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.

May-8-2015 Lab 16: Angular Acceleration

Part 1 : We want to apply a known torque to an object that can rotate, and measure the angular acceleration. we can find a measured value for the moment of inertia. 





this is set up to do this lab. we can use it to measure the angular acceleration. 
































1.Before we start the experiment, we measured each of the following:


2. we plug the power supply into the Pasco ratational sensor. there is a cable with the yellow paint or tape, connect only that cable to the Lab Pro at Dig/Sonic 1, so the computer could read the top disk.

3. Set up the computer. Open Logger Pro. Choose Rotary Motion. there are 200 marks on our top disk, so we need to set the Equation in the Sensor Settings to counts per rotation. when we collect data, we can see graphs of angular position, angular velocity and angular acceleration vs. time. However, the graph of angular acceleration vs. time is useless due to the poor timing resolution if the sensors.

4. Make sure the hose clamp on the bottom is open so that the bottom disk will rotate independently of the top disk when the drop pin is in place.

5. Turn on the compressed air so that the disks can rotate separately. Set the disks spinning freely to test the equipment. 

6. With the string wrapped around the torque pulley and the hanging mass at its highest point, start the measurements and release the mass. 

EXPTS 1,2, and 3: we are going to look at the effect of changing the hanging mass(25 g, 50 g, 75 g). 

Here is graph of Expt 1: 



Here is graph of Expt 2 :

Here is graph of Expt 3: 

For Expt 1 and 4, we are going to look at the effect of changing the radius and which the hanging mass exerts a torque (small torque pulley, Large torque pulley). 

Here is graph of Expt 4: 

For Expt 4, 5, and 6, we are going to look at effect of changing the rotating mass(top steel, top aluminum, top steel + bottom steel).

Here is graph of Expt 5: 

Here is graph of Expt 6: 

After linear fit those angular velocity vs. time graphs, we can know the angular acceleration up and down for each Expt.

Then we calculated  the average of angular acceleration of each Expt, we write down all data we collected into a form:

From those graphs, the we could know the angular acceleration up is always bigger than the angular acceleration down because we cant ignore some frictional torque in the system. So, When the hanging mass is going down, the net torque is equal to torque mass - torque friction , so angular acceleration down is smaller than the real.

Conclusion: 
From this form, we could see For Expt 1,2, and 3, when the mass of hanging mass is increasing, the angular acceleration is increasing.
For Expt 1 and 4, with same hanging mass, when radius of torque pulley is increasing, the angular acceleration is increasing.
For Expt 4, 5, and 6, with same hanging mass, when the rotating mass is increasing, the angular acceleration is decreasing.


Part 2:

For this part, we can use our data (which from part 1) to determine the moment of inertia of each of disks.

Purpose:
we want to work out the torque of friction, Because there is some frictional torque in the system, the angular acceleration of system when mass is descending is not the same as when it is ascending.

First, let's call the counterclockwise direction of rotation positive and clockwise direction of rotation negative. Newton's second law would lead us to predict that:

Here is calculation for Expt 1: 

By formula, we calculated the inertia of disk is 0.0025556 kg*m^2



By measurement, we got the inertia of disk is 0.0026563 kg*m^2


Comparing the moment of inertia by using formula and by using measurement, the uncertainty is 3.8%. 





by formula, we can get the value of frictional torque = 0.000178 N*m 











Here is calculation for Expt 2: 

By formula, we calculated the inertia of disk is 0.0026553 kg*m^2



By measurement, we got the inertia of disk is 0.0026563 kg*m^2



Comparing the moment of inertia by using formula and by using measurement, the uncertainty is 0.3%.




by formula, we can get the value of frictional torque = 0.0002639 N*m 












Here is calculation for Expt 3: 

By formula, we calculated the
inertia of disk is 0.0026653 kg*m^2



By measurement, we got the inertia of disk is 0.0026563 kg*m^2



Comparing the moment of inertia by using formula and by using measurement, the uncertainty is 0.339%.




by formula, we can get the value of frictional torque = 0.000265 N*m 

Here is calculation for Expt 4: 

By formula, we calculated the
inertia of disk is 0.002614 kg*m^2



By measurement, we got the inertia of disk is 0.0026563 kg*m^2



Comparing the moment of inertia by using formula and by using measurement, the uncertainty is 1.59%.




by formula, we can get the value of frictional torque = 0.000341 N*m 

Here is calculation for Expt 5: 

By formula, we calculated the
inertia of disk is 0.0009133 kg*m^2



By measurement, we got the inertia of disk is 0.0009102 kg*m^2



Comparing the moment of inertia by using formula and by using measurement, the uncertainty is 0.35%.




by formula, we can get the value of frictional torque = 0.0003566 N*m 

Here is calculation for Expt 6: 

By formula, we calculated the
inertia of disk is 0.005152 kg*m^2



By measurement, we got the inertia of disk is 0.005313 kg*m^2



Comparing the moment of inertia by using formula and by using measurement, the uncertainty is 3.41%.




by formula, we can get the value of frictional torque = 0.0002868 N*m 

Conclusion :

For Expt 2, 3, and 5, their uncertainties that we calculated are very small(under 1%), we could say our predict values of inertia of disk is very close to the values of inertia of disk we measured.
However, for Expt 1, 4, and 6, their uncertainties are little bit big. our data, like radius and weight, we may made some mistakes for measuring them. and, during the disk spinning, top disk may not rotate independently with bottom disk.

May-7-2015 Lab 15: Inelastic Collision

Purpose :  work out the initial speed of the small ball and its speed of uncertainty by using a small ball to hit the holder from the machine which like the picture.

set up:


Using this machine, we could give the small ball a initial speed, and the holder will hold the ball (which means they will stick together) when the ball hit the holder .  In addition, there is a red rod that will show us the highest position and angle of the holder.

After we collect all of data we need, we could calculate the initial speed of the ball by Laws of conservation of momentum and energy conservation.



Put the holder at the initial position, and put ball into the machine. then push the button, the ball will shoot out and hit the holder.

there is the graph after the ball and holder stick together, and the data we measured:

we measured the mass of ball M(ball) = 7.63 g +- 0.1 g

M(block) = 80.9 g +- 0.1 g

the changing degree of the holder:     θ = 17.5 degrees +- 0.5 degrees

the length of the string L = 0.201 m +- 0.001 m







after the ball and holder stick together, the velocity will change. so we need to know the velocity when ball and holder stick V(total) which we could calculated it by using law of  energy conservation,  Then we could know the initial speed of the ball V(ball) of ball by using law of momentum conservation.  Here is calculation:  

there is calculation to find ball's uncertainty of initial speed:

Conclusion :

For this lab, we measured the length of string, mass of ball and holder, and we collected the changing of degree of the holder after we shot the ball. Then we used the laws of conservation of momentum and energy conservation to work out the initial speed of the ball is 4.9545 m/s, and its uncertainty is only 3.22%. that's mean Our result is pretty good.

May-6-2015 Lab 14: Collisions in two dimensions

the Purpose of this lab: look at a two-dimensional collision and determine if momentum and energy are conserved.

           -steel ball with steel ball

           -steel ball with aluminum ball

we got the experiment equipment like the picture.

Open the program, go to set up the camera by following those steps:

Camera settings (continued):

Lab setup:

Before we start the lab, we need to level the table first. 

Then gently set the stationary ball on the leveled glass table. 

Aim the rolling ball so that it hits the side of the stationary ball. 

The balls should ideally roll of at some decent angle from one another. 

after we got the video of two ball collision, we need to do something on the video: Grab the point which two ball just collided to rotate the axes, set origin and add point series for the way of ball path. 

First, we do steel ball collide with steel ball. what we got from the experiment is : 

We measured the mass of steel ball m(sb) = 0.07 kg.

after linear fit all of those point, we could have the velocities at x-axis (horizontal with initial velocity) and y-axis(vertical with initial velocity ) before and after two ball collision. 

Here is the calculation:

before two ball collision:

we got the velocity at x-axis V(x) = 0.8474 m/s (which is the slope of green points) and V(y) = 0.0062 m/s (which is the slope of red points) of first steel ball. 

second ball is at rest.

from the data we collected, we calculated the momentum of two balls before collision p(x) = 0.059 kg*m/s and the momentum of two balls after collision p(2x) = 0.056 kg*m/s at x-axis. 

in addition, we calculated the momentum of two balls before collision p(y) = 0.004 kg*m/s and the momentum of two balls after collision p(2y) = -0.003 kg*m/s at y-axis. 

here is the calculation of energy:

conclusions:  p(x) and p(2x) are pretty close, p(y) and p(2y) are also very close, so we could say the momentum is conserved in this two steel balls of two-dimensional collision. 

however, the initial kinetic energy KE(i) is bigger than final kinetic energy KE(f). there were some energy lost from this collision. so we could say the energy is not conserved in this two steel balls of two-dimensional collision.

Right now, we do the steel ball collide with aluminum ball. we measured the mass of aluminum ball m(ab) = 0.02 kg. there is what we got from the experiment:

after linear fit all of those point, we could have the velocities at x-axis (horizontal with initial velocity) and y-axis(vertical with initial velocity ) before and after two balls collision.

Here is the calculation:

before two ball collision:

we got the velocity at x-axis V(x) = 0.6071 m/s (which is the slope of green points) and V(y) = -0.0034 m/s (which is the slope of red points) of the steel ball. 

the aluminum ball is at rest.

from the data we collected, we calculated the momentum of two balls before collision p(x) = 0.0425 kg*m/s and the momentum of two balls after collision p(2x) = 0.0444 kg*m/s at x-axis. 

we calculated the momentum of two balls before collision p(y) = -0.00024 kg*m/s and the momentum of two balls after collision p(2y) = 0.0016 kg*m/s at y-axis. 

here is the calculation of energy: 

conclusions:  p(x) and p(2x) are pretty close, p(y) and p(2y) are also very close, so we could say the momentum is conserved in this two steel balls of two-dimensional collision. 

however, the initial kinetic energy KE(i) is smaller than final kinetic energy KE(f). there were some energy lost from this collision. so we could say the energy is not conserved in this two steel balls of two-dimensional collision.