## Sunday, December 16, 2012

### Hooke's Law and the Simple Harmonic Motion of a Spring Lab

The purpose of this lab is to find the force constant of a spring and to also study the motion of a spring with a hanging mass when vibrating under the influence of gravity.

INTRODUCTION:
We first need to understand how to calculate the force of a spring before performing this lab. We look at Hooke's Law which can be described as a spring stretched a distance, x, from its equilibrium position will exert a restoring force that is directly proportional to the distance. The force can be written as:

F = -kx

where k is the spring constant and depends on the stiffness of the spring. The minus sign means that the Force is opposite the direction of the displacement of the spring.

When a mass hangs from a spring, is stretched, and let go, it will oscillate about the equilibrium point of when the mass hangs. From Newton's Second Law and with calculus we can find the period, T (in seconds), from the equation: where m is the mass supported by the spring
This equation will be altered to solve for k:

Now that we know what our equations we are going to be working with, we need to see where we are going to get these values from: In this picture we can collect a good amount of information. The x in Hooke's Law can be found in the picture on the left. When the spring is left alone, the spring will not be stretched out at all. When we hang a mass on it, the spring will stretch and whatever the difference is will be the x value (in meters). In the picture on the right, we can see how we will start the second part of the lab which is finding k in an oscillating motion. The mass will be pulled down and oscillate around the equilibrium point of the hanging mass on the spring.

PROCEDURE:
To begin this lab, we will first need to gather the proper materials:

• spring
• masses
• weight hanger
• meter stick
• support stand with clamps
• motion detector
• Lab Pro Interface
• wire basket
We set up our support rods the same way as the picture above and set the motion detector directly underneath the hanging spring inside the wire basket to protect the detector. We hung the spring on the rod and measured how high off the ground the bottom of the spring was.We hung a 350 g mass on the spring and measured how far the bottom of the spring was from the ground with the meter stick. We found the difference in the length of the spring and recorded the value. We then did it for 450 g, 550 g, 650 g, 750 g, 850 g, 950 g, and 1050 g. We then calculated the normal force of the masses which is equal to the weight force since the (Fnet)y was 0. We will use these values to find the spring constant, k, with the equation:

k = F/-x

With these values we were able to make a graph and were able to solve for k doing a linear fit since the k value is the slope. We plotted the points of the Force applied on the spring vs the change in the length in the spring. We could see that the plot created a linear graph so we performed a linear fit of the points. The slope was -24.5 kg/s^2 and this was our spring constant.

Next, we opened up the file Hooke's Law in Logger Pro because now we are going to stretch the spring with a mass, in equilibrium, 10 cm and allow it to oscillate. We recorded the position of the oscillating mass with the motion detector and allowed it to oscillate for 10 seconds. This gives us the time necessary to perform 5 cycles because in this part of the lab we will be finding an average time, T (sec), per period of the oscillation. We first hung 350 g on the spring and then stretched it 10 cm down and let it go and run for 10 seconds. We then performed this test for all the same masses used in the previous test. This is the graph of the oscillating mass of 1.050 kg on the spring. From it we performed a sinusoidal fit and could now find the amplitude as well as  the period. The amplitude can be found from the A value given in the Auto Fit of the graph which was 0.07 m or 7 cm. The period of the graph can be found using the equation 2PI/B where B is the value given in the graph. The B value is 1.28 sec/cycle. The value we found for the period from using an average was 1.29 sec/cycle.

With the information obtained above, we were able to make a T^2 vs mass graph. From this graph we can confirm equation 2 because we can compare it to the equation of a line y=mx+b. The y would be T^2 in this situation because it changes with respect to x, the slope, m, can be found from (4PI^2/k) because it will always be constant. The x value, which is the variable of the function is the mass since the mass is what is changing. The b value should be 0 because this line should be proportional to every single T^2 vs mass all the way to 0.
We can find the value for k from equation 2 since we know the slope of the graph.

CONCLUSIONS:

In this lab I learned how we could find the spring constant of a spring from 1. Hanging a mass and measuring the displacement of the spring and 2. Having a mass oscillate on the spring.
One factor that could have caused a difference in our values for k is that the oscillating mass could have lost some potential energy causing the spring to oscillate slower. We can further study spring constants by studying a spring that compresses and finding its constant.

## Tuesday, December 11, 2012

### The Ballistic Pendulum Lab

The purpose in this lab is to use a ballistic pendulum to find its initial velocity of a projectile using the conservation of momentum as well as the conservation of energy.

INTRODUCTION:
We will be shooting a steel ball into the bob of a pendulum at a certain height which is where the bob will get stuck at. With this information we will be able to determine the initial velocity of the bob once it receives the moving ball. This picture depicts what is going to happen when the ball shoots into the cup. The ball with mass, m, is shot with an initial velocity, V0, into a cup with mass, M, and the pendulum then rotates and gets stuck on the rubber at a certain height, h.

We can calculate the Kinetic Energy of the bob and ball at the bottom and set it equal to the potential energy at the top since they are equal to eachother. There will be no Potential Energy at the bottom because the height is 0 and the top will not have any kinetic energy because it will have no velocity. From the law of conservation of energy, we must have the same total energy in the initial and final positions. This combines both the conservation of energy equation with the conservation of momentum and shows how we can find v0, which is the initial velocity that the ball is shot with.

We can also find the velocity at which the ball is shot by shooting the ball as a projectile. Instead of shooting the ball into the pendulum, we can move the pendulum out of the ball's shooting range and let it hit a piece of carbon paper for us to measure. This shows what information we can gather from the ball being shot as a projectile. The change in y is the height at which the ball was shot at, the change in x is the distance gone in the x direction, the mass of the ball is m, and the carbon paper is setup for the ball to land on. When it hits the paper a mark will be made on the paper and this will allow us to measure the x direction.
We can find the initial velocity of the ball using kinematic equations. You can see there is a negative in the square root and there should never be a negative in the root but in this case, our change in y is going to be negative because it is starting from a higher point and falling to hit the ground. This will give you a positive number in the square root making it possible to do the calculations.

PROJECTILE:
Now that we have gone through what this lab is asking to find, we can perform the lab. To perform this lab the materials we needed were:
• Ballistic pendulum
• carbon paper
• meter stick
• clamp
• box
• triple beam balance
• plumb
We set up the Ballistic pendulum near the edge of the table in order to clamp it down. We don't want the apparatus to move because then it will alter our results. We then got the metal ball and shoved it onto the rod until it clicked to engage the trigger. This will set the spring to give the ball a constant initial velocity. Once the  arm is left hanging down and not moving we can press the trigger which allows the ball to shoot into the cup of the pendulum allowing it to go into motion. Once the pendulum reaches the rubber part of the apparatus it will get stuck in a notch. The notches were marked every 10 notches and this is where we will take an average height from. We did 9 trials and recorded the heights

. We took an average of the height the pendulum moved. We measured the height difference from where it was shot from to where it landed and came up with 11.2 cm.
Once finishing the 9 trials, we then weighed the mass of the pendulum and the ball. The mass of the ball was 0.0567 kg, the mass of the pendulum was 0.2153 kg, and the average height the ball traveled was 0.112 m.
With these given values we can calculate what the initial velocity of the ball was from the equation we used when combining the law of conservation of energy and the law of conservation of momentum.

For the next part of the lab, we performed the projectile test of the ball. We set up the apparatus to shoot a projectile and land on a piece of carbon paper so we can measure the distance in the x direction. We recorded 5 trials and averaged them.

Our average distance traveled in the x-direction was 2.9128 m. The distance traveled in the y-direction, which was the height the apparatus was set at, was -1.02 m because it fell that distance. We can now use the equation to find v0 from the kinematics equations.

Now that we have 2 values for v0 we can calculate what the percent difference of them are.

CONCLUSION:
In this lab, I learned that you can find the initial velocity of an object by combining the laws of conservation of momentum as well as energy. I knew you could find v0 from kinematics by finding a value of time because we have done that before in previous work. With the two different methods in finding v0 we had a 10.8% difference which was a close value. One thing that could have contributed in a difference was on the ballistic pendulum when it reached the rubber notches. The friction there could have affected it's maximum height it could have reached. I think the method of shooting the ball as a projectile was the best method because in the method where we solved for the energy, there were more sources of error. The impact of the ball hitting the cup could have affected its height as well as the friction along the rubber notches. When the ball moved as a projectile the only thing that could have altered it was air drag. Since it had a small cross sectional area it could practically be negligible in our situation given.

### Inelastic Collisions Lab

The purpose of this lab is to analyze the motion of two low friction carts in an inelastic collision. We will also verify that the law of conservation of momentum is obeyed with this collision.

The materials needed for this lab are:

• Computer with Logger Pro
• Lab Pro
• Motion Detector
• Horizontal Track
• Two Carts
• 500 g masses
• Triple beam balance
• Bubble level
INTRODUCTION:
To understand what we are doing this lab we have to know what momentum is. Momentum can be described by the equation p=mv and SI units are (kg m/s). We are going to have a track set up with 2 carts with masses m1 and m2. The mass of m1 is 499.8 g or .4998 kg, m2 had a mass of 504.1 g or .5041 kg, and the mass of the bar that will be placed on the carts is 494.6 g or .4946 kg. The picture shows m1 being pushed into a stationary m2 on the track. This is how the lab is going to be set up and performed. There will also be a point when we add a bar with nearly 500 g mass that we put on the carts to see what happens.
The law of conservation of momentum states that in a perfectly inelastic collision, which is, when two objects stick together after the collision, your initial momentum should be equal to your final momentum. The initial momentum can be found of both objects. v2 is equal to 0 because it is not moving. The equation for momentum as pointed out before is p=mv. In the final momentum, we add the masses together because it is moving together as one unit. The V is the velocity of the two masses that are moving together.

PROCEDURE:
We set up the track with the carts and motion detector exactly as the picture shown above. We used a bubble level to make sure the track was perfectly level. We used the adjustable feet to make the track level. Once setting up the track we had 2 carts with m1=.4998 kg and m2=.5041 kg (we use our mass in kg because that is what the units are for momentum). We started Logger Pro and opened the Mechanics folder and opened the file Graphlab for our data collecting with the motion detector.

We checked the motion detector to make sure it was working properly. Once we did that we made predictions on what our graph should look like. The graph shows the position vs. time graph of the cart moving by itself, colliding with the other cart, and then showing both carts moving together as time goes on. This graph shows that the cart prior to collision will have a higher velocity than the two carts together. This makes sense because according to the law of conservation of momentum, the momentum should be equal before and after the collision. Since the mass will be greater after, it should move with a lower velocity.
The process of performing the lab was:
1. One person push the Start Collecting button on the LoggerPro Software.
2. Have another person ready to push the first cart into the second cart once the motion detector starts clicking.
3. Let both of the carts move towards the end of the track and end the collection of data.
After collecting the data for the first run, we selected a small area of the graph right before the collision and made a linear fit. This is going to be the initial velocity for the initial momentum. We did the same for right after the collision. This is the final velocity for the final momentum. We performed 2 more trials for this setup.

An Extra Mass On Cart 2
Once completing this set of trials, we then added a mass onto the second mass which was 494.6 g. The total mass of the bar and the cart added up to 998.7 g or .9987 kg since we are working with masses in kilograms. We did the same thing as far as collecting data for this set of trials. We performed 3 total trials with this scenario.
We also made predictions of what the position vs. time, velocity vs. time and acceleration vs time graphs look like: The position vs. time graph would be similar to the one at the beginning. The difference though is that this time, now that there is an extra mass on the cart2 which is going to be after the collision will cause the position to increase slower because more mass is moving at a slower rate. This makes sense because according to the law of conservation of momentum says that if you are going to add more mass to an initial mass with an initial velocity then the new velocity will be significantly less. As the object keeps moving, the velocity will slowly decrease because there is a little bit of friction between the cart and track which is why it slows down a little compared to the initial push to the point of collision. After the collision, when both carts are attached to each other, the masses will slow down a little bit faster. This is why we are finding our velocities right before and right after collision. The starting velocity and the ending velocities are not completely accurate. The accelerations are just describing what the velocities are doing. Since the cart slows down just a little bit before the collision, it will have a small acceleration in the negative direction. After the collision the mass is larger and the friction will be greater than before which is why the masses slow down faster having a greater acceleration in the negative direction.

Extra Mass Moved onto First Cart:
We moved the extra 494.6 g weight on the second cart onto the first cart to see what would happen with a larger mass on the initial momentum. We performed 3 trials just as the previous tests. We made a linear fit just before and just after the collision and made the following table:

Finding Average % Difference:
We next have to find the average percent difference of all 9 trials to see how well the law of conservation of momentum was obeyed in our experiment.

The average was 21.7%. According to that percent we didn't do so well in proving the law of conservation of momentum. The reason is because of the 2nd trial of Test 2. We had a 138% difference on that trial. If we don't take that trial into account and average the other 8 tests we had an average difference of 7.19%.
Based on what we found, without the second trial of the second test, our average difference is very close to verifying the law of conservation of linear momentum with our difference being less than 10%.

% Difference of Kinetic Energy:
When looking at kinetic energy, the equation we need to know is K=(1/2)mv^2. Since we know the final and initial masses as well as velocities we can calculate what the difference is in kinetic energy from before and after the collision. We made a table of the % difference of each trial's kinetic energy.

\
Next we wanted to look at theoretical problems:
1. When mass, m, collides with an identical mass, m, initially at rest.

2. When mass, 2m, collides with mass, m, initially at rest

3. When mass, m, collides with mass, 2m, initially at rest

CONCLUSIONS:
I learned that momentum will always be conserved in a perfect inelastic collision. I also learned that even though Kinetic energy uses the same values for initial and final masses and velocities, it is not conserved. We saw this with our % differences in our momentums compared to our % differences of Kinetic Energy. However, total energy will be conserved, which is the sum of initial kinetic and potential energies are equal to the sum of the final potential and kinetic energies. Therefore if momentum is conserved, kinetic energy will not be conserved. Some possible sources of error obtained in this lab could have been from not having a perfect inelastic collision occur in our 5th trial. We also are taking an average velocity at a moment just before a collision and just after so we won't necessarily be able to get "perfect" numbers giving us an error.

## Wednesday, November 28, 2012

### Balance Torques and Center of Gravity

Our purpose in this lab is to study the rotational equilibrium of a meter stick and to also determine its center of gravity with a system of masses.

The materials needed for this lab are:

• Meter stick
• 3 mass holders
• masses for the holders
• knife edge clamp
• 3 mass clamps

INTRODUCTION:
When an object is not moving, it has Fnet=0. We are looking at balancing a meter stick on a knife edge clamp with masses hanging on each side. We aren't working in straight line motion when working with this particular lab, we are looking at rotational motion. The force in rotational motion is called torque.

Torque can be calculated with the following equation: Since torque is a force and we are balancing the meter stick, the Tnet=0.
The Force is the weight force which is perpendicular to the lever arm. The angle should be 90 degrees. The lever arm would be the distance from the center of gravity to where the mass, or force, is at. Since our value for theta is 90, sin(90)=1 so we don't have to take consideration of the angle in this case.

PROCEDURE:
To begin the lab we first set the meter stick on the knife edge clamp and set it to where the meter stick was perfectly balanced. This will be where the center of mass is. We found that to be at 49 cm on the meter stick.
We measured out two different total masses of 100 g or more with the masses, mass hanger, and mass clamps because you must take into account the mass of the hanger and the clamps. They affect the torque as well since it is added to the meter stick. Our first mass was 220.6 g and our second was 171.5 g. We set our second mass 40 cm away from the center of gravity and tried to balance the meter stick with the first mass on the other end of the meter stick. That ended up being 31.16 cm away from the center of gravity. We calculated our values to see how close to zero for a Tnet we obtained. We calculated our Tnet value and compared it to how close to 0 we could get with our measurements we obtained.
Next we put the two masses in different locations but on the same side. We added the third mass clamp with hanger and masses that was 270.8 g on the opposite side and moved it until the meter stick was balanced again. We also found how close we got to a Tnet of 0 by solving. We set the same weights from the first test on the same side and added one to the left and balanced it out. We calculated our Tnet and compared it to what it should have been.
For our next step, we replaced the third mass with an unknown mass. We left masses 1 and 2 on the right side at 30 cm and 15 cm respectively from the center of gravity. We adjusted the unknown mass to balance the meter stick. Our distance from the center of gravity was 30.2 cm. We solved for what the mass should be and then measured the mass to see how close to the actual we obtained. We calculated the percent error we had. We changed our third mass on the left side with a new unknown mass. We found it's mass by using our equations. We compared the mass we obtained with the actual mass of the unknown mass and the mass clamp.

Next, we put 200 g (including mass clamps and hangers since they aren't part of the ruler and are causing the ruler to have torque) at the 90 cm mark and find a new balance point on the meter stick. We found that point to be 78 cm, which is 12 cm away from the mass hanging. The center of gravity of the meter stick is 29 cm away. We can say that the center of gravity of the meter stick is where the weight force is located because that's the point where the mass is concentrated. The new center of gravity will be at the 78 cm mark. We calculated what the mass should be and compared it to the actual mass of the meter stick. The mass of the clamp that holds the meter stick is irrelevant because it is keeping the meter stick in equilibrium. We found the new center of gravity with a 200 g mass hanging at the 90 cm mark on the meter stick. We solved to find what the mass of the meter stick is and compared it to the actual mass.
We now had to picture a similar scenario with the 200 g weight still at 90 cm but now we add an additional 100 g at the 30 cm mark. We had to calculate to find it's center of gravity and where it should be on the meter stick. We found it to be at 65.5 cm once we did our calculations. I then went ahead and made it the center of gravity and checked to see if that center would give us a net torque of 0 or how close it would be. We calculated to find the center of mass of the meter stick when a 200 g mass is hanging at the 90 cm mark and a 100 g mass is hanging at the 30 cm mark. We calculated our center of mass to be at 65.5 cm on the meter stick.

CONCLUSION:
In all of our calculations we had a very small percent error throughout the entire lab. This is because there is a very small window, as far as length is concerned, that we can obtain to keep our meter stick in perfect balance. Also when we were calculating our percent error, we weren't comparing the net torques because our theoretical value was 0 and we cannot have a 0 in a denominator. Instead we compared the theoretical length of where we should have put our mass and where our masses actually were on the meter stick. We weren't able to apply the last scenario of finding a center of mass to the actual meter stick to see if 65.5 cm was actually the center of mass when 200 g is hanging at 90 cm and 100 g is hanging at 30 cm. We did use calculations to see how close to 0 we got for a torque net but weren't able to compare to any tests. We weren't able to prove the last scenario to be true. I calculated what the Tnet would be with what we found if 65.5 cm on the meter stick is really the new center of mass with the given masses at the given lengths.

## Tuesday, November 13, 2012

### Human Power Lab

In this lab we are determining the power output of ourselves from walking up stairs.

In order to perform the lab we are going to need:

• two meter metersticks
• a stopwatch
• kilogram bathroom scale
INTRODUCTION
Power can be described as the rate at which work is done which can also be translated as the rate at which energy is converted from one from to another. To show this we look at the equation:

Change in PE = mgh

where:
• PE is the potential energy
• m is the mass of the object working
• g is the acceleration of gravity
• h is the vertical height gained
We can use this equation to find the change in potential energy which we need because the equation for  power output is:

Power = (change in PE) / (change in time)
where:
• change in time is the time it takes to climb the vertical height This is the unit analysis of the lab we are doing. The change in potential energy is read as (kgm^2/s^2) which is the same as Newton meters (Nm) which is equal to Joules.

PROCEDURE:

We first started this lab out by weighing ourselves on a kilogram bath scale. We measured our mass in kg. Once we weighed ourselves, we then went to the stairs that were down the hall from the lab and measured the height from the floor of the first floor to the floor of the second floor. We wanted the height in meters. This is a sideview of the stairs to show the height we climbed.
We each started at the bottom of the stairs and were timed how long it took to reach the top stair. We performed two trials each for our data.

Once we all collected our two time trials we then calculated our value for our power output. We calculated our value in watts.
Once I found my average Power output in watts, I then solved for it in Horse Power.

DISCUSSION:
I calculated a % difference for what my values were compared to the rest of the class.
The values I obtained for Power were greater in watts and in Horsepower by more than 10% for each. This meant that I output more power than the average of the class going up the stairs.

Answers to Questions:
1. It is OK to use your hands and arms on the handrailing to assist you in your climb because we are calculating the power our whole body outputs to get us up the stairs as fast as we can. If we use our hands and arms then we are outputing more power to get ourselves up the stairs.\

2. Some problems that can affect the accuracy of this experiment are human error in the stop watch because when told to go the watch isn't started when the person begins to move, it begins when someone says go. People's reaction times are sometimes slower than others. You can also obtain error from the way we measured the height of the stairs. We measured from the floor of the first floor to the half way point where the direction the stairs go is turned around, and then from there to the floor of the 2nd floor.

Conclusion:
I learned in this lab how to find the power output of a person. The process we learned is when something is changing in height. We were able to calculate our power going up the stairs. I output more power than the class average which means I was working harder since our height was all the same and Work can be expressed as m*g*h. I also learned how to apply the % difference when appropriate. We couldn't solve for a percent error because we weren't given an accepted value or a theoretical value to compare from. We only had others' values to compare to. This is why we used percent difference because we could only compare values obtained from fellow classmates.

Answers to Follow Up Questions:
1. They would both produce the same amount of work because they have the same mass, working against the same acceleration due to gravity, and traveling the same height. However, Hinrik would output more power than Valdis because Hinrik does the work faster and Power is a comparison of the amount of work done in a certain amount of time. The longer it takes a person to work, the less power output they will have.

2.

3.

4.

## Monday, October 15, 2012

### Centripetal Force

In this lab we are verifying Newton's Second law of Motion for the case of uniform circular motion. To better understand this, we are going to try and show that the force of circular motion to cause a spring to stretch is equal to the force of a mass hanging to stretch the spring the same amount.

The materials needed for this lab are:
• Centripetal force apparatus
• Metric scale
• Verneir Caliper
• Stop Watch
• Slotted weight set
• Weight hanger
• Triple beam balance
INTRODUCTION
Before starting this lab we need to understand a couple concepts first. The centripetal force apparatus rotates a known mass in a circular path with a known radius. When we time the motion for a number of revolutions we can find the distance traveled and calculate the velocity. We can use Newton's Second Law to determine the velocity with the equation:

F = (mv^2)/r

m: mass of the object
v: velocity
r: radius
F: centripetal force

This is derived from the equation F = ma and in uniform circular motion the value for acceleration, a, is given by:
a = (v^2)/r

PROCEDURE
To set up this lab we first measured the mass of the weight. We then put a centripetal force apparatus on the table and leveled it by adjusting the legs appropriately. Next, attached the weight to the end of arm of the apparatus on a string and let it hang until it stopped moving. We adjusted the post to where the weight was directly above the post. Once the post was tightened down, we attached the spring to it.

Once everything was set up, we then measured the radius of the apparatus from the center of the rotating pole to the string where the weight hung from. This is going to be our radius, r, for the equations we use.

After having the measurements we spun the apparatus until the spring stretched far enough to where the tip of the weight reached the post. We timed how long it took the apparatus to go around 50 complete revolutions. We took 3 trials of this and put our measurements on to a table.

The next part of the lab is to now find the Force that is required to stretch the spring the same distance that spinning the apparatus did. In order to do that you must attach a string to the weight opposite of the spring and hang a mass hanger over the pulley of the apparatus. Once doing that, begin adding slotted weights to the hanger until the weight is over the post just as it was when we were spinning the apparatus. Record the mass and calculate the force that was required to stretch the spring.

Repeat this experiment except this time around, add a 100 g slotted weight to the hanging weight.

DATA ANALYSIS
All the data that was collected we put into tables. In order to do so we first had to do some calculations. We had to find the linear speed, the centripetal force, the force of the hanging mass, and the percent difference.

First we calculated the linear speed for each trial and found an average.

Once we had and average velocity, we could then calculate for our centripetal force with the given equation F = (mv^2)/r. We also solved for the force of the hanging mass

These were the calculations for the mass of 0.4492 kg. We also did the calculations for the mass of 0.5492 kg. The table represents all of our calculations.

DISCUSSION
In this lab I learned how to calculate the centripetal force of an object in a circular motion. I also learned that the centripetal force to stretch the spring with a weight attached moving in a circular path is equal to the force required to pull the string. This makes sense because the faster you spin something, the force on that object that is pulling it away is greater.
Some sources of error that could have occurred in this lab is that we may not have given a completely constant velocity on while rotating the apparatus. Also there were a couple of times where the weight wasn't completely over post; sometimes the spring was not stretched enough and sometimes it was stretched too much.