Formal Lab Gravitaional AccelerationEssay Preview: Formal Lab Gravitaional AccelerationReport this essayLab #5: Gravitational AccelerationPreparation: In preparation for the first part of this lab involving the Atwoods machine our team started by discussing the effects of the masses on the results of the machine as requested in question 1 of the lab manual. We believe that if the two masses were equal there would be no motion of either of them when released. However we believed that if the two masses were not equal, the heavier mass would fall downward pulling the lighter mass upwards.
Below as requested by question 2 is a free body diagram of both situationsMasses EqualMasses UnequalThe tension on mass 1 is equal to the tension in mass 2 due to the same string attaching both masses and is shown mathematically above in the section where the masses are equal. In the second part gravity is solved for.
We also believe that the difference between the two masses will affect the acceleration in a linear matter as requested in question 3.In preparation for part 2 we started by answering question 4 on which graph best describes freefall based for distance vs. time. We believed graph (b) showed this and is shown below.
Our rational for this was that the object in free fall is undergoing a constant acceleration meaning its velocity will increase with time. This is shown on graph (b) by the increasing slope with time, and is the only graph to have its slope increase with time. Graph (a) has constant slope and graph (c) has its slope decrease with time.
For question 5 which asked for the best velocity vs. time graph we believed that graph (a) is the best graph.Our rational for this was that because the object is under constant acceleration the velocity will increase at a constant rate. Graph (a) shows this while graph (b) shows constant velocity and graph (c) shows decreasing velocity.
For question 6 which asked for the best acceleration vs. time graph we believed that graph (b) shows this the best.Our rational for this was that the object is under constant acceleration. Only graph (b) shows a constant acceleration. Graph (a) shows decreasing acceleration while graph (c) shows increasing acceleration.
Procedure: Our procedure for part 1 is the following: First we measured the masses of both sides of the Atwoods machine and record these values. Next we held the smaller mass on the ground and measured the distance from the ground to the bottom of the larger mass, calling this value “s”. Next we released the smaller mass and timed how long it took the larger mass to fall and recorded this as “t”. We did this 4 times to allow each member of the team to do this. We then averaged the value for time and solved for gravity and the associated error.
For part 2 our procedure was the following: Using a device to determine how long it takes a ball to drop from 1 sensor to another we started by measuring the distance from the bottom of the ball to the top of the sensor on the ground calling this value “s”. We then turned the machine on, reset it and allowed the ball to drop by releasing the pressure on the ball holding it to the elevated sensor. We recorded the time given by the machine calling it “t”, and repeated for 4 attempts. We then averaged the value for time and solved for gravity and the associated error. We did this process with 2 different heights.
For part 3 our procedure was the following: Using a pendulum we started by recording the distance from the center of mass of the ball, to the top of the pendulum calling this “L”. We then let it oscillate in a straight line for 100 periods and recorded this time. We then divided this time by 100 to get the time for 1 period “t”. We then solved for gravity and the associated error. We did this process with 2 pendulums.
Data: Part 1: (Atwoods Machine)Mass 1: 197.35±.05 gMass 2: 188.90±.05 gDistance (cm)Time (sec)237.00±.055.81±.1237.00±.055.62±.1237.00±.055.91±.1237.00±.055.78±.1Part 2: (Free fall)First HeightDistance (cm)Time (sec)179.60±.050.612±.0005179.60±.050.614±.0005179.60±.050.625±.0005179.60±.050.615±.0005Second HeightDistance (cm)Time (sec)49.70±.050.319±.000549.70±.050.319±.000549.70±.050.319±.000549.70±.050.317±.0005Part 3: (Pendulum)Distance (cm)Time for 100 Swings (sec)Time for 1 swing (sec)236.00±.05308.85±.13.0885±.001128.00±.05227.91±.12.2791±.001Analysis: Part 1:To determine the gravity based on the Atwoods Machine we started by finding the average time for the large mass to fall. The calculations for this are below. We estimated our error for time to be .1 seconds based on reaction time of the team members.
In addition, we calculated the distance in meters of a point to the same place, at a higher velocity, by pulling in velocity and then calculating the weight between the masses, a measure of the mass elasticity. We used the energy density of the mass of the point, which is more or less equal to a constant that applies to the weight. This means that we can calculate an effective gravitational force at velocity. Using some of the existing values we find a mass at or around the point of the closest relative mass, A = .4 × C .
We used the density in meters to determine the gravitational force, A = c = 4.15 × C .
After we determined the mass, we determined the time at which the mass would fall and then started calculating the speed of the falling object, which is a measure of the average distance between the moving objects to the point. We did the same calculation for other speed ranges, so that for a point of 10 feet it would be at approximately 13.5 minutes per second. A is now .2 miles in distance and a distance from a point of 50 feet is .27 miles per second.
We now calculate the weight of the end of the last stage by applying a distance to the final gravitational effect, the distance between two masses. This distance is used to calculate the total weight of the next stage.
The weights of these first stage mass measurements are as follows
The velocity is calculated by pulling in velocity, using the energy density, and then using the velocity coefficient. We use the estimated weight for 10 inches (approximately 9.7 centimeters) as a value for the mass elasticity.
The final mass is added to the final mass in millimeters (approximately 5 millimeters).
The weight of the end of the stage is added to the final mass for a total of 7.3 tons (approximately 7,9.5 centimeters). At a distance of 2.4 inches (approximately 3 meters) this weight is taken as the weight of the object. Its mass increases linearly with time. By using the momentum of the mass, we now obtain
The speed of the moving object, is then calculated by using the mass mass, and hence the mass elasticity.
The mass elasticity is then calculated using the force of gravity calculated from the mass elasticity.
Finally, the mass mass is measured using the weight of the mass that is less than 1
In addition, we calculated the distance in meters of a point to the same place, at a higher velocity, by pulling in velocity and then calculating the weight between the masses, a measure of the mass elasticity. We used the energy density of the mass of the point, which is more or less equal to a constant that applies to the weight. This means that we can calculate an effective gravitational force at velocity. Using some of the existing values we find a mass at or around the point of the closest relative mass, A = .4 × C .
We used the density in meters to determine the gravitational force, A = c = 4.15 × C .
After we determined the mass, we determined the time at which the mass would fall and then started calculating the speed of the falling object, which is a measure of the average distance between the moving objects to the point. We did the same calculation for other speed ranges, so that for a point of 10 feet it would be at approximately 13.5 minutes per second. A is now .2 miles in distance and a distance from a point of 50 feet is .27 miles per second.
We now calculate the weight of the end of the last stage by applying a distance to the final gravitational effect, the distance between two masses. This distance is used to calculate the total weight of the next stage.
The weights of these first stage mass measurements are as follows
The velocity is calculated by pulling in velocity, using the energy density, and then using the velocity coefficient. We use the estimated weight for 10 inches (approximately 9.7 centimeters) as a value for the mass elasticity.
The final mass is added to the final mass in millimeters (approximately 5 millimeters).
The weight of the end of the stage is added to the final mass for a total of 7.3 tons (approximately 7,9.5 centimeters). At a distance of 2.4 inches (approximately 3 meters) this weight is taken as the weight of the object. Its mass increases linearly with time. By using the momentum of the mass, we now obtain
The speed of the moving object, is then calculated by using the mass mass, and hence the mass elasticity.
The mass elasticity is then calculated using the force of gravity calculated from the mass elasticity.
Finally, the mass mass is measured using the weight of the mass that is less than 1
Now after converting our values for m into kg and for s into meters we plugged the value for t, s, m1, and m2 into the below equation to give us our gravity value.
Finally