5-Nov.-14: Moment of Inertia

PURPOSE:

To verify the sum-of torques equation, Στ=Iα, by finding the moment of inertia of a certain apparatus and directly applying the equation to predict how long it would take for a cart tied to our apparatus to accelerate down a 1-meter ramp.



APPARATUS:

  



  




Equipment -

Large metal disk on a central shaft, m = 4.840-kg
Calipers to measure the dimensions of the apparatus
Camera
1-meter ramp
String
Stopwatch
Rolling Cart, m = 500-g
Tape and marker (or any means of marking a specific point on the disk)


EXPERIMENT:

1. Using calipers and a ruler, we measured the apparatus' dimensions, which included:
   the radius of the cylinder, R1
   the radius of the disk, R2
   the width of the disk, w
   the distance from the surface of the disk to one end of the cylinder, x
   
   
   
2. We then calculated the moment of inertia for the entire apparatus.
   I = 0.021-kg*m^2
   
   
   
3. Using a piece of tape we marked a point at the edge of the disk.
   (red-dotted piece of tape on apparatus picture)
   We gave the disk an initial torque to get the apparatus spinning and measured the deceleration of the disk by recording the disk as it slowed to a stop.
4. We gave the disk an initial torque to get the apparatus spinning and measured the deceleration of the disk by recording the disk as it slowed to a stop.
5. To calculate the acceleration of the disk, we used logger pro to track the position of the red-dotted tape frame by frame. Tracking this specific point resulted in a graph (not included*) which gave us the acceleration tangential to the edge of the disk.
   a = -2.385-cm/s^2 = -0.02385-m/s^2
   
   We converted the tangential acceleration into rotational acceleration using the following equation which expresses the relationship between the two accelerations: a = r*αUsing r = R2 = 0.1002-mα = -0.238-rad/s^2
   
   *Faulty flash-drive
6. The apparatus spun to stopped due to a frictional torque. We were able to calculate this frictional torque with our calculated value for I and α
   
   Στ=IαΣτ=(0.021-kg*m^2)(0.238-rad/s^2)τf = 5.00*10^-3
7. We then attached a 500-gram cart to our apparatus with a string, with one end of the string tied to the cart and the other end tied and wound around the small cylinder part of the apparatus (R1). We also set secured a ramp near the apparatus at an angle of 40°(θ).
   
   
8. We derived a new expression for acceleration for an apparatus which now has a tied mass to it.
   (Note: In the calculations below, r_s is equal to the previously defined R1 and m is the mass of the cart.)
   
   
   
   We calculated the new acceleration to be a = 0.0326-m/s^2
9. We were able to use a kinematics equation to find a value for the time it would take for the cart to move down the ramp using our new found value for acceleration.
   
   
   
   t = 7.83-s is our prediction value.
10. We then released the cart on the ramp and let it roll toward the ground while simultaneously timing how long it took for the cart to reach the ground.
    We did this three times so that we could average the times, thus obtaining a more accurate time measurement.
    
    1) t = 7.53-s
    2) t = 8.03-s
    3) t = 7.97-s
    avg. t =7.84-s
    
    t = 7.84 is our measured value.
11. We then measured the percentage of how far off we were from the actual measured value and found that we were off by 0.128% 

CONCLUSION:

We verified the sum-of torques equation, Στ=Iα, by finding the moment of inertia of a certain apparatus and directly applying the equation to predict how long it would take for a cart tied to our apparatus to accelerate down a 1-meter ramp. We compared our prediction with the value we actually measured and we were within 0.128% of the actual value.

A Great Day for Physics.

(11/05/14, 6:58am)

29-Oct.-14: Angular Acceleration

Purpose:

To understand the effects on the angular acceleration when various changes are applied to our apparatus.
This is accomplished by applying a known torque to our apparatus and measuring the angular acceleration.



Apparatus:

  






Experiment:

We began with the physical measurements such as the dimensions of our disks, pulleys, and the mass of any relevant parts.

The radius and mass of the top steel disk - 63.3-mm, 1356-g
The radius and mass of the bottom steel disk - 63.3-mm, 1348-g
The radius and mass of the top aluminum disk - 63.3-mm, 466-g
The radius and mass of the smaller torque pulley - 12.5-mm, 10.0-g
The radius and mass of the larger torque pulley - 25.0-mm, 32.0-g
The mass of the hanging-mass supplied with the apparatus - 24.5-g

Using logger pro we set up the Pasco rotational sensor by changing the sensor settings to 200 counts per rotation.

To measure the acceleration, we analysed the graphs** of the angular acceleration of the disk as the hanging mass moved downward, then we analysed  it as the hanging mass moved upward. We took the average of the two values for each experiment and used these values for comparison.
The resulting measurements are as follows (variations underlined) -

Expt.#1, α = 0.6239-rad/sec^2
Hanging mass only, with small pulley and top steel disk.

Expt.#2, α = 1.255-rad/sec^2
Doubled hanging mass, with small pulley and top steel disk.

Expt.#3, α = 1.884-rad/sec^2
Tripled hanging mass, with small pulley and top steel disk.

Expt.#4, α = 1.221-rad/sec^2
Hanging mass only, with large pulley and top steel disk.

Expt.#5, α = 3.457-rad/sec^2
Hanging mass only, with large pulley and top aluminum disk.


**Graphs for this activity were lost due to a flash-drive malfunction.



Conclusion (observations):

When we doubled the mass of the hanging mass from 24.5-g to 50-g, we observe that the value for α is approximately doubled from 0.6239-rad/sec^2 to 1.255-rad/sec^2.

When we tripled the mass of the hanging mass from 24.5-g to 75-g, we observe that the value for α is approximately tripled from 0.6239-rad/sec^2 to 1.884-rad/sec^2.

When we doubled the radius of the torque pulley from 12.5-mm to 25.0-mm, we observe that the value for α is approximately doubled from 0.6239-rad/sec^2 to 1.221-rad/sec^2.

When we changed the top disk from steel to aluminum, we decreased the mass of the disk by factor of about 3 from 1356-g to 466-g. We observe that the value for α is approximately tripled from 1.221-rad/sec^2 to 3.457-rad/sec^2.

A Great Day for Physics.

(10/29/14, 7:02am)

15-Oct.-14: Conservation of Momentum in Two Dimensions

Purpose: To verify that momentum is conserved in two dimensions with a marble collision system, using a steel-to-steel ball collision and a steel-to-marble ball collision.



Apparatus:

  






Equipment -

Two small steel balls, each with mass 67.1-g
A non-steel ball, like a marble, with mass 60.3-g
The glass surface (pictured above) with a camera secured above the surface to capture the collision trajectories.



Experiment:

The experiment began by setting a steel ball near the center of the glass table.
We then rolled a second ball toward the resting steel ball in the center so that the two object would collide.
For the first part of the experiment we collided two steal balls -
for the second part, we collided a steal ball with a marble ball.

We then used the video capture with logger pro to track the position of both balls.
By tracking the position of the balls with logger pro we can calculate the initial and final velocities of each ball before and after they collide.

After finding the mass of each ball with their different initial and final velocities, we were able to analyzed the collision to see if the momentum was conserved.
This was done by calculating the components in the x direction and the y direction and comparing the initial to the final values.


Our Calculations -






Graphs* 
The graphs of our recorded data along with the video captures were lost.



Conclusion:

When the components of the momentum before the collision are nearly equal to the components of the momentum after the collision, momentum is conserved.

A Great Day for Physics.

(10/20/14, 3:50pm)

13-Oct.-14: Impulse and Momentum

Purpose:

To verify the Impulse-Momentum Theorem with an elastic and inelastic collision between two masses.

Impulse Momentum Theorem - The change in momentum of a particle during a time interval equals the impulse of the net force that acts on the particle during that interval.

J = Δp

F•t = mΔv

Apparatus:

Equipment -
A rolling cart* - with a mass of 435-g and one cart fixed to a pole
a track - to guide the rolling cart into a collision,
a motion sensor - to find the velocity of the rolling cart,
a force sensor* - resting on the rolling cart to measure the force of the impact,
some clay - to create an inelastic collision in which the masses "stick"
a nail - attached to the rolling cart to penetrate the clay.
a clamp - to secure the clay in the desired position.
a known mass - 500 grams to rest on the rolling to do the experiment with a "different mass."

*combined mass of cart and sensor: 435-g

  

  





Experiment:

To do before the experiment:

• Leveling out the track - this required stacking sheet of paper beneath the track.
• Calibrating the force sensor -  accomplished by using the program's calibration option and hanging a known mass to tell the program how much mass it is experiencing.
• Adjustments for accurate data - We fixed the force sensor to the blue cart in a horizontal position by screwing the sensor to a metal plate that had been previously secured to the cart. we then needed to zero the sensor, which was done through the sensor's computer program.
• Adjusting the motion sensor - We switched the sensor into its "narrow beam" mode for our experiment and ensured that the beam was point in the direction of our cart, parallel to the track.
• Setting the amount of collections - We set the motion sensor to 50 measurements per second.

There were three parts to this experiment,

 A) An elastic collision

 B) an elastic collision with a different mass

 C) an inelastic collision

We note that, in this case, the velocity of the fixed mass is always zero - which leads us to the fact that its change in momentum is also zero.

PART A - Elastic Collision:

1. We gave the rolling cart an initial push.
2. The cart collided with a fixed mass and bounce back toward its origin.
3. During this collision, the motion sensor created a position-vs-time graph, which was then used to calculate the carts velocity.
4. The force sensor simultaneously recorded the force of the impact and the time duration of the impact.

Sensor data:
m_cart = 435-g

V_initial = 0.260-m/s

V_final = -0.223-m/s 

Calculating the change in momentum:

PART B - Elastic Collision #2:

1. We placed a 500 gram mass onto the cart.
2. We gave the rolling cart an initial push.
3. The cart collided and reacted just as it did in part A.
4. The motion sensor gave us new data for velocity.
5. The force sensor gave us new data for the force and time.

Sensor data:
m_cart = 935-g

V_initial = 0.337-m/s

V_final = -0.246-m/s

Calculating the change in momentum:

PART C - Inelastic Collision:

1. We attached a nail with its pointed end directed in the same direction as the carts motion. The nail was secured to the force sensor's "sensor point" using some tape.
2. We gave the cart an initial push
3. The cart, now with a nail, collided into a ball of clay without bounce back.
4. The motion sensor gave us an initial velocity only - the clay stopped the cart.
5. The force sensor gave us new data for the force and time.

Sensor data:
m_cart = 435-g

V_initial = 0.728-m/s

V_final = 0-m/s

Calculating the change in momentum:

Integrating the values for force that our sensor recorded gives us a value for the impulse of the collision.

PART A:

J = -0.2405-N•s

Calculated Δp = -0.210-N•s

PART B:

J = -0.7538-N•s

Calculated Δp = -0.562-N•s

PART C:

J = -0.2779-N•s

Calculated Δp = -0.317-N•s

Conclusion:

The Impulse data and the change in momentum we calculate, with some uncertainty, are about equal.

This verifies the Impulse momentum theorem which states: J = Δp

A Great Day for Physics.

(10/20/14, 7:17am)

8-Oct.-14: Magnetic Potential Energy

Purpose:

To verify that conservation of energy applies in a magnetic energy/air-track system.

Note: The idea we are applying is that at every point, the sum of the PE_magnet and the KE_cart is constant.



Apparatus:

Equipment -
 - A nearly-frictionless track (achieved with an air-track),
 - a cart (m=0.351-kg) that fits comfortably on the air-track,
 - two magnets (one for the cart, and one for the end of the track)

The two fixed magnets have the same polarity and are both at the same height perpendicular to the track - This force is very effective since the track is nearly without friction.
By raising one end of the air-track the cart will end up at some equilibrium position, where the magnetic repulsion force between the two magnets will equal the gravitational force component on the cart parallel to the track.

Track and Air Source:

   

Leveling the track:

   


The two magnets repelling each other:






Experiment:

1. We began the experiment by leveling the air-track which was accomplished by adding individual pieces of paper under one side of the air-track, while simultaneously monitoring the angle (using a smartphone app.)
   This was done so we know where we'd be measuring h from.
2. We let "r" represent the distance between the two magnets and "h" be the height of one end of the track. This relationship helps us find the values for potential energy.
   It's assumed that the relationship between the F_magnet and r takes the form of a power law: F=Ar^n.
3. We stacked books beneath one end of the track to change the slope of the track and measured the separation distance "r" at that angle.
   This step was repeated for a total of 8 measurements
4. Calculating the values for F_magnet required finding the component of the cart's weight parallel to the track. Which led to the equation: F_magnet =mgsinθ
5. Plotting our data allowed us to find the unknown variables A and n in the power law.
   A = 0.0003785, n = -1.792
6. We then attached a motion sensor at the end of the track where one of the magnets was fixed.
7. Recording the position of the cart gave us the velocity of the cart. From the velocity we find the kinetic energy of the cart.
8. We added the PE and KE to find the total Energy.
9. Finally, on the same graph we plotted the potential, kinetic, and total energy to determine if energy is conserved.

Conclusion:

The energy in magnetic system is conserved.

A Great Day for Physics.

(10/8/14, 7:06am)

6-Oct.-14: Conservation of Energy

Purpose:

To verify the conservation of energy in a mass-spring system.



Apparatus:

Our set-up included -
a supported/hanging spring (m=0.064-kg),
a weight (m=0.300-kg), and
a motion detector

Things to measure: 

1. KE of mass (from motion sensor)
2. Gravitational PE of mass (measure the height above the sensor)
3. Elastic PE in spring - 1/2k(stretch^2)
   k - Hang mass. Get stretch Δy
4. Gravitational PE of spring
   -Find expression for GPE of the spring
   -Choose a representative bit of spring, with mass dm (piece of mass)
   The little bit of GPE it has is, d(GPE)=dmgy 
5. KE of spring
   - choose representative piece of mass dm

Experiment:

1. We first measured the spring constant, k, by measuring the distance between the bottom of the spring and the ground when there is no mass attached to the spring
   Next, we measure between the same points, but with a mass of 300-g attached to the spring.
   
   Using Hooke's Law : F=-kΔxwe find k=11.76-N/m
2. We then derived equations to calculate the Gravitational PE of the spring and the KE of the spring.
   
   
   ------------------------------------------------------------------------------------------------------------------
   
3. We pulled the mass, which was attached to the hanging spring, downward to give the spring a vertically oscillating motion and recorded the mass' position a certain time period which gave us the velocity needed for our calculations. (graph below)

After finding a way to calculate each energy in our apparatus -



- we graphed each with a calculated column in LoggerPro. We note that when calculating the total energy of the system through the graph, it closely resembles a straight, horizontal line. This shows that with some uncertainty, energy is conserved.





Conclusion:
The total energy is reasonably constant - this shows that the energy is conserved in our mass-spring system.

A Great Day for Physics.

(10/6/14 7:02 am)