PHY 1032 (General Physics Lab I)

Purpose

To compare the moments of inertia calculated using two different methods, and to verify that angular momentum is conserved in an interaction between a rotating disk and a ring dropped onto the disk.

Hypothesis

If a weighted ring is added to the disk, the moment of inertia will be the same as the disk without the weighted ring. The angular momentum before the ring is dropped on the disk during part two will be greater than the angular momentum after the ring is dropped.

Labeled Diagrams

See attached sheet.

Data

Part 1

Mass of disk (M): 1.500 kg
Radius of disk (R): 0.114 m
Radius of shaft (r): 0.006 m
Mass of ring (m): 1.420 kg
Inner radius of ring (R₁): 0.054 m
Outer radius of ring (R₂): 0.064 m

Disk Alone

Disk plus ring

Part 2

Graphs

Part One (Disk Alone)

Part One (Disk Plus Ring)

Part Two

Questions

Part 1

1. In your data table in Part 1, you have two values for the moment of inertia. One is found from the theoretical equation for moment of inertia that is introduced in the Theory section and other is an experimental value obtained using Newton’s 2^nd law for rotational motion, τ = Iα, in conjunction with the definition of torque, τ =rF. How well do your two values agree with each other? What is the percent difference? Which do you think is likely a better way to calculate a value for moment of inertia?

The values are extremely close, as the percent difference for the disk alone is 9.5% and the disk plus the ring is 0% (they are of equal value). I think the better way to calculate the moment of inertia is to use I = ½ MR², as it is a more elegant equations that takes into account less variables. The other equation takes more variables into account, mainly for calculation torque, which I feel leads to increased error.

Part 2

1. How do your values for the angular momentum before and after the ring is dropped onto the disk compare? What is the percent difference?

The angular momentum before the ring is dropped onto the disk is greater than the angular momentum after the ring is dropped onto the disk. The percent difference is 11.7%.

2. Does there appear to be an inverse relationship between moment of inertia and angular velocity?

No, there appears to be a direct relationship between moment of inertia and angular velocity. As the angular velocity decreased, so did the moment of inertia.

3. How well do your results support the theory of conservation of momentum? What are the limitations of the experimental setup?

The results somewhat support the theory of conservation of momentum. The percent difference is 11.7%, which I suppose isn’t a huge discrepancy, but it could be better. The limitations of the experimental setup were that it is difficult to drop the ring on the spinning disk perfectly. We were able to drop the ring into the grooves of the disk, but there was still some wiggle room in those grooves. The ring would need to fit in the grooves like a puzzle piece in order to be positioned dead center to yield the least amount of error.

Conclusion

Lab Summarized

During the first part of the lab, the moments of inertia for a spinning disk with and without a weighted ring on top were calculated using two different methods. The force coercing the disk to spin was a 300 g weight attached to the shaft of the disk using a string a pulley system. The weight was allowed to free fall and the resulting graph of velocity versus time was used to find the final angular velocity by taking the mean of the segment after which the string had completely unraveled from the shaft. The angular acceleration was found from the slope of this graph up to that point.

These values, along with the force of kinetic friction, found by determining the minimum force needed to get the disk spinning, were used to find the moment of inertia. The moment of inertia was also calculated a second way, using the radii and masses of the disk and ring.

The second part of the experiment was performed much like part one of the experiment using the disk alone, only this time shortly after the string had unraveled, the ring was dropped onto the spinning disk. Using the angular velocities and moment of inertias, determined much like they were in part one, the angular moments before and after the ring were dropped were calculated and compared.

The percent differences between the two different calculations of the moments of inertia in part one were quite low. Using the disk alone, the percent difference was 9.5% and with the disk plus the ring, the percent difference as 0%. Under perfect conditions, the values should have been equal. The angular acceleration and final velocity for the disk along were greater than that of the measurement for the disk plus the ring, which could be expected, due to the extra mass. The percent difference between the angular momentums before and after the ring was dropped in part two was 11.7%. They should have been equal under ideal conditions.

As stated previously in the questions, some of this error is most likely due from the ring not being placed dead center around the spinning disk. If the disk was dropped at and angle and did not make complete contact with the disk at the same instant, this could have also caused error. If the shaft was not properly lubricated, this would have caused error throughout the experiment. Lastly, if the string ever caught a snag while unraveling, this would have also contributed to the error.

Equations

I =τ/α
I = ½ MR2
L = I_fω_f
I_iω_i = I_fω_f

Purpose

To utilize two different methods of determining the initial velocity of a fired ball, namely a ballistic pendulum and treating the ball as a projectile, and then compare these two calculated values. The loss of kinetic energy from firing the ball into the pendulum is also an area of interest.

Hypothesis

The initial velocity determined by firing the ball into the ballistic pendulum should theoretically be equal to the initial velocity determined by firing the ball as a projectile.

Labeled Diagrams

See attached sheet.

Data

Part One

v= 6.23 m/s α_v=0.0187 m/s

Part Two

Y=1.003 m v= 8.75 m/s α_v=0.120 m/s

Questions

1. Compare the two different values of v average. Calculate the percent difference between them. State whether the two measurements agree within the combined standard errors of the two values of v average.

The average initial velocity for the ballistic pendulum was 6.23 m/s while the average initial velocity for the projectile determination was 8.75 m/s. This is a percent difference of 33.6%. It should have been expected that these two values would be equal. The two measurements also do not agree within the combined standard errors of the two values for v average, as the standard errors only total 0.1387 m/s, and the average velocities fall out of that range.

2. Calculate the loss in kinetic energy when the ball collides with the pendulum as the difference between ½ mv² (the kinetic energy before) and ½ (m + M)V² (the kinetic energy immediately after the collision). What is the fractional loss in kinetic energy? Calculate by dividing the loss by the original kinetic energy.

The average kinetic energy before the collision is 1.35 J and the average kinetic energy immediately after the collision is 0.272 J, so the loss of kinetic energy is 1.08 J. The fractional loss in kinetic energy is 0.8.

3. Calculate the ratio M / (m + M) for the values of m and M in Part 1. Compare this ratio with the ratio calculated in the previous question. Express the fractional loss of kinetic energy in symbol form and use equations from the Theory section to show it should equal M / (m + M).

The ratio M / (m + M) is equal to 0.8. This ratio is exactly the same as the fractional loss of kinetic energy.

The fractional loss of kinetic energy equals ( ½ mv² – ½ (m + M)V² ) / ( ½ mv² ).

Conclusion

During part one of the experiment, a ball was fired into a ballistic pendulum to ultimately determine its initial velocity. This process was repeated five times in order to obtain average values to work with in order to eliminate error. By massing the ball and the pendulum, recording the initial and final heights, the values for V and finally v could be calculated. It was found that the average initial velocity of the ball was 6.23 m/s.

During part two of the experiment, the same ball was fired as a projectile instead of into a ballistic pendulum. The ball was fired from a table horizontally to the ground. A piece of carbon paper was used to capture the spot where the ball first struck the ground. Height and horizontal distance the ball traveled were then measured in order to determine the initial velocity of the ball. The average initial velocity of the ball was 8.75 m/s.

As far as the accuracy of the results from the lab, the percent difference between the average velocities calculated is 33.6%. This is a fairly significant difference, which suggests that there sources of error during the procedure. The notched part of the ballistic setup could have had finer groves to yield more accurate measurements. The major contributor of error, however, was most likely from the distance measurements from the projectile part of the lab. One positive to come from the results was that the fractional loss in kinetic energy was identical to the mass ratios from the ballistic pendulum setup, which is theoretically expected.

Equations

∆KE = ½ mv²

½ (m + M)V² = (m + M)gh

mv = (m + M)V

V = (2gh)^0.5 v = (m + M) (2gh)^0.5 / m v = ∆x / (2∆y / g)^0.5

Purpose

To determine the relationship between force, mass, and acceleration using a cart attached to a pulley with varying weights.

Hypothesis

If the mass of the weights attached to the pulley is increased, the force exerted on the cart and the acceleration of the cart will also increase.

Labeled Diagrams

See attached sheet.

Data

Graphs

See attached sheets.

Questions

1. Is the graph of force vs. acceleration for the cart a straight line? If so, what is the value of the slope?

Yes, the graph produces nearly a straight line; the correlation for a linear fit is 0.9787. The value of the slope is 0.6129 N/(m/s^2).

2. What are the units of the slope of force vs. acceleration graph? Simplify the units of the slope to fundamental units (m, kg, s). What does the slope represent?

The units of the slope of force vs. acceleration are N/(m/s^2). This simplifies to kg. The slope represents the mass of the pulley.

3. What is the total mass of the system (both with and without extra weight) that you measured?

The total mass of the system without the extra weight was 0.6759 kg and the mass of the system with the extra weight (300 g) was 0.9916 kg, which seems to make sense. 0.9916 kg is almost exactly 300 g more than 0.6759 kg.

4. How does the slope of your graph compare (percent difference) with the total mass of the system that you measured?

The slope of the graph without any added weights was 0.6129 kg, which is a 9.78% difference. The slope of the graph with the added weights was 0.8939 kg, which is a 10.36% difference.

5. Are the net force on an object and the acceleration of the object directly proportional?

Yes, as the net force is increased, the acceleration is also increased.

6. Write a general equation that relates all three variables: force, mass, and acceleration.

F = ma

Conclusion

Lab Summarized

The overall goal of the lab was to determine and show the relationship between force, mass, and acceleration. The goal was achieved using a cart and pulley system with varying weights to measure force and acceleration. The forces and accelerations collected were then graphed against each other the construct a linear fit line, whose slope showed the mass of the system (the cart, sensor, and any added weights). This value could then be compared to the mass calculated from the force of the free hanging system. The force and acceleration from each trial run could also be analyzed to show any relationship between the two values.

The data collected seemed to show a direct correlation between force and acceleration. Thus, the stated hypothesis was confirmed that if the force was increased, the acceleration would also increase. The compared values for the masses of the cart systems were about 10% different in each case. For the trial without any added weight, the calculated value of 0.6759 kg is 9.78% different from the extrapolated value of 0.6129 kg. In regards to the trial with the added weight, the calculated value of 0.9916 kg is 10.36% different than the extrapolated value of 0.8939 kg. This error could have been caused by a number of factors. For instance, the air resistance from the weight on the pulley dropping could have caused error, and any possible friction from the track could have attributed to this, too. It could also be thought that if the pulley did not drop straight downward, i.e. it was swaying at all, this would have further error. Lastly, if the rope was not completely taught when the system was put in motion, that could have caused error as well.

Equations

F = ma

a = 9.8 m/s^2

N/(m/s^2) = kg