blog 18

Last blog. 

The objective of this lab was to find the period of a semicircular pendulum rotated at its top and bottom.

We cut out a small semicircular disk of light material and measured its dimensions, then we rotated each around the two required orientations and measured the period.

In order to come up with a predicted value for period, we first had to find the moment of inertia around both axises. We did this by first finding the moment around its flat side axis, then using the parralel axis theorum to figure out its Icm. From I center of mass we could then use the parallel axis theorum again to find its moment of inertia of its curved side. Before all of this however, we had to integrate to determine the location of the center of mass.

After we had a value for each moment of inertia, we could calculate a theoretical period at small angles for each orientation by using T=I(alpha) calculations. All relevant calculations are as follows

As we can see, our predicted value around the curved axis was 0.707s, compared to the actual value of 0.72, and around the other axis 0.72s, compared to 0.722. Giving us an error margin of 1.9 and 1 percent off (respectively),  This small errors is probably due to some amount of friction in our pivot points, or the fact that this is an approximation assuming small angles, or some small measuring error when we found the dimensions of the pendulum.

blog 17

Mass spring osculation 

This lab concerns a mass spring system that displays simple harmonic motion, the goal of the lab is to find the period of the spring as a function of spring constant and the mass at the end of the spring. 

In order to collect more data, we were one of 4 groups, each with a different spring with a different spring constant, each group would suspend the same amount of functional osculating mass, from the equation (M osc = M hanging + M spring/3)

Before we could measure period, we had to measure the spring constant of our spring. We treated it as an ideal spring that obeyed hookes law. So when the spring has a mass hanging on it and the system is in equilibrium:

Fnet=0=mg-kx

where x is the amount the spring stretches when the weight is added to the spring

mg=kx

k=mg/x

k=0.1(9.8)/0.067=14.64

What follows is our derivation of the theoretical relationship between period, spring constant, and mass

We then ran the experement 4 times with different hanging masses each time. The chart turned our as seen below

When our theoretical values were compared to our actual observed values for period, we found them to be extremely accurate. With this done, we turned to the other groups observations.

remembering that our theoretical T was calculated to be

T=2pi(m/k)^(1/2)

we predicted that 

1. as m goes to infinity, T will go to infinity in a relationship T=m^(1/2)

2. as k goes to infinity, T will approach 0

To verify, we plotted the other groups periods against thier values for  k and m. Giving us the following two graphs

As we can see, both our predictions played out. With T decreasing by a power law as k increases and increases by a power rule as M increases. There was a small amount of error between or predicted and observed values, but this is probably due to measuring errors or small inconsistencies in the springs used in the experiment. 

blog 16

Conservation of linear and angular momentum 

In this lab, we seek to prove that angular and linear momentum are conserved in a system.

The system in question is a ball which rolls without slipping down a ramp and collides inelastically with a disk that spins horizontally with no friction. The motion of this system is in 2 parts

First, the ball rolls down the ramp, and in doing so its GPE is converted to rotational and translational KE

Mgh=1/2mv^2+1/2I(omega)^2

Second, the ball collides with the spinning disk, linear momentum of the ball is converted to rotational momentum of the disk and ball.

m(v/r)=I(disk+ball)(omega)

In order to find the actual initial velocity of the ball once it exits the ramp, we let the ball go down the ramp and land on the floor. We measure the distance from the edge of the ramp (superimposed on the floor) that the ball travels before hitting, and the height of the edge of the ramp. We then use kinematics equations to determine the exit velocity.

In order to find inertia of the disk we use a hanging mass attached to it in the same way that we did in earlier experements with this aparatus. We then measure angular acceleration and use T=I(alpha) equation to find  I.

Our calculations are below

As we can see, there is a large discrepancy between the theoretical initial exit velocity of the ball and the actual velocity. This indicates that the ball does not roll without slipping. Our theoretical value for omega final is 2/43 rad/s

We then run the experement and measure angular velcocity, which turns our to be 1.87 rad/s.

The large difference between our actual and expected values for omega are probably due to the issue of the ramp not allowing the ball to roll without slipping.

Blog 15

Ruler with inelastic collision 

Our goal in this lab is to derive a function of max height of a meterstick swing throught the vertical after an inelastic collision (in this case, with a piece of clay).

Our meterstick had a hole drilled at 0.2 meters from its end, which was mounted on an axis so that it would swing through the vertical. At the bottom of its path, a piece of clay is taped in place which will collide and stick to the meterstick as it passes its lowest point. In order to collect data, we filmed the process and used loggerpro to chart its position with respect to time. A screenshot of this process is below

In our trial, we found our max height to be 0.157 m.

In order to verify this, we now had to derive a theoretical maximum height. The motion of this rod has 3 parts.

First as it falls from its initial height to the bottom of its swing, gravitational PE is being converted to rotation KE.

Mgh=(1/2)I(omega)^2

In the second part, we have an inelastic collision or

I(omega)i=I(meterstick+clay)(omega)f

and finally, we have rotational KE being converted to gravitational potential energy

(1/2)I(meterstick+clay)(omega)^2=Mgh

The inertia of the meterstick had to be determined using the parallel axis theorem,

I=Icm+M(d)^2

Our calculations are shown below

As we can see, our final height is theoretically 0.131. Which is about 16 percent off of our experimental value.

This is a large discrepency, but unfortunately we had no time to go back and re-do the experement. We suspect that this error is due to a scaling issue when we plotted the data from our video.

blog 14

The goal of this lab is to determine the moment of inertia of a triangle with different orientations.

To do this, we again use a frictionless spinning disk apparatus as we did in our previous moment of inertia experiment. This time, however, we attach a small metal triangle to the top of our spinning object as shown below.

Before we ran an experement to determine moment of inertia of our triangle, we first derived its theoretical moment of inertia as shown below

We arrive at a value for inertia of the triangle about its center of mass

with the 0.15 m side down= 0.184

and with the other side down =0.098

To determine a value experimentally, we used the F=ma and T=I(alpha) forms of newtons second law as shown above to determine an equation for center of mass as a function of alpha. Running the experiment on the first orientation of the triangle, we get a value of I of

I=0.1843

which is only about 1 percent off of our predictied value. Due to time constraints, we were unable to test the other orientation of the triangle experimentally, However from our first result it seems probable that our model is fairly accurate, with the small discrepancy between our expected and acutual values probably due to the small amount of friction in the pulley and spinning mass apparatus.

blog 13

Moment of inetia 

In this lab, we had a large disk shaped object with two small cylinders attached on either side of its center of mass, pivoted on a (presumably) frictionless axis so that it spins vertically. Our goal was to figure our the acceleration of a cart attached to a string wrapped around this object which is released to run down a fictionless ramp.

Pictured, our apparatus without cart attached

Since we were going to be using torque=m(alpha) forms of newtons second law, it becomes immediately necessary to calculate the moment of inertia of our spinning object. We are given the mass of the object, but we cannot use a simple 1/2mr^2 equation to determine I, because of the small side cylinders.  The calculation is shown below

M=4.615

dimentions of large disk

R=0.101m

H=0.015m

V1=(pi)R^2(H)= 4.809*10^-4

And of each small cylinder

R=0.0157

H=0.0516

V2=(pi)R^2(H)= 3.647*10^-5

Total volume of the system=Vt=V1+2(V2)

Therefore, mass of each component (where M1 is the large cylinder and M2/M3 are the small side cylinders):

M1=M*(V1/Vt)=3.915

M2 and M3=M*(V2/Vt)=0.35

From these we can determine the moment of inertia for the system

moment of inertia system = moment of inertia large cylinder-2(moment of inertia small cylinder)

I=1/2(M1)R1^2+2(1/2M2R2^2)

I = 2.008*10^-2

Using this value we can now approximate alpha by plugging into an appropriate equation

ma=sin(x)mg-t   t=m(g-a)   t=m(sin(x)g-alpha(r))

t(r)=I(alpha) 

a=r(alpha)

mr(sin(x)g-alpha(r))=I(alpha)

mgrsin(x)=alpha(I+mr^2)

alpha=mgrsin(x)/(I+mr^2)

plugging our values for m, g, r and I, we get the value

alpha=-1.253 rad/s^2

Now that we have a theoretical baseline, we must run the experiment to see how close we came in our prediction. We run the experiment and timed the result. Giving us a value for alpha of

alpha=-1.266

This value is only 1.04 percent off of our predicted value, Which means that our model is fairly accurate. The small amount of error is probably due to small errors in measuring the dimensions of our spinning apparatus, and the fact that we treated the ramp as frictionless.

blog 12

Rotation 

In this lab, we have a system which pumps air between two metal disks, allowing the one on top to rotate with nearly 0 friction. We set up a system with a string wrapped around this top disk, which then goes over a pulley to a hanging mass, as pictured below Our goal was to determine how certain variables within this system affect the angular acceleration of the spinning disk.

We run 6 trails during this experiment. The first three use a top disk (the one that will spin) made of steel, with a pulley of equal diameter, but increasing weight of the hanging mass (0.0245, 0.0495, 0.0995 respectively). For the next two experiments, we use the same hanging mass and a larger diameter pulley, but change the material of spinning disk from steel in trail 4 to aluminum in trial 5. Finally, in trial 6, we use 2 steel disks stuck together as our rotating object. In all cases we measured the acceleration of the system by way of a motion detector mounted below the hanging mass. When the mass reached the bottom of its string, the inertia in the disk cause it to start to rise, so in our graphs for acceleration we ended up with positive and negative values for acceleration. We chose to call these a up and a down, and averaged them in the following equation.

a_avg=((a_up^2)^1/2+(a_down^2)^1/2)/2

We then equated a_avg to average angular acceleration by the equation

alpha = a_avg/r

where alpha is angular acceleration and r is the radius.

In order to figure our how angular acceleration was theoretically going to be affected in each situation, we had to derive an equation for angular acceleration. To do this, we used T=I(alpha)

and F=Ma equations as shown below.

T=I(alpha)

T=t(r)

I=1/2mR^2

Ma= Mg-t

a=alpha(r)

from these we get

Mr(alpha)= Mg-t

t=M(g-alpha(r))

Mr(g-alpha(r))=I(alpha)

Mrg=alpha(I+Mr^2)

alpha=Mrg/(I+Mr^2)

From this, we can see that in theory, the larger the hanging mass the greater the alpha, and the higher the inertia of the spinning mass the lower the alpha. Finally, the larger the r the greater the alpha. 

Below we have the data from our 6 trials

As we can see, our predictions hold out fairly well. Trail 1, 2 and 3 show increasing alphas and increasing mass. Trials 4 and 5 show that the lower moment of inertia spinning disk of the 2 (aluminium) has a larger alpha than the steel disk trial. In addition, when comparing trails 1 and 4, the only difference is the size of the radius of the pulley, and the trail with the larger pulley (trial 4) has a higher alpha. Finally, in trail 6, with the highest moment of inertia for any trial and the lowest hanging weight, we have our lowest value for alpha.