Activity Simple Harmonic Motion for Solid Ring (June 1) & Physical Pendulum Lab (June 3)

PURPOSE:  To determine the period of a a solid ring swinging at a pivot and the period of an isosceles triangle about two different pivots. 

PART 1:

To find the period of the pendulum of a solid ring, we found angular freqency, ω, by calculating angular acceleration using Torque = Inertia * Angular acceleration.  We made angular acceleration look like the equation: α = - (ω^2) θ, rearranged from Newton's Second Law of Motion, F = ma.  After substituting for the value for ω^2 and solving for ω, we used T=2π/ω and solved for the period.

Image of our solid ring with small radius, r, large radius, R, Ravg (distance from center to pivot)

We set the pivot of our system to have radius Ravg, which was directly in between the large radius, R, and smaller radius, r.  The net torque of the system was only the perpendicular component of its weight, which was mg*sinθ, when the ring was held at angle θ.  This force was a distance Ravg from the pivot, giving us Torque = (- mg*sinθ)(Ravg).  The inertia of the solid ring at its center of mass was found previously as (1/2M(R^2+r^2), but since we moved the pivot a distance Ravg from the center of mass, the moment of inertia about the ring's new pivot was [1/2M(R^2+r^2 + M(Ravg)^2].

Plugging this into our equation for T=Iα and solving for α:

After solving for α we set up the equation on the right hand side to make it look like α = - (ω^2) θ, and using the Taylor series expansion for sinθ, sinθ is approximated to just θ if θ is small enough.  Therefore, we knew that the quantity within the parenthesis was equal to ω^2.  We then used this to solve for T using T=2π/ω.

We then measured the solid ring to find R, r, and calculate Ravg:

When we plugged in our data to our equations, our period was 0.718 seconds (see equations above).

We then did the experiment, hanging the ring hung on its pivot, while the other end of the ring moved through a motion sensor that measured the period of the pendulum.  Our experimental value for period, T, was 0.72 seconds.  Our percent error for found values was 0.28%.

Set up used to measure the period of oscillation.

PART 2:

We then derived for expressions for the period of an isosceles triangle with height, H, and base, B, oscillating about its apex, and then oscillating about the midpoint of its base.  After deriving these expressions, we verified them by using a cutout of an isosceles triangle and oscillating it about the two points.

We first derived an expression for a small thin bar, dm, of the entire triangle.  Setting it equal to the proportion of the area of this thin bar over total area, we solved for dm.  We then found the moment of inertia of this thin bar, treating it as a thin rod rotating about its axis with the equation 1/12ML^2.  Each thin bar was a distance, y, away from the pivot.  Using this, we added up the moments of inertia of all these thin bars from 0 to H to find the moment of inertia of the entire triangle oscillating about its apex.

Moment of inertia of isosceles triangle oscillating about apex.

We then used dm to solve for the center of mass of the triangle in order to apply this to the parallel axis theorem:

Center of mass is 2/3 of the height from the apex.

Using the parallel axis theorem, we substituted for our known moment of inertia oscillating about the apex and also substituted distance from apex to center of mass (2/3*H) to find the inertia oscillating about its center of mass.

Once we knew the moment of inertia about the center of mass, we could again apply the parallel axis theorem to find the moment of inertia oscillating about the center of the base of the triangle (a distance 1/3 away from the center of mass).

Moment of inertia of triangle oscillating about the midpoint of the base.

We used our expressions for moments of inertia in our torque equations, then made each equation resemble α= - ( ω^2 ) θ.  By finding ω^2, we were able to solve for the period.  We measured our triangle to have a base of 0.14 m and height of 0.147 m.  Plugging these values into our equations we solved for theoretical periods of the oscillating triangle.

Theoretical period for triangle oscillating about its apex: 0.69081 seconds.

Theoretical period for triangle oscillating about the midpoint of its base: 0.60238 seconds.

We set up the triangle in both positions, with the other end passing through a photogate:

Period for the triangle oscillating about its apex: 0.695362 s.

Period for the triangle oscillating about the midpoint of its base: 0.607886 s.

Theoretical period for triangle oscillating about its apex: 0.69081 seconds.

Experimental period for triangle oscillating about its apex: 0.695362 seconds.

Error: 0.65%

Theoretical period for triangle oscillating about the midpoint of its base: 0.60238 seconds.

Experimental period for triangle oscillating about midpoint of its base: 0.607886 seconds.

Error: 0.91%

CONCLUSION:

For both parts of our experiment, we were able to use an expression for angular acceleration of a physical pendulum swinging with simple harmonic motion to calculate its angular frequency and then find the period of the system.  For part 1, we used a net torque equation, and for part 2, we used moments of inertia of the shape oscillating about 2 different pivots, and we also used the parallel axis theorem.

In part 1, our theoretical time for period was 0.718 seconds, while our experimental time was 0.72 seconds, giving us an error of 0.28%, which shows that our calculated data for the period of a pendulum was very precise to the experimental result.

In part 2, our percent erros were 0.65% for the triangle oscillating about its apex and 0.91% for the triangle oscillating about the midpoint of its base.  In all 3 cases, the error was less than 1%, which just shows us that there is very high precision in the periods of pendulums that oscillate with simple harmonic motion.

2015-May-20 Lab 19: Conservation of Energy/Conservation of Angular Momentum

PURPOSE:  To use conservation of energy and conservation of angular momentum to predict how high a meter stick will rise, pivoted about its end, after hitting and sticking to a clay blob.

PROCEDURE:  

We set up a system in which a meter stick rotated about (close to) one end (2.5 cm away from the end of the meter stick).  The meter stick was held up to its horizontal position, released from rest, and swung down to hit a clay blob so that the clay blob would stick to the end of the stick and the stick-clay system would continue moving to a maximum height.

Diagram of stick-blob system.

In the time that the stick is released from its horizontal position until just before it hits the clay blob, energy is conserved.  Then, from right before the meter stick hits the clay, angular momentum is conserved through the inelastic clay-stick system.  Finally, energy is again conserved in the stick-clay system as it continues to rotate together to its maximum height, when kinetic energy becomes zero (angular velocity is zero).

We used our measurements to calculate angular velocity just before the stick hits the clay blob, when it is in its vertical position.  Then we used this ωfinal as ωinitial in our conservation of angular momentum equation (equation 2) to solve for ωfinal just after the meter stick hit the clay.  Again, this ωfinal became ωinitial in our last conservation of energy equation, for initial angular velocity of the system as it rotated to its maximum height.

For gravitational potential energy, we set zero at the pivot, so when the meter stick was in its horizontal position its center of mass was at zero, and when the meter stick was in its vertical position its center of mass was 0.475 meters below GPE = 0, so -0.475 meters.  

For our first conservation of energy equation:

On the left hand side, gravitational potential energy is at 0 and the stick is at rest so it has no kinetic energy.  On the right hand side when the meter stick is in its vertical position, its center of mass is at height -0.475 meters.  The moment of inertia of a rod about its center of mass is 1/12*mass*length^2, and using the parallel axis theorem we moved the pivot 0.475 meters from the center of mass, so we must add mass*distance^2 (shown above).  Solving for angular acceleration, we find that it is 5.477 rad/s just before impact with the clay blob.

Our next equation is conservation of angular momentum:

Initial angular momentum equals final angular momentum, or Iωinitial = Iωfinal.  Moment of inertia of the stick is the same as in the previous equation.  Our initial angular velocity comes from our previous equation.  After the collision, the moment of inertia of the stick is the same but we must add the inertia of the clay, which is the mass of clay * distance of clay from the pivot, squared (equation for inertia of a particle).  Solving for our new final angular velocity of both the stick and clay system, we get 2.733 rad/s just after collision.

For part 3, we use conservation of energy again.  Final kinetic energy is zero because we want to find the maximum height that the meter stick rotates, which will be when angular velocity is zero.

Our inertia on the left hand side of the equation is the same as the inertia of the clay-stick system from the previous equation.  However, we have new potential energies.  We can get the gravitational potential energy of the system by finding the GPE of each individual part and adding them together.  On the left hand side of the equation, the meter stick is vertical, so its height is -0.475 meters (0.475 meters below the pivot, where GPE=0).  The clay is at the very end of the meter stick, so its height is -0.975 meters.  On the right hand side of the equation when the stick-clay system is at its highest, it will be at an angle θ.  The vertical height of the stick's center of mass at angle θ will be -0.475cosθ, or the vertical component of the stick's center of mass at this angle, and the vertical height of the clay at angle θ will be -0.975cosθ, or the vertical component of the distance the clay is from the pivot at this angle. 

Plugging all this into our equation for conservation of energy, we find that when the stick-clay system reaches its highest point it is at an angle of 45.36 degrees from the vertical.  Taking the cosine of this angle gives us a vertical distance from our pivot towards the table, which is 0.7026 meters.  If we subtract this from 0.975 meters, we can see that the end of the meter stick reached a height of 0.272 meters above the ground, or 27.2 cm.

Using video capture to plot the origin and maximum y distance the stick-clay system reached, we found it to be 0.290 meters above the ground, which was within 6.2% of our calculated height.

CONCLUSION:

Conservation of energy and conservation of angular momentum allowed us to calculate the angular velocity before the stick and clay collided and just after they collided, to find the gravitational potential energy when angular velocity was at zero at the stick/clay's highest point.  We also used moments of inertia, the parallel axis theorem, and sum of individual components of GPEs to calculate our prediction.  In the end, we had a 6.2% error with our experimental result, but this could have come from a number of factors.  The height at which we released the meter stick may not have been perfectly horizontal, and the video freeze frame may not have stopped exactly at the highest point on our grid.  The angle that the camera was set up may not have been positioned directly in front of the rotating system, which could cause distances to be skewed because of the angle it was at.  Another important factor could be that the clay blob's center of mass may not have been exactly at the end of the meter stick.

2015-05-13 Lab 17: Moment of Inertia of a Uniform Triangle

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 using the parallel axis theorem.

For this lab, we first calculated the moment of inertia of a thin triangular plate about its edge, and used the parallel axis theorem to verify the moment of inertia around its center of mass.

PROCEDURE:

We used a device that rotates a disk on a cushion of air, which has a string attached to a pulley on its center.  The string hangs over a frictionless pulley to a hanging mass.  By measuring the angular acceleration of this system (using logger pro), we were able to find the moment of inertia of this system.

Setup of system without triangular plate added.

System with triangular plate mounted about its center of mass, in vertical position.

We measured angular acceleration in both the ascending and descending positions of the mass to get the average of the two, because of the frictional torque in the system.  This was found by taking the slopes of the velocity vs. time graph in both directions.

Initial system without the triangular plate.
The slope of the velocity vs. time graph gives us angular acceleration in both the up and down directions of the mass.

We then repeated this experiment, this time adding the triangular plate first with the tall side vertical (we will call this vertical position), and then with the short side vertical (horizontal position).  Both times the plate was mounted around its center of mass.

Velocity vs. time graph for the triangular plate in its vertical position.

Velocity vs. time graph for the triangular plate in its horizontal position.

From these 3 experiments, we derived the following data:

We used an equation derived from the previous lab to find moments of inertia in each system.  This equation uses hanging mass, radius of pulley, and average acceleration, which canceled out the small frictional torque in the system.

We measured the hanging mass to be 0.025 kg and radius of the pulley to be 0.025 m.  For the system with no triangle, moment of inertia was 0.0009078 kg m^2: 

For the system in which the triangle was mounted on its center of mass in the vertical position, moment of inertia was 0.001179 kg m^2:

For the system in which the triangle was mounted on its center of mass in the vertical position, moment of inertia was 0.001448 kg m^2:

We were then able find the difference of moment of inertia of the system with no triangle with the triangle in each position.  Moment of inertia of the triangle in the vertical position minus moment of inertia of the system with no triangle was 2.712 x 10^-4 kg m^2, giving us the moment of inertia of the vertical triangle by itself:

Doing the same for the triangular plate in the horizontal position, we find that its moment of inertia is 5.402 x 10^-4 kg m^2: 

After experimentally finding the moments of inertia of the triangle in 2 perpendicular orientations around the center of mass, we used our calculations to verify the results.  Using calculus, we derived the moment of inertia for the triangle, and evaluated it with the axis at its edge because the limits of integration were simpler.

Calculations for determining the moment of inertia for triangular plate.

Moment of inertia around the edge of the triangular plate: 1/6Mb^2.

After finding this, we used the parallel axis theorem to calculate what the moment of inertia is about its center of mass.  Since we found the moment of inertia around its edge, we can subtract the mass of the triangle times the distance squared from the edge to the center of mass (Mh^2).  We measured the mass of the triangle to be 0.462 kg.

In the triangle's vertical position, the base was 0.0987 meters and the distance from the center of mass to the tall vertical edge was 0.0331 meters.  Plugging this in, we found the moment of inertia around the center of mass to be 2.44x10^-4 kg m^2.  

Our experimental moment of inertia for the triangular plate in vertical position was 2.712 x 10^-4 kg m^2.  This gives us an error of uncertainty of 10.02%.

In the triangle's horizontal position, the base was 0.149 meters and the distance from the center of mass to the short horizontal edge was 0.0489 meters.  Plugging this in, we found the moment of inertia around the center of mass to be 6.05x10^-4 kg m^2.  

 Our experimental moment of inertia for the plate in the horizontal position was 5.402 x 10^-4 kg m^2.  This gives us an error of uncertainty of 10.71%.

CONCLUSION:

By finding the difference between the moment of inertia of a rotating system and then the moment of inertia of a triangular plate mounted about its center in two positions, we were able to find the moments of inertia of the triangular plate itself.  Then, we calculated the moment of inertia about the triangular plate's edge using calculus, and plugged all of this into the equation for the parallel axis theorem to compare our mathematical results for moment of inertia about the triangle's center to our experimental results for moment of inertia about the triangle's center.

In the vertical position, our experimental value was 2.712x10^-4 kg m^2 and our theoretical value was 2.44x10^-4 kg m^2, giving us an error of uncertainty of 10.02%.  In the horizontal position, our experimental value was 5.402 x 10^-4 kg m^2 and our theoretical value was 6.05x10^-4 kg m^2, giving us an error of uncertainty of 10.71%.

The value of these numbers represent the quantity that it takes for the triangle to spin around a fixed axis.  Our results show that it takes a higher value for the triangle to spin around its center of mass in the horizontal position, which makes sense because the mass is distributed further out away from the axis.  Our errors in theoretical and experimental values could have been made from a number of factors: the mass of the triangular plate may not have been completely in uniform density; the air being released to spin the rotating disks may have fluctuated, causing acceleration to fluctuate; there also could have been error in our measurements.

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

PURPOSE:  To find the moment of inertia in a system and calculate its frictional torque using anguelar acceleration; to set up a system that can use tension in both a torque equation and a force equation to predict linear acceleration of a cart.

PROCEDURE: 

We first found the inertia of a system made up of three circular disks, one large metal disk on a central shaft with two smaller disks on each side, by finding the moments of inertia of each individual disk and adding them together.

• By finding the width and diameter of each disk, we were able to calculate the volume of each one, and then add the three to find total volume.
• We then took the ratio of the volume of one disk to total volume, and set it equal to the ratio of the mass of that disk to total mass (given), to find the mass of that particular disk.

By finding the masses of each individual disk, we were able to calculate moments of inertia for each disk, then add them to get the moment of inertia for the system.  Inertia for a solid cylinder around its axis is 1/2mr^2:

Total inertia for the system is 0.02048 kg m^2

Next, we found the frictional torque of the system by setting frictional torque equal to inertia of the system times its angular deceleration.  To find angular deceleration, we used video capture on the spinning system, and marked each time a piece of tape made one full rotation.

Set up of video capturing angular acceleration.

Set the origin to the axis of rotation and marked points (at the top) when piece of tape rotated each time around.

These points gave us a graph of position (theta) vs. time (seconds), which we placed a curve fit to to give us the equation of position, theta = -0.516t^2 + 10.59t - 26.84.

By taking the derivative of this twice, we were able to find angular acceleration:

Then, multiplying inertia for the system by angular acceleration (0.02048 kg m^2 x -1.0322 rad/s^2), we get a frictional torque of -0.0211 N m.

We then used this data to connect a cart to the system by a string, and predicted how much time it would take for the cart to travel 1 meter from rest.

Set up for the second part of our experiment.

We did this by setting up a free body diagram of the cart attached to the system.  We found the sum of forces in the x-direction, assuming the track was angled at 40 degrees with the horizontal.  We also found the equation for torque, which was torque of the string (distance r from the axis), minus frictional torque, which was equal to inertia times angular acceleration.

Solving for T (tension) in each system, we then set the two equations equal to each other to find linear acceleration.  *Note: we replaced angular acceleration in the torque equation with linear acceleration divided by radius.  *Since the string was wrapped around the smaller cylinder, we used that radius.  *Mass of cart was measured as 0.497 kg.

Linear acceleration came out to be 0.01788 m/s^2, which we then plugged into a kinematics equation, with distance as 1 meter, to find the time.

Prediction for the time it takes the cart to travel 1 meter down a ramp angled at 40 degrees.

After finding these calculations, we ran the experiment by releasing the cart from rest and timing (with a stopwatch) how long it took for the cart to travel 1 meter down the ramp.  We also made sure the ramp was angled at 40 degrees with the ground.  

We ran our experiment twice, with the first time resulting in 11.86 seconds, the second in 10.58 seconds, and an average of 11.22 seconds.

Our error of uncertainty was (10.576-11.22)/11.22 x 100% = 5.74%

CONCLUSION:

This experiment allowed us to find the total inertia of a system by adding up individual moments of inertia.  We then found the angular acceleration of the system, and used the equation Torque = Inertia x angular acceleration to calculate frictional torque.  By connecting a cart rolling down a ramp to the system, we solved for the tension force in two equations: one for the sum of forces in the x-direction of our cart and one for the sum of torques in the rotating system.  This allowed us to solve for linear acceleration, and then predict the time it took the cart to move 1 meter.

Our error of uncertainty came out to be 5.74%.  The second trial was a lot closer to the theoretical time, and perhaps if we had done a third trial our results may have been close again. making our average time more accurate.  Error could have resulted from timing the experiment, because it was difficult to stop the stopwatch at the exact moment the cart passed 1 meter.  It also could have been from any slight error in measuring the initial dimensions of each cylinder.  Another source of error was from marking the position of the tape each time it made a revolution in the video, because the frames that the video stopped at did not always match up to exactly one full rotation.

2015-May-04 Lab 16: Angular Acceleration

PURPOSE:  To observe how various factors affect angular acceleration, and then use data collected from a system of angular acceleration with a hanging mass to verify our experimental moment of inertia with our known equation for the moment of inertia of a disk.

SET UP:

Compressed air blows into a system of two large disks rotating one on top of the other, independently of each other.  On top of these disks is a pulley (of interchangeable sizes), with a string wrapped around it with a hanging mass tied to the other side, hanging over a frictionless pulley at the edge of the system (see below).

When the system is in motion, a sensor counts marks on the sides of the disks to graph angular position and angular velocity.

PROCEDURE:

We ran 6 experiments, with various changes in the experiment in order to observe how these changes affected angular acceleration.

• In the first 3 experiments, we used a top steel disk, bottom steel disk, small pulley, and only varied the hanging mass.  Experiment 1 used 25 g, 2 used 50 g, and 3 used 75 g.
• In the 4th experiment we hung the 25 grams back on, and only changed pulley from a small one to a large one.
• In experiment 5 we hung 25 grams, kept the large pulley, but changed the top disk from steel to aluminum (smaller mass).
• In experiment 6 we hung 25 grams, kept the large pulley, and replaced the steel disk back on the top, but this time allowed the top and bottom steel disks to rotate together.

For each experiment we ran the sensor to graph an angular position vs. time graph and an angular velocity vs. time graph.  We used the angular velocity graph to measure angular acceleration as the mass moves up and down by finding the slopes of angular velocity as the mass moved up and down. 

Our graph from experiment 2:

Slope of 5.715 rad/s^2 is angular acceleration while the mass is moving down, and -7.460 rad/s^2 is angular acceleration while the mass is moving up.

We take the average of the magnitudes of the two angular accelerations because there is some frictional torque in the system, which causes all the downward accelerations to be slightly less than the upward accelerations.  We take the average to balance this out a little more.  After finding the slopes of the angular velocity vs. time graphs for each experiment, we came up with the following data:

CONCLUSIONS from Part 1:

In the first 3 experiments when we only changed the hanging mass, acceleration increased by the same factor that we increased the mass.  Experiment 2 had two times the mass of Experiment 1, and angular acceleration was around 2 times as large.  Experiment 3 had 3 times the mass of Experiment 1, and angular acceleration was around 3 times as large.

In Experiments 1 and 4 when we only changed the size the pulley but had the same hanging mass of 25 grams, angular acceleration decreased with a larger pulley.  This makes sense because it would take longer for the pulley to make a full rotation.

In Experiments 4, 5, and 6, we changed the mass of the rotating body.  In Exp 4 with a top steel disk of 1.361 kg, angular acceleration was 2.228 rad/s^2.  Exp 5 had a top aluminum disk which was about 1/3 the mass of the steel disk (0.466 kg), and angular acceleration was about 3 times as much as with the top steel disk (6.261 rad/s^2).  This also makes sense because a larger mass would not rotate as quickly, which leads us to Exp 6 using both the top steel and bottom steel disks together with a combined mass of 2.709 kg.  This is roughly double the mass of the disk rotating in Exp 4, and its angular acceleration is roughly 1/2 of that in Exp 4 (0.922 rad/s^2).

*** One other part we modified to Part 1 of this lab was to add a motion sensor to record the linear velocity of the hanging mass as it moved up and down (shown below):

Only ran the motion detector for Experiment 1.

By doing this, we were able to compare angular acceleration of the system to linear velocity of the system, by multiplying angular acceleration by the radius of the small pulley (used in Exp 1).  This gave us the following graphs:

The first two graphs are angular position and velocity vs. time, and the second two graphs are linear position and velocity vs. time.  We found the slopes of each graph with the mass moving down and then down.  If we multiply the magnitude of angular acceleration in the down direction (3.301 rad/s^2) by the radius of the small pulley (0.013 m), we get a linear acceleration of 0.043 m/s^2, which is close to the experimental linear acceleration from the motion detector of 0.042 m/s^2.  We can do the same for the magnitude of angular acceleration in the up direction (3.620 rad/s^2) multiplied by 0.013 m, to get a linear acceleration of 0.047 m/s^2, while the experimental value was 0.046 m/s^2.

PART 2:

We can use our data collected and measurements to calculate moment of inertia for the system.  We used Newton's second law to find the tension, T, in the hanging mass.  We also used the equation for torque and plugged in tension to this equation (also changing linear acceleration to angular acceleration):

Moment of inertia using data we collected in the experiment from Part 1.

For Experiment 1, we plugged in our data:  on the right hand side is the equation for moment of inertia of a disk, using the measurements of the disk.  

However, our percent error shows we were extremely far off when our answers should have been the same. This could have been from a misread from any of the measurements in the disk or the pulley.

Continuing with the calculations for the rest of the experiments, we see that our percent error is much less in the last 3 experiments.  What's different about the last 3 experiments is the use of the large pulley instead of the small pulley, so our error may have been caused from a misreading in the measurements of the small pulley.

CONCLUSION:

By using Newton's second law, the torque equation, and conversion of linear to angular acceleration, we were able to come up with an equation for the moment of inertia of a rotating system with a hanging mass.  This equation used the mass of the hanging object, radius of the pulley in which it was spinning, and the angular acceleration of the mass.  This should have given us the same moment of inertia if we used the moment of inertia for a disk, 1/2MR^2.  However, our results were off by a significant value.  Our errors were much too large to have been caused by a small source of uncertainty.  We must have had made a huge blunder in measurements, possibly by reading the vernier caliper wrong.  Other small sources of error could have been caused by fluctuations of air in the compressed air tube, causing the disks to rotate at a not completely constant speed.  Also, there was friction in the pulley which caused our angular accelerations to vary.  An error could have been caused when the mass was hanging, and if we taped multiple masses in a way that made the mass sway just a little bit.

Despite our errors, our group understood clearly the way we came about deriving an experimental equation for the moment of inertia of our system and how it can compare to the moment of inertia of a disk.