June-05-2015 Lab 20: Physical Pendulum

Purpose:  Derive expressions for the period of various physical pendulums and verify our predicted periods by experiment.

Part 1: predict the oscillation period of steel ring, and compare the result with the actual oscillation period by experiment.

We use the setup with a photogate (like the left picture) to determine the actual oscillation period of the steel ring.

At first, We measured the radius of outside ring is 0.06955 m.

the radius of inner ring is 0.06355 m.


the mass of steel ring is 0.405 kg.




we calculated the system of moment of inertia is 0.0035911 kg * m^2





We know net torque = moment of inertia times angular acceleration which from Newton's 2nd law.  Here is calculation:

we came up a equation. When the angle is very small, sinθ = θ.
then we got a equation of simple harmonic motion.

we got
angular speed = sqrt(mg(R+r)/2I)

and we know angular speed = 2pi/period

We predicted a period is 0.7235 s.

Then we use the set up like the left picture to measure the period.

We got the actual oscillation period 0.7218 s.















Here is calculation for uncertainty:

It's pretty nice to get only 0.2% uncertainty off.


Part 2:  Derive expressions for the period of the various physical pendulums.

Procedure: 
1. At first, we cut out or find objects of the appropriate shapes. Measure the appropriate dimensions of each one.   We measured base B = 10 cm, height H= 15.6 cm, mass of triangle m = 4 g, radius of semicircular plate R = 11.75 cm, mass of semicircular plate M = 11 g.

2. Attach a thin piece of masking tape to each object. Attach a pivot to the appropriate location in each object.
3. Use the same setup of part 1 with a photogate to determine the actual oscillation period of each object.
4. Compare the actual and theoretical values for the periods using the actual measured dimensions for the various objects.
  
Derive expressions for the period of the various physical pendulums:
a) Isosceles triangle, base B, height H, oscillating about its apex. 

We got the actual period T' = 0.7001 s from the program by experiment .

To determine the theoretical values for the periods, we need to find center of mass of the isosceles triangle :

After we got the position of center of mass, we could work out the moment of inertia of triangle oscillating about its apex I(top) = 0.000050338667 kg*m^2:

Then, it's easy to calculate the period T by using our measured dimensions from Newton's 2nd law:

Uncertainty of period is only 0.3% and less than 1%.

b) Isosceles triangle, base B, height H, oscillating about the midpoint of its base:

We got the actual period T' = 0.6039 s from the program by experiment.

We already know the position of center of mass and the moment of inertia of triangle oscillating about its apex I(top). We could calculate the I(cm) = 0.000007074667 kg*m^2 from the parallel axis theorem:

In addition, we could know the moment of inertia of triangle oscillating about the midpoint of its base I(bot) = 0.000017890667 kg*m^2 by using the parallel axis theorem: 

Here is result for the periods using the actual measured dimensions:

Uncertainty of period is only 0.3% and less than 1%.

c) a semicircular plate of radius R, oscillating about the midpoint of its base:

We got the actual period T' = 0.7433 s from the program by experiment.

To determine the theoretical values for the periods, we need to find center of mass of the semicircular plate :

After we got the position of center of mass, we could work out the moment of inertia of semicircular plate oscillating about the midpoint of its base I(top) = 0.000075934375 kg*m^2:

Then, we could predict for the period T using measured dimensions from Newton's 2nd law:

Uncertainty of period is only 0.4% and less than 1%.

d) a semicircular plate of radius R, oscillating about a point on its edge, directly above the midpoint of its base:

We got the actual period T' = 0.7433 s from the program by experiment:

We already know the position of center of mass and the moment of inertia of semicircular plate oscillating about the midpoint of its base I(top). We could calculate the moment of inertia of center of mass I(cm) = 0.0000048578781 kg*m^2 from the parallel axis theorem:

In addition, we could know the moment of inertia of semicircular plate oscillating about a point on its edge, directly above the midpoint of its base I(bot) = 0.000098892926 kg*m^2 by using the parallel axis theorem:

Then, we could predict for the period T using measured dimensions from Newton's 2nd law:

Uncertainty of period is only 0.1% and less than 1%.

Conclusion: 
This lab is to determine the actual oscillation periods of those five objects first, then using the knowledge we learned at class to calculate the periods with measured dimensions. At last, compare the actual period with calculated period for each object.  For those 5 objects, all of uncertainty is pretty low, they are less than 1%. We could say we did a successful experiment.

June-04-2015 Lab 19: Conservation of Energy/Conservation of angular momentum

This experiment goes like:
We release a meter stick, pivoted at or near one end, from a horizontal position. Right when the meter stick reaches the bottom of its swing it collides inelastically with a blob of clay. The meter stick and clay continue to rotate to some final position. Like the picture:

Purpose: 
Measure the mass of meter stick and clay, and length of meter stick to work out how high the clay-stick combination should rise by using Laws of Conservation of Energy and Conservation of angular momentum. Then capture the the experiment on video and compare our actual results with our predictions.

The calculation of Predict: 
1. We measured the length of meter stick L = 1 m, the mass of meter stick M = 77 g. the mass of clay m = 30 g.

Because we did not pivot the meter stick actually at one end (There is 2 cm away from one end). We had to work out the new stick of moment of inertia first. After calculated, we got the new stick of moment of inertia I(stick) = 0.02416 kg*m^2.


2. Before and after collision :
Before collision: We release the meter stick from a horizontal position. At the moment that before meter stick reaches the bottom of its swing to hit the clay, the meter stick will have a angular speed w(0). We could calculate w(0) by using laws of conservation of energy.
At first, we set the position of zero Gravity Potential Energy at middle of pivot point between bottom which is at 48cm on meter stick. So meter stick have M*g*0.48 GPE before we released it. Because rotating objects have rotational Kinetic Energy, we know GPE all transfer to rotational Kinetic Energy which is GPE = KE from laws of conservation of energy.
we got
           M*g*0.48 = 1/2 * I(stick)*w(0)^2.
We know I(stick) from step 1.  We calculated w(0) = 5.81 rad/s.

After collision: After meter stick hit the clay, clay and meter stick will stick together. The mass of system will be different (mass of system = m(mass of clay) + M(mass of meter stick)), the system of moment of inertia will be different (I(system) = I(stick) + m*(L-0.02)). So the angular speed is not w(0) any more. There is a angular speed of system w(f) that we could calculate it by using laws of conservation of angular momentum. Here is calculation:

We got w(f) = 2.6499 rad/s.

3. Then the clay-stick combination continue to rotate together to a final position. There is a angle θ between the position of combination at bottom and the final position of combination. We set the position of zero GPE at bottom of combination.

From Law of conservation of energy, we got
       1/2 * I(system) * w(f)^2 = M*g*(0.48 - 0.48cosθ) + m*g*(0.98 - 0.98cosθ)

We could calculate cosθ. Once we know cosθ, we can work out the h' which is the highest position of clay-stick combination. Here is calculation:

We got h'= 0.28024 m.

4. We already have our result with prediction. Now we need to capture the experiment on video.
Here is the Set up:

One meter stick, pivoted at near one end.

One clay which could stick with things if that thing hit it.

A laptop which connect with a video to capture the whole experiment.


We measured the length of meter stick L = 1 m, the mass of meter stick M = 77 g. the mass of clay m = 30 g.






We released that meter stick from horizontal position, then we captured the whole process of experiment by video. After processed the video by using the advanced program in the laptop, We easily got the height of the clay-stick combination rose:

We got h' = 0.279 m from this experiment.

There is uncertainty by Comparing our actual result with our prediction:

Conclusion:

We came up with a prediction for how high the clay-stick combination should rise by using laws of Conservation of Energy or angular momentum which states total energy or angular momentum of a system remains constant.    Then we obtained the actual result from the experiment.    After compared our actual result with our prediction, there is 0.4% uncertainty.   For this experiment, I think there are some energy lost when meter stick collided with clay. That means the total energy of that system are not perfectly constant. Air resistance will also cause energy lose. In addition, we may not release the meter stick from a perfectly horizontal position. All of things could effect the actual result.

May-17-2015 Lab 18: Finding the moment of inertia of a uniform triangle about its center of mass.

Purpose of this lab:  We need to determine the moment of inertia of a right triangular thin plate around its center of mass, for two perpendicular orientations of the triangle.
Theory:

Part 1: 

For this kind of perpendicular orientations of the triangle, we mount the triangle on a holder and disk. The upper disk floats on a cushion of air.

A string is wrapped around a pulley on top of and attached to the disk. The tension in the string exerts a torque on the pulley-disk combination. By measuring the angular acceleration of the system we can determine the moment of inertia of the system.


Note : Because there is some frictional torque in the system.(the rotating disk is not truly frictionless and there is some mass in the frictionless pulley, which we can approximate as a frictional torque.)

So we have to measure the angular acceleration of disk without triangle to determine the moment of inertia of disk without triangle I'.


Here is the angular velocity vs. time graph of disk without triangle:

Here is the angular velocity vs. time graph of disk with triangle:

After we collected data, we got :

Newton's second law would lead us to get the formula:

Then we could calculate the moment of inertia. Here is calculation:

We calculated the I(disk with triangle) = 0.0028551 N*m.
We calculated the I'(disk without triangle) = 0.0026006 N*m.
We calculated the I(around cm) = I(disk with triangle) -  I'(disk without triangle) = 0.0002545 N*m by collecting data.

Then we measured the mass of triangle M = 455 g. 

a = 9.8 cm.

b = 14.85 cm.


For this kind of perpendicular orientations of the triangle, a is radius.

By definition, We got I(parallel axis) = 1/6 * M * a.



From approach-theory:   
I(parallel axis) = I'(around cm) + M*(d(parallel axis displacement))^2.

We got the I'(around cm) = 0.000242768 N*m by measuring. 

We calculated the uncertainty between the moment of inertia what we calculated and what we measured is 4.8%.

Part 2: 

For this kind of perpendicular orientations of the triangle,
repeat all of steps from part 1.





After we collected data, we determined the moment of inertia. Here is calculation:

For this kind of perpendicular orientations of the triangle, b is radius. 

By definition, We got I(parallel axis) = 1/6 * M * b.

From approach-theory:   
I(parallel axis) = I'(around cm) + M*(d(parallel axis displacement))^2.

We got the I'(around cm) = 0.000557432 N*m by measuring. 

However,
We got the I'(disk without triangle) = 0.0026006 N*m from part 1.
We calculated the I(disk with triangle) = 0.00314948 N*m.

We calculated the I(around cm) = I(disk with triangle) -  I'(disk without triangle) = 0.0005488 N*m by collecting data.
For part 2, We calculated the uncertainty between the moment of inertia what we calculated and what we measured is only 1.53%.

Conclusion: 

For each part (different perpendicular orientations of the triangle) , we found the moment of inertia of a right triangular thin plate around its center of mass by using formula and our measurements. we compared the results between the what we measured and what we predicted to test the data we collected. For part 1, 4.8% uncertainty is higher than 1.53% uncertainty from part 2. That's mean our measurements is good.  I think the only reason to make different result is that we may made big mistake for determining the angular acceleration with triangle on part 1.  
 

May-16-2015 Lab 17: Moment of inertia and frictional torque

For this lab, Given a large metal disk on a central shaft:

To do this lab:
1.  Make appropriate measurements of the rotating part of the apparatus and determine its moment of inertia.

here is our measurements.

w(1) and w(3) are widths of the cylinders. w(2) is width of large disk.

R(1) and R(3) are radius of the cylinders. R(2) is radius of large disk.

Two cylinders have same mass. Total mass is 4.615 Kg.







Before to determine moment of inertia of the rotating part, we have to find the mass of each rotating parts.    Here is calculation for finding mass of each rotating parts:

Then, we could calculate the moment of inertia of whole system:

2. Next step, we need to spin the apparatus and Use video capture to determine its angular deceleration as it slows down. Calculate the frictional torque acting on the apparatus.(because friction exists and can affect the result.)

Before using video capture to take video, we need to stick a tape on the edge of large disk. Because we have to track the apparatus to determine its angular deceleration when it's spinning. 




















After we got the video, we need to set the center of the large disk is the origin first and follow the mark position and point it out.

Then we got the graph.

Red point shows us the mark position on X axis. V(x) could be found.

Blue point shows us the mark position on Y axis. V(y) could be found.



To find the angular deceleration, we have to set up a new column which called Total Velocity to be (V(x)^2 + V(y)^2)^0.5.
There is the graph of V(total) vs. time. After we linear fit it, we will get the linear deceleration a = -0.0533 m/s^2.

Then, we can calculate the frictional torque right now. Here is calculation:

 Using formula, we got angular deceleration alph = - 0.533 rad/s^2

the frictional torque = moment of inertia of system *
angular deceleration

we got the frictional torque = -0.01063 N*m







3. we are going to be connecting this apparatus to a 500-gram dynamics cart. the cart will roll down an inclined track for a distance of 1 meter. Calculate what the time for the cart which from rest to travel 1 meter should be with the actual angle. 

We measured the actual angle is 48 degrees.

we measured three times (7.15 s, 7.19 s, 7.26 s) for how long cart should take for traveling 1 meter from rest, and we got the average time is 7.2 seconds.

After Drawing a force diagram:

We could know three equations. then plug the data we got into those three equations: 

We calculated the acceleration a = 0.0366 m/s^2.
plug this acceleration into the formula  d = 0.5 * acceleration * time^2 by newton laws. (d = 1 m.) then we got t = 7.4 s.
Then we calculated the uncertainty of time between the time we calculated and the time we measured is 2.7%.

Conclusion : 

we found the moment of inertia and frictional torque by step 1 and 2. then we were going to test the data that we found and compare the result we calculated with the result we measured.  we did this lab carefully. However, we have to say that the 2.7% uncertainty is little big for us. Even our average time is not more than 4% off from what we calculated.  I think all measurements are good but the cart, because the mass of cart is not perfectly 500-gram. it may be around 500-gram. But we used 500-gram to calculate.