# Theoretical Acceleration of an Atwood machine

• andrewdavis23
In summary, the descending mass of the pulley system (atwood machine) is 60g and the ascending mass is 55g. DataStudio was used to measure the slope of the velocity over time (read by the spinning spokes of the pulley) and the experimentally given acceleration was found to be 0.316m/s^2. The theoretical acceleration was then calculated using the equation g=a (m1+m2) / (m2-m1). The incorrect equation used in the first post was 9.8m/s^2 while the correct equation was 0.8m/s^2. If it is not right, the equation was originally g=a (m1+m2)
andrewdavis23

## Homework Statement

The descending mass of the pulley system (atwood machine) is 60g and the ascending mass is 55g. Using DataStudio to measured the slope of the velocity over time (read by the spinning spokes of the pulley) the acceleration (constant) is experimentally given as 0.316m/s^2. Now they want to know the theoretical acceleration so you can find the % error.

## Homework Equations

They ask you to solve for (m2-m1)g in the same data table in the lab book. (m2-m1)g = (65-55)9.8 = 49N

## The Attempt at a Solution

I used f=ma and other equations and I keep on getting 9.8m/s^2 as the acceleration. This makes sense to me because the objects are acted on by gravity, but the main confusion is: WHY IS THE ACCELERATION MEASURED AS 0.316m/s^2 and not 9.8m/s^2?

andrewdavis23 said:
I used f=ma and other equations and I keep on getting 9.8m/s^2 as the acceleration. This makes sense to me because the objects are acted on by gravity, but the main confusion is: WHY IS THE ACCELERATION MEASURED AS 0.316m/s^2 and not 9.8m/s^2?
9.8 m/s^2 is the acceleration of a body in freefall--where gravity is the only force acting. But the masses in an Atwood's machine are not freely falling--they constrained by strings.

Show how you derived the acceleration.

I actually tried again, figuring m2a2=m2g-m1g, plugged in the numbers 60g*a2=60g * 9.8m/s^2 -55g * 9.8m/s^2. I solved for a2 and got 0.8m/s^2, does that seem right?

If its not I originally used the equation g = a (m1+m2) / (m2-m1) and solved for acceleration and got 9.8m/s^2 and I'm sure that, that equation is meant for the atwood machine.

andrewdavis23 said:
I actually tried again, figuring m2a2=m2g-m1g, plugged in the numbers 60g*a2=60g * 9.8m/s^2 -55g * 9.8m/s^2. I solved for a2 and got 0.8m/s^2, does that seem right?
No.

If its not I originally used the equation g = a (m1+m2) / (m2-m1) and solved for acceleration and got 9.8m/s^2 and I'm sure that, that equation is meant for the atwood machine.
That's the correct equation, but the acceleration is a, not g. g is a constant! (Which is equal to 9.8 m/s^2, of course.)

I tried the correct equation and got 0.8m/s^2 for a (inputing g=9.8m/s^2). This has to be right then...so thank you very much for the help!

andrewdavis23 said:
I tried the correct equation and got 0.8m/s^2 for a (inputing g=9.8m/s^2). This has to be right then...so thank you very much for the help!
No, not right.

In post #3 you had two equations. The first one, which I think you used to get your answer, was not correct. The second equation is the correct one. Use it to solve for the acceleration.

I think I see what was throwing me off. We wrote down m2=15g and m1=10g, but we also had a base weight on each piece of 55g (to "slow down the system"). So when I go to put it in the equation for acceleration [ a=((m2-m1)g)/(m2+m1) ] I need to make m2=(15g+55g) and m1=(10g+55g) because while the difference of masses (m2-m1) is still 5g, the sum of the masses (m2+m1) changes from 25g to, the correct; 135g which is what was making my theoretical acceleration so different from the experimentally record acceleration.

## 1. What is an Atwood machine?

An Atwood machine is a simple mechanical device used to demonstrate the principles of acceleration and forces in physics. It consists of two masses connected by a string or rope, with one mass hanging over a pulley. The device was invented by British mathematician George Atwood in the 18th century.

## 2. How does an Atwood machine work?

In an Atwood machine, the two masses are connected by a string or rope that passes over a pulley. When one mass is released, it begins to accelerate downwards due to the force of gravity. This causes the other mass to accelerate upwards, as the string is pulled over the pulley. The acceleration of the two masses is related by the ratio of their masses.

## 3. What is the theoretical acceleration of an Atwood machine?

Theoretical acceleration refers to the predicted or expected acceleration of an Atwood machine, based on the known values of the masses and the force of gravity. It can be calculated using the formula a = (m1 - m2)g / (m1 + m2), where m1 and m2 are the masses of the two objects and g is the acceleration due to gravity (9.8 m/s²).

## 4. How does the theoretical acceleration differ from the actual acceleration of an Atwood machine?

The theoretical acceleration is based on ideal conditions, where there is no friction or air resistance. In reality, these factors will affect the actual acceleration of an Atwood machine, causing it to deviate from the theoretical value. The actual acceleration can be measured through experiments or simulations.

## 5. How is the Atwood machine used in real-world applications?

The Atwood machine is commonly used in physics classrooms to demonstrate the principles of acceleration and forces. However, it also has real-world applications, such as in elevators and cranes, where it is used to lift and lower objects using pulleys and counterweights. It is also used in some scientific experiments and research studies.

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