acceleration at the top of an object’s path will be 10m/s² because gravity does not stop working.