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

Purpose: To determine the force of a spring and to study the motion of a spring and mass when vibrating under influence of gravity.

2. Make a plot of the downward force applied to the spring verses the displacement of the spring.

Spring vs. Displacement Graph

3. Create a table with data.

Data Table

Graph for m = 1050g

Conclusion: This lab was very difficult. We were supposed to use Hook's Law for this lab. I learned that the force and the displacement are inversely proportional. If one is increasing the other is decreasing. To make this lab better, it would be good to have learned Hook's Law first and then tried the lab. Some sources of error would be that the masses on the spring would get too close to the motion detector so it wouldn't read all of it.

The Ballistic Pendulum

Purpose: To use the ballistic pendulum to determine the initial velocity of a projectile using conservation of momentum and conservation of energy.
Part 1:

Ball shot into cylinder data

 5. From these data calculate the initial velocity v using equations V = (2gh)^1/2 and mv = (M+m)V.

mv = (M+m)(2gh)^1/2

v = [(M+m)(2gh)^1/2]/m

v = [(.1994+.0573)(2*9.8*.096)^.5]/.0573
v = 6.145 m/s

Part 2:

Ball shot not into cylinder data

2. From the values of deltaX and deltaY calculate Vo by the use of equations deltaX = Vox*t + .5ax*t^2 and deltaY = Voy*t + .5ay*t^2.

deltaX = 2.7319 and deltaY = -1.006

Vox = deltaX/(deltaY/-4.9)^.5

Vox = 2.7319/(-1.006/-4.9)^.5

Vox = 6.029

3. Find the percent difference

[(6.145 - 6.029)/6.029]*100 = 1.92% difference

Conclusion: This lab was to have us find velocity for a projectile object. I learned that it is best to have the same person launch the ball each time. Otherwise it is one more variable that isn't controlled. The error in our lab could have come from having different people launch the ball. To make this lab better we can keep as many things consistent as we can.

Inelastic Collisions

Purpose: To analyze the motion of two low friction carts during an inelastic collision and verify that the law of conservation of liner momentum is obeyed.

Mass bar 1: .497 kg
Mass bar 2: .495 kg
Cart 1: .501 kg
Cart 2: .517 kg

2. Does it provide a reasonable graph of the motion of the cart?
- The graph showed a reasonable graph for the motion of the cart closest to the detector and then again when the carts collided and stuck.

3. What should these graphs look like?
- The graphs of position vs. time should be increasing throughout the whole graph. At the point of collision, the slope is small that before the collision.

- It is a good approximation to get the velocity right before and after the collision if momentum is conserved. P of C1 proportional to P of C1 + C2.

Trials with different mass

5. The velocity vs. time graph is all constants at different values during certain time intervals. The acceleration vs. time graph is a constant zero for the entire graph.

7. For most of the trials we got around 20% difference. There was a few that we were about to get around 10% difference. In the final set of trials, we got one trail around 6% and another at just under 4%. The law was obeyed fairly well. We could have done better. However, there are sources of error to take into account.

Kinetic energy lost

8. Kinetic energy = .5mv^2











9a) A mass, m, colliding with an identical mass, m, initially at rest.

deltaK/Ki = (Kf-Ki)/Ki = Kf/Ki - 1

9b) A mass, 2m, colliding with a mass, m, initially at rest.

 deltaK/Ki = (Kf-Ki)/Ki = Kf/Ki - 1 

Kf/Ki - 1  = .5(2m)Vf^2/.5mVi - 1 = 2Vf^2/Vi - 1

Vf = Vi/2

2Vf^2/Vi - 1 = 2(Vi/2)/Vi - 1 = -.5

9c) A mass, m, colliding with a mass, 2m, initially at rest.

deltaK/Ki = (Kf-Ki)/Ki = Kf/Ki - 1 

Kf/Ki - 1  = .5(m)Vf^2/.5(2m)Vi - 1 = Vf^2/2Vi - 1

Vf = Vi/2

Vf^2/2Vi - 1 = (Vi/2)/2Vi -1 = -.75

Conclusion: In this lab we learned about inelastic collisions and conservation of momentum. I learned about the kinetic energy and how to find the loss of kinetic energy when there is a collision. One source of error would be friction. We did not take into account the friction on the track. Also, depending on where the war was in relation to the motion detector, it may not have read it. The closer the carts get to the motion detector the harder it is to read it. To make this lab better, we would have to do more trials. Also, do all the calculations right after the trials so that we can use the best results.

Balanced Torques and Center of Gravity

Purpose: To investigate the conditions for rotational equilibrium of a rigid bar and to determine the center of gravity of a system of masses.

Clamp 1: .02305 kg
Clamp 2: .0212 kg
Clamp 3: .0195 kg
Clamp 4: .0215 kg
m actual = .1469 kg
Meter stick: .091 kg
Balance point: .486 m

The meter stick balanced

2) C2 + .1 kg at .1 m                  C3 + .15 kg at .762 m

Meter stick with the two masses balanced

g*length*(C2+mass) = g*length*(C3+mass)

(9.8)(.386)(.0212+.1) + (9.8)(.276)(.0195+.15)

.45848N = .45846N

% difference = .0002%

- It is important to add the mass of the clamp into the calculations because they are not small enough to be negligible.

3) C3 + .15 kg at .19 m              C2 + .1 kg at .38 m                   C4 + .2 kg at .769 m

Meter stick balanced with three masses

g*length*(C3+mass) + g*length*(C2+mass) = g*length*(C2+mass)

(9.8)(.296)(.0195+.15) + (9.8)(.106)(.0212+.1) = (9.8)(.283)(.0215+.2)

.4917N + .1259N = .6143N

.6176N = .6143N

% difference = .0033%

4) C3 + .15 kg at .19 m              C2 + .1 kg at .38 m                 C4 + m at .86 m

Meter stick balanced looking for mass of meter stick

g*length*(C3+mass) + g*length*(C2+mass) = g*length*(C4+m)

(9.8)(.296)(.0195+.15) + (9.8)(.106)(.0212+.1) = (9.8)(.374)(.215+m)

.4917 +.1259 = (9.8)(.374)(.0215+m)

.6176 = (3.6652)(.0215+m)

.1685 = .0215+m

.147 = m

% difference = .07%

5) BP at .486 m                          New fulcrum at .78 m       C2 + 200 at .9 m

New fulcrum work and picture

- The clamp holding the meter stick shouldn't be included in the mass because it is holding it balanced not cause more or less mass on it to cause it to change the balance point.

Work and picture for masses at set points

Conclusion: This lab we worked with torque. We were able to find an equilibrium for different scenarios with a meter stick. I learned that there are several ways to add the clamps into the amount of mass on the clamp. Also, I learned that torque is force*lever arm. For us it was simple because there wasn't an angle to deal with. Some of the source of error would be the balance support. The piece supporting the stick wasn't exactly level by itself. Another is that there was a hole drilled into the stick. This changes where the balance point should be. Also, the table and the block of wood that the system was on could not have been balanced. To improve this lab, we would need a better meter stick and to make sure everything was level and working correctly.

Human Power

Purpose: To determine the power output of a person.

Vertical distance between the ground floor and the second floor: 4.29 m
Time 1: 13.47 s

Time 2: 13.31 s   

average time: 13.39 s

F = 618 N

delta PE = mgh

delta PE = 618*4.29

delta PE = 2651.22

Power = 2651.22/13.39

Power = 198 watts

Horse power = 198*.00134102209

Horse power = .2655

Stair diagram

Questions:

1. Is it okay to use your hands and arms on the hand railing to assist you in your climb up the stairs? Explain why or why not.

- It is not okay because it can change the time it takes you to climb the stairs. By changing the time, it changes the power exerted. 

2. Discuss some of the problems with the accuracy of this experiment.

- One problem could be the measurement of the stairs. It could have been off a little. Also, the time. Since the person climbing the stairs wasn't timing themselves, they might not have started walking at the exact time the person started the timer. Along with that, when stopping, the timer may not have stopped exactly when they got to the top of the stairs.

Conclusion questions:

1. Two people of the same mass climb the same flight of stairs. Hinrik climbs the stairs in 25 seconds. Valdis takes 35 seconds. Which person does the most work? Which person expends the most power? Explain your answers.

H: (600)(4.29)/25 = 108.14w(.00134102209) = .145 hp

V: (600)(4.29)/35 = 77.47w(.00134102209) = .104 hp

- Work doesn't depend on time so the work is the same. Hinrik exerts more power because it was a shorter amount of time. Power = work/time

2. A box that weighs 1000 newtons is lifted a distance of 20 meters straight up by a rope and pulley system. The work is done in 10 seconds. What is the power developed in watts and kilowatts?

(1000)(20)/10 = 2000w(1kw/1000w) = 2kw

3. Brynhildur climbs up a ladder to a height of 5 meters. If she is 64 kg:

a) what work does she do?

work = mgh

work = (64)(9.8)(5)

work = 3136J

b) what is the increase in the gravitational potential energy of the person at this height?
- It is depended on the potential energy

c) where does the energy come from to cause this increase in PE?
- Chemical energy

4. Which requires more work: lifting a 50 kg box vertically for distance of 2 m, or lifting a 25 kg box vertically for a distance of 4 meters?

A: work = (50)(9.8)(2) = 980J

B: work = (25)(9.8)(4) = 980J

- The work is the same.

Conclusion: In this lab i learned how much power it takes to climb the stairs. I learned that the amount of time it takes is what controls how much power it is. One problem could be the measurement of the stairs. It could have been off a little. Also, the time. Since the person climbing the stairs wasn't timing themselves, they might not have started walking at the exact time the person started the timer. Along with that, when stopping, the timer may not have stopped exactly when they got to the top of the stairs. To make this lab improved, it would be smart to have each person time themselves.

Motion in One Dimension with Air Drag

Purpose: To analyze how changing force affects motion in one dimension.

Introduction Questions:
1) By unit analysis, show that the above equation (Vnew = Vold +aavg*delta(t)) is valid.

m/s = m/s + (m/s^2)*s = m/s

2) Why do we use aavg in equation (1)?

- We assume that acceleration is changing

3) Come up with an analogous equation relating Ynew to Yold.

Ynew = Yold + deltaY

4) What is the benifit of choosing a small delta t?

- For more accuracy

Questions: Fd = -kv (where k is a proportionality constant)

1) Draw a detailed motion diagram of the object falling down.

1a) Now draw a force diagram for the object falling down. Include vectors for all forces, and write a statement of Newton's Second Law. Solve for the acceleration.               Force Diagram

- For every action there is an equal and opposite reaction.

Fnet = Fd - Fg

ma = Fd - Fg

a = (Fd - Fg)/m

a = (-kv - mg)/m

a = (-kv)/m - g

1b) Give the condition for the object at terminal velocity (hit: a = 0). Using the condition solve for k.

0 = (-kv)/m - g

g = -kv/m

mg = -kv

k = -mg/v(terminal)

1c) Substitue k into your expression for the acceleration from part a.

a = -(-gm/v(terminal))v/m - g

a = gv/v(terminal) - g

a = -g(v/v(terminal) +1)

2a) What are the assumptions?

- t = 0 s

- g = 9.8 m/s^2

- v = 100 m/s

- r = 1000 m

- v (terminal) = 40 m/s

- delta t = .1 s

2b) What is v-halfstep?

- Taking the average velocity at two times to find the actual velocity

2c) What is a-halfstep?

- Taking the average acceleration at two times to find the actual acceleration

3) Using the graph paper provided, draw scaled graphs of position vs. time, velocity vs. time, and acceleration vs. time for the object, first assuming no air drag. Make predictions on another sheet of paper about how position and velocity would change if you include air drag.

The position vs. time graph that we predicted is very similar to the actual position vs. time graph. However, the velocity vs. time graph that we predicted is no where near the actual graph. I believe that we accounted for drag wrong and we predicted the graph without drag wrong too.

Now look at a drag force that is dependent on the square of the velocity. Assuming a drag force, Fd = |kv^2|, find the new formula for the acceleration.

Conclusion: This lab was complicated. Even though we did it more as a class, it was still difficult. There could have been a lot of different kinds of error. The fact that most of us didn't really understand it, is cause for a lot of error. I think the only way to do better on this lab is to do it over again.

Centripetal Force

Purpose: To verify Newton's second law of motion for the case of uniform circular motion.

 Table of Trials and Times

The table above shows the five trails for two different masses for the bob. The radius was the same for both masses. To find the time we spun the bob around in a circle 50 times and that gave us our time. The frequency was found by dividing the revolutions (50) buy the time. The time was different for each trial. To find the velocity it is (2pi*50*radius)/time. To find acceleration, it's just velocity/time. To find force we did the (bob mass*velocity^2)/radius.
Once we did that, we found the hanging mass. To find that we hung weight from the bob. The weight was determined by where the weight leveled out the bob to the radius. After found the hanging mass we could find weight, which is just hanging mass*gravity. Then we found the difference between the weight and the force for each trial. As you can tell, for the first bob mass we have under 6% difference for each. Then for the second bob mass, we have under 6.5% difference.

This is the force diagram for the spring. The spring is what the bob is connected to. There are only two forces acting on it. The force of static friction and the force of tension. The forces are equal.

This is the diagram for the hanging weight that held the bob in place at the radius. There are two forces acting on it. The force of tension and the force of gravity. These two forces are equal.















Conclusion: I learned that the hanging weight should be the objective of the force that we found. For sources of error, we could have timed things wrong, measured things wrong, or any other possible human error. To improve we could do more trials and take the best five. When adding mass to the hanging bob, the centripetal force stays the same, but the radius the bob is at changes.