Phys4AS17cceron

Wednesday, June 14, 2017

June 7, 2017 Physics Pendulum Lab

Physics Pendulum Lab

By: Chris Ceron, John Choi, Amy Chung

June 7, 2017

Goal of Lab:
Derive expressions of the period of oscillation of different objects. Compare the experimental values to theoretical values.

Theory/Introduction:
If we know the moment of inertia about a specific axis we can use what we know about simple harmonic motion to find the period of oscillation. The period is given by:

$T = \frac{2\pi}{\omega}$

where omega is angular frequency.

Omega is a quantity we can find by using Newton's 2nd Law torque equations.
The general torque equation for this lab looks like:

$mg(y_{cm})sin\theta = I_{axis}\alpha \Rightarrow \alpha = \frac{mgy_{cm}}{I_{axis}}\theta$

$\omega^2 = \frac{mgy_{cm}}{I_{axis}}$

The moment of inertia calculation for semi circle.

Using the data from above, we can use the parallel axis theorem to find the Inertia at the top of semi circle.

The moment of inertia for the apex of the isosceles triangle.

The pictures above show the moment of inertia derivation for the two different shapes. We use the equations found in our pre-lab to calculate the moment of Inertia around a certain point on the object

The isosceles triangle we used.

Apparatus and Procedure:

The setup shown with the semi circle

We set up a stand with a bar coming off the side to hang the object on a paper clip. We use a photogate to measure the period of the object. On the object, there is a thin strip of paper that passes through the photogate to accurately find the period.

We collect data, and give the object a small push in order to have it oscillate back and forth. We take this period and we compare it to the calculated value.

Data:

The radius of the semicircle was .105 m.
The base and height of the triangle were 0.145 m and 0.160 m respectively.

The data we obtained via LoggerPro and the photogate setup was 0.694 secs for the semicircle and 0.715 sec for the triangle.

The period for the semi circle.

The period for the isosceles triangle.

Calculated Data:

Calculations for Period using the equations mentioned in the Theory/Introduction section

The calculation for the period using the Inertia derived above.

Conclusion:

The percent error for both cases are extremely small, both <1% (0.29% for the semicircle and 0.55% for the triangle). This shows that our derivations to find the period were accurate. Sources of error and uncertainty come from potential friction at the apex where the paper clip swings back and forth as well as usual sources such as distance measurements for radius, base and height.

May 22, 2017 Lab 17: Moment of Inertia of a Uniform Triangle About Its Center of Mass

Lab 17: Moment of Inertia of a Uniform Triangle About Its Center of Mass

By: Chris Ceron, John Choi, Amy Chung

May 22, 2017

Goal of Lab:
Determine the moment of inertia of a right triangular thin plate around its center of mass with two different orientation, one with the long side facing down and the other with the short side facing down (Fig. 1 and 2 show the set up and triangle).

The triangle we used.

Theory/Introduction:
In order to find the moment of inertia of the thing right triangle plate, we first find the inertia about the edge and use the parallel axis theorem to derive an equation for the inertia at the center of mass. The shift in distance is found using the x center of mass position:

$\frac{dM}{M} = \frac{dA}{A} \Rightarrow dM = \frac{2M y dx}{BH}$

the relation between x and y can be found using the linear line, y = -(H/B)x + H

$dM = \frac{2M(1-\frac{x}{B})dx}{B}$

the x center of mass equation is derived by:

$x_{cm} = \frac{\frac{2M}{B}\int_{0}^{B}(x-\frac{x^2}{B})dx}{M} = \frac{B}{3}$

and the inertia about the edge can be found using:

$x_{cm} = \frac{2M}{B}\int_{0}^{B}(x^2-\frac{x^3}{B})dx = \frac{MB^2}{6}$

so using the parallel axis theorem we find that the inertia about the center of mass is:

$I_{new} = I_{cm} + M\left ( \frac{B}{3}\right)^2 \Rightarrow I_{cm} = \frac{MB^2}{6} - \frac{MB^2}{9} = \frac{MB^2}{18}$

In our experiment, there is a weight of the pin that holds the triangle, hence it has a inertia value. Since we want to isolate the triangle, we first find the moment of inertia of the system and subtract if by the moment of inertia of the pin, leaving us with the moment of inertia of the triangle.

Since the setup isn't like physics problems where pulleys are frictionless and massless, we need to take into account the frictional torque within the system. The derivation is done exactly the same as from previous lab on angular acceleration. The final derivation for Inertia came out to be:

$I = \frac{mgr}{\frac{|\alpha_{down}|+|\alpha_{up}|}{2}} - mr^2$

Once we find the experimental and theoretical data, we can check the % error and discuss reasons for discrepancy in data.

Apparatus and Procedure:

The apparatus with just the pin.

The setup looks identical to lab 16 for angular momentum but the disk now has a pin that holds a thin triangular plate. We connect the system to LoggerPro and find the angular acceleration going up and going down by finding the slope of the angular velocity reading on the graph.
We first find the inertia value of just the pin since the mass that it has also contributes to total inertia. Then we find the angular acceleration up and down values on one side of the triangle, we repeat the process with the triangle on its other side and for each of the inertia values we calculate, we subtract the inertia of just the pin.

Data:
The triangle we used had height 149.2 mm and base 98.2 mm. The radius of the pulley was 49.8 mm.

The mass of the hanging mass was 0.02457 kg.

The mass of the triangle was 0.462 kg.

	\|α down\| (rad/s^2)	\|α up\| (rad/s^2)
Pin	6.536	5.967
Triangle Base	5.258	4.674
Triangle Height	4.223	3.784

Alpha down for pin.

Alpha up for the pin.

Alpha down for the triangle with base down.

Alpha up for the triangle with base down.

Alpha down for the triangle with height down.

Alpha up for the triangle with height down.

Calculated Data:

Calculation of Inertia:

$I = \frac{0.02457*9.8*0.02745}{\frac{6.536+5.967}{2}}-(0.02457*0.02745^2) = 0.00104 kg*m^2$

	Pin	Triangle Base	Triangle Height
Inertia (kgm^2)*	0.00104	0.00131	0.00163

Triangle base = 0.00027

Triangle height = 0.00059

The theoretical inertia for base is found by:

$I = \frac{M_{triangle}B^2}{18} = \frac{0.462*0.0982^2}{18} = 0.000248 kg*m^2$

and for the height side down:

$I = \frac{M_{triangle}B^2}{18} = \frac{0.462*0.1492^2}{18} = 0.000571 kg*m^2$

Conclusion:

	Experimental	Theoretical	% Error
Triangle Base	0.00027	0.000248	8.87%
Triangle Height	0.00059	0.000571	3.3%

Looking at the % error we found, the results were accurate. Sources of error could be from small uncertainty values in our measurement of mass, base and height. Our graphs did not have good correlation values for angular acceleration. There is also some small friction present in the pulley that the string goes over that we ignored.

Sunday, June 11, 2017

May 31, 2017 Lab 19: Conservation of Energy/Conservation of Angular Momentum

Lab 19: Conservation of Energy/Conservation of Angular Momentum

Chris Ceron, Amy Chung, John Choi

Goal of Lab:
Test the conservation of energy and conservation of angular momentum by colliding a meter stick on pivot against a clay ball on the ground.

Theory/Introduction:
With the absence of external forces, angular momentum is conserved much like linear momentum. Conservation of angular momentum is labeled with an L and the components are as follows:

$\vec{L_{0}} = I\vec{\omega_{0}}$

$L_{f} = I\omega _{f}$

I is the inertia of the object, it could be a system of objects where the total inertia is the sum of all the individual inertia about the axis of rotation.

The conservation of energy states that these two components are equal:

$\vec{L_{0}} = \vec{L_{f}}$

We view our ball of clay as a point mass, making its inertia:

$I_{point\:mass} = ML^2$

Where L is the distance away from the axis of rotation.

We use the conservation of energy to find the angular velocity of the system. If we assume we start the meter stick from a horizontal position our change in height for the meter stick is known.

$m_{stick}g\Delta h = \frac{1}{2}I_{stick}{\omega_{0}}^2 \Rightarrow {\omega_{0}} = \sqrt{\frac{2m_{stick}g\Delta h}{I_{stick}}}$

Once we have the initial omega, we are able to solve for the final omega after the collision using the conservation of angular momentum.

Since the axis of rotation for the meter stick is not precisely at the end (for our experiment it was at the 2 cm mark on the stick), we use the parallel axis theorem to find a moment of inertia about a new axis:

$I_{new\:stick} = I_{cm\:stick}+m_{stick}{r_{stick}}^2$

$I_{total}= I_{new\:stick}+I_{clay}$

now applying this to the conservation of angular momentum:

$I_{new\:stick}\omega_{0} = I_{total}\omega_{f} \Rightarrow \omega_{f} = \frac{I_{new\:stick}}{I_{total}}\omega_{0}$

gives us a value of omega final.

This omega value can be used in our energy equations where the initial energy is rotational and transfers to gravitational potential:

$\frac{1}{2}I_{total}{\omega_{f}}^{2} = g(m_{clay}r_{clay}+m_{stick}r_{stick})(1-cos\theta)$

The only unknown within this equation is the value for theta so we can isolate and solve for it:

$cos\theta = 1 - \frac{I_{total}{\omega_{f}}^{2}}{2g(m_{clay}r_{clay}+m_{stick}r_{stick})}$

cosine inverse gives us the angle for which the meter stick travels up after sticking with the clay.

$\theta = cos^{-1}\left (1 - \frac{I_{total}{\omega_{f}}^{2}}{2g(m_{clay}r_{clay}+m_{stick}r_{stick})}\right )$

using this angle, we can find the height the clay rises by isolating the distance the clay is away from the axis of rotation and the 1 - cos theta portion:

$h_{clay} = r_{clay}(1-cos\theta)$

Apparatus and Procedure:

Apparatus used (meter stick has already collided with ball of clay)

We set up our apparatus using a stand attached to a pivot point and put the pin through the hole on the meter stick(2cm mark of the meter stick). Using Amy's slow motion capture on iPhone recording at 240 frames per sec, we took a video of the meter stick colliding with the clay. Using LoggerPro's video analysis program, we set a reference measurement, and found the highest point the clay reaches. Our reference point was the position of the clay before the collision.

Data and Calculations:

The point on the meter stick that the axis of rotation went through was the 2 cm mark (0.02 m).

	r distance from axis (m)	Mass (kg)
Meter stick	0.48	0.02357
Clay	0.98	0.1444

Above is the measured data required for calculation. Below is the LoggerPro capture:

The video capture as well as the height reading of the clay blob.

The height recorded was 0.4310 m.

The parallel axis theorem for the meter stick:

$I_{new\:stick} = m_{stick}(\frac{1}{12}(1)^2+(0.48)^2) = 0.31373*m_{stick}$

The conservation of energy to find omega initial:

${\omega_{0}} = \sqrt{\frac{2(0.1444)g(0.48)}{0.0453}} = 5.476\:rad/sec$

and for omega final calculations, we need the total system inertia after collision:

$I_{total} = 0.31373*m_{stick}+0.02357*(0.98)^2 = 0.0679 kg*m^2$

${\omega_{f}} = \frac{0.0453}{0.0679}(5.476) = 3.654\:rad/sec$

Using these values, we found theta to equal:

$\theta = \cos^{-1}\left (1- \frac{0.0679(3.654)^2}{2g(0.02357(0.98)+0.1444(0.48))} \right ) = 60.05\degree$

and to find the max height of the clay using the angle above:

$h_{clay} = (0.98)(1-\cos60.05\degree) = 0.491 m$

Conclusion:

Calculating for percent error between our experimental value and theoretical gives us:

$\%_{error} = \left | \frac{0.491 - 0.431}{0.431} \right | * 100 \% = 13.92\%$

This is a large margin of error that was the result of some sources of uncertainty. We neglected the friction at the axis of rotation. This causes the calculated height much higher than the experimental. We also made some other assumptions such as the clay being treated as a point mass and energy being conserved when it truly isn't (energy can be lost through friction experienced by the collision and at the pivot point during the collision). Uncertainty in our distance from axis of rotation measurements probably also played a small role as the measurements we came up with were rather rough estimates.

Tuesday, May 30, 2017

May 22, 2017 Lab 18: Moment of Inertia and Frictional Torque

Lab 18: Moment of Inertia and Frictional Torque

Chris Ceron, Amy Chung, John Choi

Goal of Lab:
Derive the frictional torque of a large metal disk and evaluate this value is by timing the movement of a dynamics cart down a near frictionless track.

Theory/Introduction:

When we give our large metal disk an initial velocity ,the frictional torque is what brings the disk to a stop. Sum of torques are expressed in the general format of:

$\Sigma \tau = I\alpha$

I = Inertia of the large metal disk
$\alpha$ = angular acceleration or deceleration

That means once we find the inertia of the disk, we can find the frictional torque since this is the only torque acting on the system. The large metal disk can be modeled as three different shapes, two small radius cylinders on the side and one big metal disk in the middle.

$I_{system} = I_{cyl \: 1} + I_{disk} + I_{cyl \: 2}$

Apparatus used in lab

The summation of these values gives us the total inertia of the system. Inertia for a solid cylinder spinning about its center axis is given by:

$I_{cylinder_{center}} = \frac{1}{2}MR^2$

To find the mass of each piece of the system, we took the rations of the total mass:

$M_{1} = \frac{V_{1}}{V_{1}+V_{2}+V_{3}}*M_{total}$

Once we have this calculated data, we spin the apparatus and use our phone's slow motion capture to derive the angular deceleration of the system. The method we used to derive angular deceleration involved finding out the tangential deceleration first and using the relationship between tangential acceleration and angular acceleration.

To evaluate our deceleration value, we a cart to the disk with some string and allow the cart to roll down a frictionless track going a distance of 1 meter. We timed this and compared it to the time we got using our derived value of angular deceleration to compare how accurate our value was.

Apparatus and Procedure:

Calipers used for messurements

For the first part of the lab, we calculated the total inertia of the system by using the procedure outlined in our introduction. The total mass of the disk was given on the side of the disk, and using our ratio of volume to the total mass we were able to find the mass of the 3 different pieces. We use different precision vernier calipers to find the appropriate measurements

For the side thickness, we used a smaller caliper measurement that protrudes from one end.

The larger caliper we used for the diameter measurements are even more precise than the smaller caliper.

For the second part, we gave the disk an initial spin and used Amy's phone recording at 240 frames/sec to record the deceleration of the disk. Once we had this video, we used LoggerPro to analyze the video. We chose to advance every 8 frames so the video capture on LoggerPro was at 30 frames/sec.

The dot plots of the rotation slowing down.

Our method of deriving angular acceleration:

1. Use LoggerPro's video analysis to setup different aspects needed for calculation.
2. Plot a dot for every frame on the blue tape portion of the disk.
3. Find the linear distance between each dot using the distance formula.

$\Delta x = \sqrt{{(x_{2}-x_{1})^2}+(y_{2}-y_{1})^2}$

4. Find midpoint velocity between each delta x:

$v = \frac{\Delta x}{t}$

5. Plot this midpoint velocity against time and do a linear fit on the graph.

6. The slope of this line is the tangential deceleration.

7. The relationship between tangential acceleration and angular is:

$a_{tan} = r\alpha$

Using the equation above, we divided the tangential acceleration by the radius to find alpha.

The setup for the cart running down the track connected to the disk.

For the third part of this lab, we test how accurate our derived value of angular deceleration is. We set up our apparatus in a manner that allows for the string to always be parallel to the track, otherwise the torque from the tension force changes as the cart rolls down. We allow the cart to go a distance of 1 meter and time how long it takes. We then compare this value to our calculated value for time.

We found two different ways to calculate for time.

Method one:
Use rotational kinematics where theta was found using the relationship:

$\Delta x = r\theta$

Here delta x is 1 meter, so theta came out to be 63.69 rad. Since we know the initial angular velocity is 0, we can use the formula:

$\theta = \omega_{i}t + \frac{1}{2}\alpha t^2, \omega_{i} = 0 \Rightarrow t = \sqrt{\frac{2\theta}{\alpha}}$

Another method involves using forces and torques. From our setup we can set up two equations:

$\Sigma \tau = Tr -\tau_{f} = I\alpha$

$\Sigma F = mgsin\theta - T = ma$

Manipulating the first equation we can obtain:

$Tr-\tau_{f} = I\frac{a}{r}$

$T-\frac{\tau_{f}}{r} = I\frac{a}{r^2}$

Adding this new equation with the sum of forces equation cancels out the value of T which brings us to:

$mgsin\theta + \frac{\tau_{f}}{r} = a(m+\frac{I}{r^2}) \Rightarrow a = \frac{mgsin\theta + \frac{\tau_{f}}{r}}{m+\frac{I}{r^2}}$

Using this tangential acceleration value, we can use kinematics again to find the time:

$\Delta x = v_{0}t + \frac{1}{2}at^2, v_{0} = 0 \Rightarrow t = \sqrt{\frac{2\Delta x}{a}}$

We can compare this time value to the one we found by just timing the descent of the cart 1 meter down the track.

Data and Calculated Data:

The mass of the disk system we used was 4.928 kg.

The mass of the cart we used was 0.5249 kg.

We can find the volume of each part using this data:

$Volume = \pi r^2h$

where r is radius and h is the thickness.

Examples of mass and inertia calculations:

$M_{1} = \frac{1.2497*10^{-5}\pi}{\pi(1.2497*10^{-5}+1.2757*10^{-5}+1.5139*10^{-4})}*4.928 = 0.3416kg$

$Inertia_{cyl \: 1} = \frac{1}{2}(0.3416)(0.0157^2) = 4.210*10^{-5} kg*m^2$

Data we obtained/measured:

	Thickness (m)	Diameter (m)	Radius (m)	Volume (m^3)	Mass (kg)	Inertia (kgm²)*
Cylinder 1	0.0507	0.0314	0.0157	1.2497 * 10^-5π	0.3416	4.210*10^-5
Cylinder 2	0.0511	0.0316	0.0158	1.2757 * 10^-5π	4.139	0.01973
Disk	0.0159	0.19528	0.09764	1.5158 * 10^-4π	0.3487	4.352*10^-5

*Note* the uncertainty in the diameter measurement and for disk thickness was +/- 0.00002 m (from the bigger caliper). The uncertainty for the thickness of the two cylinders was 0.0001 m (from the smaller caliper).

So Inertia of the system will be 4.210*10^-5+ 0.01973 + 4.352*10^-5= 0.01982.

The graph we obtained by plotting the midpoint velocity with time.

The tangential deceleration value we found was -0.1485 m/s^2 (not the -0.01077 on the graph above).

With the relationship between tangential deceleration and angular deceleration we find angular deceleration to be -1.5209 rad/s^2.

For the timed descent down the track, we record the time to be about 7.1 seconds.

The angle of our track was 48.0 degrees from the horizontal.

The time calculated using our rotational kinematics approach was:

$t = \sqrt{\frac{2*63.69}{1.5209}} = 9.152 secs$

and for the force/torque approach:

$a = \frac{0.5249*9.8*sin48^{\circ} + \frac{0.01982*(-1.5209)}{0.0157}}{0.5249+\frac{0.01982}{0.0157^2}} = 0.02351 m/s^2$

$t = \sqrt{\frac{2*1}{0.02351}} = 9.223 sec$

So the calculated times are very close to each other regardless of method (small discrepancy between the two values comes from uncertainty in the different measurements).

Conclusion:

We had an issue in our lab where the time we recorded for the cart was different from the calculated time above. This would be fine if our method of finding angular acceleration was incorrect or presented flaws, but even the professor said our method for finding angular acceleration and frictional torque was valid. The error calculation for these values comes out to be:

$\%_{Error} = \left | \frac{9.152 - 7.1}{7.1} \right | * 100 = 15.16 \%$

Which is a rather large % error. Of course the 7.1 second value we have for expected is a rather rough estimate (due to human error when observing when the cart has moved a distance of one meter. Our value of angular acceleration could have been more accurate if we advanced the videos by a number less than 8 frames. Our inertia value matched up very closely to the expected value. This is largely due to the precision in the equipment used for our measurements. The only potential source for this great and error seems to be from our data gathering for finding angular deceleration. It could potentially be the video capture wasn't aligned as well as possible with the rotational axis, we used too little frames. and potential human error from our plot of the dots being as very rough eyeballed estimate for the location on the video.