June 3rd, 2015; Physical Pendulum Lab

Purpose:

Derive an expression for period of oscillation for 3 different geometric shapes (also finding moment of inertia) and compare the value to the experimental data.

Part 1: Finding metal ring's period of oscillation

Experiment:

Finding predicted value for "T":
- Using T= 2π / ω, we have to find ω to find T. In simple harmonic motion, ω^2 comes from 
α = - constant * θ, where constant = ω^2.

- Moment of inertia of this geometric shape around its center is (1/2)M (R^2 +  r^2)

- We use net torque equation as this experiment is related to rotational motion. The rigid body diagram (not shown here but similar to those of the triangle and semicircle) gives us mgsinθ, the force perpendicular to the distance from the pivot to center of mass. That distance, therefore is R-average = (outer radius + inner radius)/2. 

- Use Parallel Axis Theorem to shift axis from the center to where distance of shift = R-average.

The experiment
- Weigh the ring's mass and use caliper for diameter.
- The ring has a hole that could go in the pivot point.
- Open "Pendulum Timer.cmbl" to set up photogate to record the period of the metal ring. Attach a thin stick note to the bottom of the ring so that when oscillation takes place, the photogate could detect from the stick note.
- Gently tap the ring so the oscillation occurs at a small angle displacement; collect data.

Data and Analysis:

- The period data should be consistent of one another (represented by its straight line from the graph).

- Our metal ring's period is T= 0.7196 sec. Calculating percent error gives us ~ 0.3%.

Part 2: Triangle and semicircle period of oscillation:

Experiment:

- Cut out the shape of isosceles triangle and semi circle.

- Measure the dimensions of the shapes and their mass.

- Attach masking tape to electrical connectors, one on each side so the hole allows for a pivot point roughly close to the tip of the object.Use paper clip to pivot the objects.

- Set up photogate and have a sticker (attached from bottom of the object) go through the photogate.

- Measure the period of the pendulum at pivot:

a). isosceles triangle oscillating about its apex

b). isosceles triangle oscillating about the midpoint of its base

c). semicircular plate oscillating about the midpoint of its base

d). semicircular plate oscillating about a point of its edge, directly above the midpoint of its base.

Data and Analysis:

M-triangle = 0.0093 +/- 0.001 kg; base = 0.15 +/- 0.001 m; height = 0.175 +/- 0.001 m

M- semicircle = 0.0112 +/- 0.001 kg; radius = 0.108 +/- 0.001 m

- Below is one of the four screenshots taken after period data is collected.

T (semicircle, pivot at midpoint of base) = 0.695931 s

a). and b).  

- Since we're dealing with oscillation, we can write a net torque equation (and Newton's second law) to get the equation to be in a form of: acceleration = -constant * displacement. We also know that one side of torque equation has Iα in it. So moment of inertia is needed.

-  Finding center of mass (left-hand side), and moment of inertia of triangle (pivot at apex) respectively.

- For "I", use dI(rod) around its center and use Parallel Axis Theorem To shift every dm that is a distance "y" from the new pivot point to that pivot point.

- Deriving moment of inertia for triangle pivot at middle of its base shown below,

- Coming up with an expression for "T' for triangle pivot at midpoint of base and comparing this "T" to the one from LoggerPro.

- Similar process here for "T" of triangle pivot at apex. Error this time turns out to be ~ 2.6%

c). and d).
- On left-hand side, moment of inertia of semicircle at midpoint of its base, middle: finding center of mass with setting origin at midpoint of the base, right-hand side: expression for moment of inertia of semicircle pivot directly above midpoint of base.

- Calculating period and comparing it to experimental value for semicircle pivot at midpoint of base,

- Calculating period and comparing it to experimental value for semicircle pivot at point directly above midpoint of base,

- Despite the written order of this lab, deriving an expression for the wanted value first is always a good idea before doing the experiment.

Conclusion:

From knowing the radius of the semicircle and base and height of a triangle, we could derive an equation for period. From both parts of the lab, we see that period of physical pendulum is independent on mass of the oscillating object. Our range of error and uncertainty is between 0.6 to 3%. In most cases, experimental values should be slightly larger and that brings us to the discussion of errors and uncertainty. The larger values result from the neglected friction at pivot point and the fact that the pivot is not exactly where it's meant to be ideally but above it. This unaccountable shift of distance would alter everything including the period of oscillation. Other usual uncertainty is from mass and length measurement. Nonetheless, within our range of uncertainty, our derived expression agrees with the experimental values of "T".

May 20th, 2015: Lab 19: Conservation of Energy and Angular Momentum

Purpose:

Given a clay+meter stick system, by releasing the meter stick to collide inelastically with the clay, show that there's conservation of energy and conservation of momentum in the system.

Experiment:

Data: M(stick+ clay) = 0.113 +/- 0.001 kg

m-meter stick = 0.080 +/- 0.001 kg

Coming up with predicted "h"
- Noticing that there's 3 parts to this lab to find "h",

• conservation of energy of meter stick before released to meter stick right before collision
• conservation of angular momentum of meter stick right before collision and the system right after collision
• conservation of energy of the system from right after inelastic collision to the maximum height reached.

* Energy and momentum approaches are more doable and kinematics wouldn't be possible because angular acceleration is not constant at all time.

- Note that at pivot on meter stick cannot be exactly at the edge. Using Parallel Axis Theorem gets us the pivot that, in our case, is 0.026 m away from the edge. So our "d" of shifting is 0.5-0.026= 0.474

- the three parts shown in calculation below,

- Note that GPE = 0 is set at the pivot for part 1 and 3. After finding cosθ, we can now find "h":

The experiment
- Take mass measurements of meter stick and clay(also with the paper clips and tape around it).
- Set up the apparatus. Place one farthest end of meter stick to the pivot point that's attached a standing rod.

camera seen attached to the bottom of chair

- Connect camera to Loggerpro for data collection. Place the camera such that the field of view show the collision and the system raised to maximum height.
- Test out the setup by releasing the meter stick, held horizontally, and make sure that it stays with the clay (after collision) all the way till they reach maximum height.
- For video capture setup, set scale and axis. Set scale by measuring the width of the whiteboard behind the setup. The axis should preferably be centered at where the clay is, for easier video analysis.

- Since the only data we mainly want from the experiment is the maximum height reached, when adding point series, we can just put two points, blue dots from the screenshot above.
- The lower dot won't be exactly at zero so we have to subtract between the two dots to find experimental value for "h".

Data Analysis:

- From the y-axis of point series data, we get h= 0.2419 - 0.007486 = 0.234414 m.

 - Comparing to predicted value and calculation of error and uncertainty:

- Experimental "h" should be smaller then predicted value due to other unaccounted error and certainty during calculation.

Conclusion

Our error and uncertainty include friction at pivot (we test it out with only meter stick released from one end doesn't travel to the same height at the other end), kinetic friction from the floor during the collision (from scraping of clay with the floor tends to slow system down), imperfect stability at the pivot point. The usual systematic error result from measurement equipments and poor video quality. Nonetheless, our experimental "h" is smaller than the predicted due to the errors and uncertainty described, and so it's safe to conclude that there's conservation of energy and conservation of angular momentum in the meter stick+ clay system.

May 13th, 2015: Lab 17: Moment of Inertia (at center of mass) of a Uniform Triangle

Purpose:

Determine I-cm of a triangle by using Parallel Axis Theorem and compare this to the experimental value obtained from the measured angular acceleration.

Experiment:

- Use the same tools from angular acceleration lab. Ideally, use the same kit from last lab so the measurements of rotational disks, hanging mass, and torque pulleys stay true for this lab as well.
- Measure the mass and dimensions of the triangle.

- Make sure the platform is leveled. Wrap the string-hanging mass around large torque pulley till the mass is at maximum height. On top of the pulley, screw in the metal rod, as shown below.

- Set up Loggerpro and other needed equipment just like angular acceleration lab. Make sure the sensor setting is 200 counts/revolution.
- Turn on the compressed air, so the the two disks turn independently. Release the disks and let them rotate. Collect data on Logger pro. Let the hanging mass move up and down three times so there's 3 α-up and 3 α-down. We're going to take average of these to omit any friction in the system.
- Repeat the same steps to taking data for system with the triangle in two different orientations as shown below,

               

            

Data and Analysis:

Theoretical inertia of the triangle:

- Coming up with an expression for an inertia of a triangle spinning around its edge is easier because we can take "dm" of the triangle and claim that each of that strip has dI = (1/3)*dm* x^2. Therefore we get the inertia to be (1/6)MB^2. The Parallel Axis Theorem allows us to move axis of rotation to the triangle's center of mass. Using equation for center of mass, we get the distance from center of mass of x-axis to be 1/3 of the base.

Calculation for center of mass of a uniform triangle

- Since it is a uniform triangle, we can use area density to express "dm".

- We can repeat the same process for another orientation although we have to keep the variables unchanged for each dimension. Using the derived equations, we could plug in the mass, base and height that we measure to get "I-cm" for both cases.

Experimental moment of inertia for the triangle:
Data:

Triangle's mass= 0.454 +/- 0.001  kg

base= 0.0987 +/- 0.00001 m

height= 0.14756 +/- 0.00001 m

hanging mass = 0.0245 +/- 0.001 kg

r-large torque pulley= 0.025 +/- 0.001 m

- Since the descending and ascending angular acceleration are not the same, we have to take their average so that we don't have to deal with frictional torque in the system.
- The way we obtain α is the same; take linear fit of omega vs. t graph.
- Due to the existing inertia in the rod used to support the triangle, we have to calculate it separately then subtract from the inertia of triangle+rod. α is also partially determined by the inertia of the spinning disk, so we don't have to calculate that inertia separately. The only inertia we're adding to the system is the triangle.
- Shown below, in order, α of system with rod only, horizontal, and vertical orientation:

- Take the average of these 3 cases give us,

Ascending α has (-) sign so we have to take absolute value

- We then use this obtained value for the equation derived below. 

       

- Moment of inertia is inversely proportional to α.

Conclusion:

Our error and uncertainty is rather large. Mass measurement would be one of the biggest contribution here since our equipment's uncertainty is 1 gram. As far as collecting α is concerned, friction due to the rotating disks is our most reasonable explanation for error because a slight change in α, even to 2 decimal places, change our error calculation by significant amount. Another source might also be from the amount of compressed air that was in the system. Too little air, indirectly, can cause α to be smaller. Nonetheless, within our range of error and uncertainty, our experimental calculations are close to calculations using Parallel Axis Theorem.

May 11th, 2015: Lab 18: Moment of Inertia and Frictional Torque

Purpose:

Measure the system's angular acceleration, moment of inertia, and derive an equation to find frictional torque in the metal disk system using these measurements.

Experiment:

- The apparatus is composed of 3 cylinders. On it, there's a stapled mark that indicates the total of the 3 cylinders combined (in grams). Record this number.

- Use small and large calipers to measure diameter, and height (or thickness) of the 3 cylinders.
-Use these measurements to calculate moment of inertia of each cylinder, then the total inertia.
- To determine the angular acceleration, set up video capture to record to the disk after it's given a spin and slows down on its own due to friction.
- Place a tape on tip of disk so we could add new point series later during video analysis. In Logger pro open video capture and set it to record to 15-20 seconds. Go to camera settings, "Adjustments", "Image", and adjust contrast and lighting accordingly.  Adjust the camera visibility to have the radius of the large disk in the middle. Test out by start recording video. Make sure that the tape is seen all the time.
- For the experiment, give the disk a spin, then record video. The video should still record until the disk stop on its own.

Data and Analysis:

Calculating Moment of Inertia
- Since only the total mass is given and we can assume that the cylinders have uniform volume density, we can find each of their individual mass.
- We use volume of cylinder = πR²H, and the relationship of ρ=m/v . Note that we need m1+m3 together and m2 alone. 

Calculating frictional torque (τ_f )

- After given a spin, the only torque acting on the system is frictional torque. So for the net torque equation, we can write: _Στ: τ_{f = (I-total) α}

_- For video analysis, set origin for x and y-axis (ideally at the center of the disk), set scale of the radius, add new point series and start adding points taking the tape as the point of interest. Continue adding points until the disk stops.

_{- The point series will give x and y- displacement. From this we could take the derivatives to get x and y-velocity, by going to new calculated column.}

- Since we want to relate our data to rotational motion, we could create another new calculated column for angle. From our x and y-position, use arctan("y" / "x") to get the angle.

- Up to this point, we could check to see if our data is reasonable by graphing angle vs. time and take linear fit,

- Although there are plenty of discontinuity in our graphs, we see that angular velocity decreases with time, which is explained by the fact that the disk slows down.

- Next up, create another new calculated column for tangential velocity, that is the linear velocity at the edge of disk. Using Pythagorean Theorem gets us V= sqrt[(V-x)^2 + (V-y)^2]; and we just found V-x and V-y.

- Up to this point, what we're trying to do is to get Logger pro to give us angular acceleration (alpha). And we know that ω vs. t gives us the slope= alpha. To find ω, use the relationship between rotational and translational motion, one of which is ω= V/R. where R= radius of the disk and V is the one obtained from Pythagorean Theorem.

- Our final data table should look like,

- We could also take derivatives 2 times to get alpha from our angle. But seeing that angular acceleration varies so much although it would've stayed constant, we stay with the idea of using slope of ω vs. t. 

- Graphing  ω vs. t and taking linear fit gives, α = -0. 4246 rad/sec^2,

- So _Στ: τ_{f = (I-total) α = (0.02146)(-0.4246) = -0.009118 N.m;} where negative sign indicates opposite direction of motion. We take I-total because the whole system is bound to frictional torque.

Disk-Cart system

- After finding the frictional torque, use this to derive an equation for linear acceleration for metal disks-cart system, where cart is attached by a string and wrapped around one of the smaller radius of the 3 cylinders.

- During this set up, make sure the string is parallel to the track and does not slip as it unwinds from the disk. Keep the angle(track with horizontal) unchanged throughout the experiment.

- As usual, we can obtain two equations one from rotational dynamics and another from Newton's Second Law, sum of forces. Since α= (a-center of mass )/radius, replace this and solve for "a-center of mass".

 

-From this linear acceleration, use kinematics equation to find the time it takes for the cart to travel 1 meter down the track, released from rest. Since we know distance and acceleration the cart travels, we can find Δt.

- The calculation below show how we get the equation of linear acceleration for the cart. Following the lab handout, we set our initial conditions to be when angle (of the track with the floor)= 40°, m-cart = 0.500kg, and r = 0.0158m and find that it takes 8.20 seconds for the cart to travel 1 meter down the track.    

- τ-string = TR; where R= radius of the disk the string is wrapped around. Since the whole rotating apparatus is acceleration, we take I-total to be on the other side of our net torque equation.

- Using our real data, m-cart = 0.504kg; angle = 47.8°, same radius, Δx = 0.98m, we get t =0.72s

- Experimental data, where Δx = 0.98m (the cart doesn't quite begin at the top nor get to the very bottom of the track)

• trial 1: t = 7.42 s
• trial 2: t = 7.44 s
• trial 3: t = 7.47s

- Hence, error of the last two trials range between 0.27% and 0.67%. The experimental trials could only have larger values than the calculated one, because there's always neglected friction on the track that could slow the cart down, therefore longer time it takes to get to the bottom.
- If we wrap the string around the larger radius and let go of the cart, it will take longer for the cart to get to the bottom of the incline. This observation is made clear in the derived equation. When you increase the radius, the acceleration gets smaller.

Conclusion

In this uniform rotating apparatus, we could find frictional torque using angular acceleration and calculated moment of inertia. This result could be further used to determine the linear acceleration. Assuming constant linear acceleration, we could use kinematics to solve other possible problems given. For part 1, where frictional torque has to be determined, errors include the fact there's drop frames and the metal supporter gets in the way when points series are taken. Besides the poor video quality, there's also usual measurement uncertainty in the diameters. For part 2, we also have measurement uncertainty in angle, cart's mass, and Δx. Within our range of error and uncertainty, we can conclude practicality in our data analysis.