MAY-11-2015 Moment of Inertia and Frictional Torque

Moment of Inertia and Frictional Torque

(How long until the cart hits the bottom of the track??)

Purpose:  To determine the total moment of inertia of a spinning disk apparatus about its center of mass, then in turn to calculate the frictional torque in order to determine the amount of time required for a cart (attached by a string to the spinning disk apparatus) to reach the bottom of an inclined plane 1 m long.

Procedure:  Firstly, we took the appropriate measurements of the rotating apparatus.  Depicted below is a diagram of said measurements:

(Diagram with relevant measurements for each piece of mass in cm)

The total mass (4,829 g) was stamped on the large portion of the rotating apparatus.  In order to determine the moment of inertia for this entire mass, we need to sum up the moment of inertia for each individual piece (ie, 1, 2, 3).

In order to achieve this, we utilized our measured values from above to determine the volume of each one.  We then summed these volumes to give us the total volume of the rotating mass.  We then took each pieces volume and divided it by the whole to determine the percentage of volume each piece took up.  This percentage is the same percentage of the overall mass for each individual piece.  These calculations are shown below, along with our determination for the moment of inertia of the whole mass:

(Determination of percentage of volume of each piece; derivation of I of system)

From this, we have:

Itotal = 0.0209 kg * m^2

Our next step was to determine the frictional torque exerted on this system.  In order to achieve this we first must determine the angular acceleration as the rotating mass comes to rest.  We can use this value of deceleration and multiply it by our value for the total moment of inertia of the system to determine the value of frictional torque (which is required to accurately predict the amount of time required for a cart attached to the mass via a string to travel 1m down an inclined plane).

We used LoggerPro to analyze video capture data of the rotating mass as it came to rest while starting with some initial angular velocity.  We then plotted Theta v. Time within LoggerPro by clicking a piece of tape (attached to the rotating mass) in the video capture data each time it was at the peak of rotation.  Below is the aforementioned graph:

(LoggerPro plot of Theta v. Time)

We applied a quadratic fit to the above graph.  Doing this yields the angular acceleration of the rotating mass; it is found by doubling the value of "A" found in the quadratic above.

From this, we have:

alpha = -0.3954 rad/s^2

Using our value of I and alpha, we have:

Frictional Torque = 0.00826 kg(m/s)^2

We know have all we need to predict the time it will take a cart attached to this rotating mass, via a string, to travel a distance of 1 m along an inclined "frictionless" surface.  This experimental apparatus is shown below:

(Experimental Apparatus)

Below are the diagrams and calculations necessary to predict the time it takes for the cart to travel 1 m:

(Predicting time for cart to travel 1 m along an incline @ 49.1 degrees relative to the floor)

Our predicted value is:

Theoretical Time = T1 = 7.30 s

After calculating our theoretical time, we ran the experiment using stopwatches to record the time it took the cart to travel 1 m along the track.  This value came out to be:

Actual Time = T2 = 7.35 s

Conclusion:  Our theoretical value was only of by .05 s, which is understandable since we assumed absolutely no friction in the track the cart traveled along.  Other sources of uncertainty include incorrect measurements of the rotating apparatus, which would effectively alter several of our calculations throughout the theoretical work.  Another source of uncertainty could have been encountered if the string running from the cart to the rotating mass was not parallel with the track the cart was traversing.

MAY-13-2015 Utilizing the Parallel Axis Theorem

Utilizing the Parallel Axis Theorem

(Finding the moment of inertia of a uniform triangle about its center of mass)

Purpose:  To determine the moment of inertia of a right triangular thin plate around its center of mass, for two perpendicular orientations of the triangle.

Procedure:  We first determined the moment of inertia for the object about its center of mass theoretically utilizing the parallel axis theorem.  We first determined its moment of inertia about its edge, then used Ip = Icm + Md^2 to find the moment of inertia about its center of mass.

The following are the measurements of the uniform right triangular thin plate:

Height = h = 14.80 cm

Base = b = 9.85 cm

Mass = m = 456 g

Depicted below are two theoretical derivations for the moment of inertia; the first with h in the vertical direction, and the second with h in the horizontal direction (two perpendicular orientations).

(Derivation of Icm in orientation 1)

(Derivation of Icm in orientation 2; evaluation omitted)

The value of Icm for the second derivation when evaluated is:

Icm = 5.55 X 10^-4 kg * m^2

In order to determine our experimental values of Icm for each orientation, we used methods similar to those in our "Angular Acceleration" lab.  Depicted below is the experimental setup for orientation 1:

(Experimental Setup for Orientation 1)

The first objective is to determine the moment of inertia of the system with the top steel rotating disk WITHOUT the triangle attached to the system.  Afterwards, we then added the triangular thin plate and calculated the new system's moment of inertia; the difference between these values yields the moment of inertia of the uniform triangular thin plate about it's center of mass.

Without Triangular Plate:

I = (mgr)/alpha - mr^2

Torque Pulley:
   Radius = 0.0126 m

Hanging Mass:
       Mass = .025 kg

We know the value of alpha from the following LoggerPro data:

(LoggerPro Data; no triangular plate)

This gives us:

alpha = (1.171 + 1.273) / 2 = 1.222 rad/s^2

Thus:

I of system w/o triangle = [(.025 *9.81 * .0126) / 1.222] - (.025)(.0126)^2
              = 2.52 X 10^-3 kg * m^2

Now that we have the moment of inertia for the system WITHOUT the triangular plate, we can attach it, recalculate the system's moment of inertia, and subtract our first result from the new value in order to attain the moment of inertia of the triangular plate about its center of mass for both orientations.

Triangular Plate; Orientation 1:

alpha = (1.072 + 1.174) / 2 = 1.123 rad/s^2

I of system with triangle @ orientation 1 = 2.75 X 10^-3 kg * m^2

Difference = I of triangle @ orientation 1 = 2.28 X 10^-4 kg *m^2

That result quite close to our theoretical calculation from above!

Triangular Plate; Orientation 2:

alpha = 0.9965 rad/s^2

I of system + triangle @ orientation 2 = 3.10 X 10^-3 kg * m^2

Difference = I of triangle @ orientation 2 = 5.77 X 10^-4 kg * m^2

Once again we're close to our theoretical value.

These results are summarized below:

(Summary of Theoretical v. Experimental Findings)

Conclusion:  I'm satisfied with how the experimental values turned out when compared with the theoretically calculated values determined earlier in the lab.  Sources for the minor differences between these values include measurements taken via vernier calipers, as well as the assumption of zero friction within the torque pulley system.

MAY-04-2015 Angular Acceleration

Angular Acceleration

(Determining moment of inertia using angular acceleration)

Purpose:  The first half of this lab involves determining the angular acceleration (alpha) of various disks / disk combinations.  Having these values of angular acceleration, we can discuss numerically the differences in values when different parts of the system are altered (ie additional hanging mass, larger pulley radius, material of disk, etc..).  The second half of the lab is utilizing these angular acceleration values to determine the moment of inertia of the various disks / disk combinations and comparing them to the theoretical values using the known formula of the moment of inertia of a uniform disk (1/2MR^2).  

Part 1)

Procedure:  We first took relevant measurements of the various disks, pulleys, and the mass to be hanged via a string.  The apparatus is depicted below:

(Experimental Apparatus)

Listed below are the relevant measurements recorded:

(Measurements of experimental items)

In order to determine the angular acceleration of each setup, we used LoggerPro to record the angular velocity over time.  Performing a linear fit of this graph yields the angular acceleration as the hanging mass falls and rises.  The average of these two values is our target.  Below are two graphs of the first run of the experiment, with a setup of the original hanging mass (26 g), only the top steel disk spinning, and the hanging mass connected to the smaller torque pulley via a string.  The first graph gives us the angular acceleration as the mass falls, and the second as the mass rises:

(Angular Acceleration as mass falls)

(Angular Acceleration as mass rises)

Taking the average of these two values' absolute value, we have:

Average Acceleration = alpha = 1.1215 rad/s^2

Shown below are the varying values of alpha while running the experiment with various setups:

(Data from various Exp. setups)

Conclusions:  By analyzing the table provided above, we're able to draw some numerically specific conclusions:

 * By doubling the hanging mass, the angular acceleration is approximately doubled

 * By tripling the hanging mass, the angular acceleration is approximately tripled

 * By nearly doubling the torque radius, the angular acceleration is nearly doubled

 * By changing the rotating disk from steel to aluminum (nearly reducing the mass by a third), the angular acceleration is roughly tripled.

 * By roughly doubling the rotating mass (both steel disks), the angular acceleration is nearly halved.

Part 2)

Procedure:  In this portion of the lab, we'll utilize Newton's 2nd Law in order to determine the moment of inertia of each disk setup (ie top steel disk, top aluminum disk, top steel disk + bottom steel disk).

Depicted below is a free-body diagram of the first setup (Top Steel Disk), along with relevant 2nd Law equations required for determination of the moment of inertia using our recorded values of alpha:

(Moment of Inertia for Top Steel Disk)

The above picture displays that our experimental value for the moment of inertia of the top steel disk is acceptably near the theoretical value.  Performing similar calculations for the aluminum disk and top steel disk + bottom steel disk combination yield:

Aluminum Disk:

Experimental I = Ia = 1.10 X 10^-3 kg * m^2

Theoretical I = Ib = 0.92 X 10^-3 kg * m^2

Top Steel + Bottom Steel:

Experimental I = 6.20 X 10^-3 kg * m^2

Theoretical I = 5.38 X 10^-3 kg * m^2

Sources of Uncertainty:

Once source of uncertainty could be the way the angular velocity is recorded into LoggerPro, meaning the number of indentations counted by the system as one rotation could be offset.  Another is incorrectly recording the masses of the various pulleys or disks.  Yet another source is improperly measuring the diameter of the torque pulleys, as there are "lips" that surround the actual diameter of each one, this would in turn yield incorrect radius values used in deriving the moment of inertia values from our experimental data.

APR-29-2015 Impulse-Momentum Activity

Impulse-Momentum Activity

(Is Impulse equal to the change in Momentum?)

Purpose:  To demonstrate that impulse (J) is equal to the change in momentum of an object (in this case a cart along a frictionless track).

Procedure:  We first leveled a cart track on top of our table.  We placed one cart on top of the track, and another (this one with a spring bit extended from its end) at one end of the track, with the spring bit facing inward towards the track.  We also placed a motion sensor at the end opposite the spring bit containing cart.  This setup is shown below:

(Apparatus setup w/o force sensor)

In order to determine the impulse of the coming collisions, a force sensor is required.  We attached this force sensor to the cart on top of the track.  Additionally, we attached a flashcard to the back of the cart on the track, which would allow the motion sensor to record data from the moving object.  This is depicted below, however, before running the experiments, we opted to remove the flashcard, and instead placed a small board attachment via velcro.

(Cart with force sensor and flashcard attached)

With the motion sensor tracking the object's position over time, we can determine the cart's velocity as the hook on the force sensor collides with the spring bit from the other cart, this can be used to determine the change in momentum (given that we know the mass of the cart and it's initial and final velocities).  The force sensor allows us to record the varying force over the time of the collision (which gives us the impulse by integration).

We ran 3 experiments using this equipment.  The 1st being a collision between the setup cart shown above and the spring bit cart.  The 2nd being the same as the 1st with an added amount of mass to the cart on the track.  For the 3rd experiment, we attached a needle to the end of the force sensor, and collided it with a piece of clay instead of the second cart (an inelastic collision).

EXP 1)

mass of cart system = m = 0.764 kg

The recorded data from this first collision is show below:

(LoggerPro data from Exp 1)

The top graph represents the cart's velocity over time, while the bottom represents the varying force over time.  From the velocity v. time graph, we can determine the cart's velocity just before and just after the collision occurs.

initial velocity = vi = 0.2867 m/s

final velocity = vf = -0.2394 m/s

Having these values enables us to calculate the change in momentum of the cart:

mvf - mvi = (0.764)(-0.2394) - (0.764)(0.2867) = -0.4019 kg * m/s

From the lower graph, we can utilize the integration function in LoggerPro to determine the Impulse of the collision:

J = -0.4175 kg * m/s

Our calculations for Impulse were off slightly from the value given by integrating the force v. time graph over the collision time interval.  We then proceeded to do the experiment with additional mass added to the cart system.

EXP 2)

m = 1.264 kg

Below are the graphs from the second experiment:

(LoggerPro data from Exp 2)

From the velocity graph, we have:

mvf - mvi = (1.264)(-0.3748) - (1.264)(0.3989) = -0.9779 kg * m/s

From the force v. time graph we have:

J = -0.9678 kg * m/s

Once again, our values are slightly off from one another.  This could be to sources of uncertainty, such as improper recording of mass value of the cart system, or even a improperly leveled track.

EXP 3)

For this experiment, we removed the dynamics cart and replaced it with a wooden object which we then attached a bit of clay to (in order to create an inelastic collision).  We also removed the hook from the end of the forces sensor and replaced it with a nail.  This is shown below:

(Exp 3 setup)

m = 1.28 kg

Below is the data from this collision:

(LoggerPro data from Exp 3)

From these graphs, we have:

mvf - mvi = 0 - (1.28)(0.4277) = -0.5475 kg * m/s

J = -0.4671 kg * m/s

This last experiment contains the largest error with respect to the above values.  I believe this occurred due to improper recording of the new cart system mass.  We simply recorded the mass of the nail and rubber stopper used to attach it and added it to the entire system's mass from the previous experiment, as opposed to replacing the items and then measuring the new mass as a while system (which would've yielded better results in our change in momentum calculation).

22-APR-2015 Collision In Two Dimensions

Collisions In Two Dimensions

(Is Momentum & Energy Conserved?)

Purpose:  In this lab, we're to determine whether or not momentum and energy are conserved in a two dimensional collision.

Procedure:  In order to easily collect data for this experiment, we observed a collision between two balls (metal v. metal / metal v. aluminum) on an ideal surface while recording the collision from an overhead camera.  This is shown below:

(Experiment Apparatus)

(Ideal surface for experiment)

We measured the mass of each steel ball and the aluminum ball:

m1 = m(steel ball #1) = 66.5 g

m2 = m(steel ball #2) = 66.8 g

m3 = m(alum. ball)   = 10.1 g

We then proceeded to video capture the collision from the overhead camera.  This video was manipulated within LoggerPro in order to capture the motion BEFORE and AFTER the collision.  An example of this is shown below.

(Screenshot of Video Capture from Overhead Camera)

By plotting points throughout the length of the video, LoggerPro can display these motions graphically!!  The manner in which it does this, is that LoggerPro displays the changes in position  in both the x and y directions for EACH particle it is tracking (the balls).  By performing linear fits upon these graphs, we can determine the slope of each, which gives us the velocity of each ball in x and y component forms!

Below are the graphs for the Hit-Er ball before and after the collision, followed by the Hit-Ee ball after the collision.

(Hit-er before collision)

This is the motion of m2; from this, we have:

Vox = - 0.06285 m/s

Voy =    0.3800 m/s

(Hit-er after collision)

This gives us:

Vfx = - 0.1643 m/s

Vfy =   0.2183 m/s

(Hit-ee after collision)

This graph represents the change in position vs. time of m1; which reveals:

Vfx = 0.1029  m/s

Vfy = 0.1783  m/s

We now have all we need to determine whether or not momentum and energy have been conserved!  This analysis is depicted below:

(Momentum Calculations in x and y Directions)

(Energy Calculation)

What the first photo above displays, is that (for the most part) momentum is conserved after the collision in both the x and y directions.  The reason the value are not EXACTLY equal is due to sources of uncertainty (as momentum should ALWAYS be conserved).  These sources of uncertainty include such things as the "fish-eyed" camera which recorded the interaction (upon which all of our position v. time data is built), user error in plotting points during the video analysis, an improperly leveled surface, or incorrect mass readings during measurements.

The Energy calculation percent difference is much more interesting.  Energy is lost within the system, which explains the larger deficit between the two calculated values on either side of the equation.  Energy was lost in the form of sound (when the two objects collided) as well as some heat (miniscule as it may be, it contributes to energy lost in the system).

27-APR-2015 Ballistic Pendulum Activity

Ballistic Pendulum Activity

(Using Concepts of Momentum and Energy to Determine Initial Velocity of Projectile)

Purpose:  This activity demonstrates how the combined use of the concepts of energy and conservation of momentum can be utilized to determine the initial velocity of a projectile ball launched out of the apparatus shown below.

(Ballistic Pendulum Device)

Procedure:  A small metal ball is set into the launcher.  It is then shot out into the white block shown above (an inelastic collision).  These combined masses are then lifted together to some height h at an angle theta from the vertical (measured on the device itself as shown).  All relevant data is given below:

m1 = mball =          7.63 g  +/- 0.1 g 

m2 = mholder =      80.9 g +/- 0.1 g

theta =    21.0 degrees +/- 0.5 degrees --> 0.00873 radians

L =  length of string = 19.8 cm +/- 0.1 cm

After collecting this data, we derived an equation for the initial velocity of the projectile utilizing equations for momentum and energy as shown below:

(Derivation of Initial Velocity of Projectile)

Using our values from above, we have:

Vo = 5.89 m/s

However, it's necessary to take into account the propagated uncertainty in our calculation of initial velocity.  The determination of dVo is shown below:

(Determination of propagated uncertainty in Vo calculation)

So, for our final calculation for Vo, we have:

Vo = 5.89 m/s +/- .0898 m/s