Moment of inertia and Atwood Machine

Click For Summary
SUMMARY

The discussion focuses on calculating the moment of inertia of a pulley in an Atwood Machine experiment. The slope of the linear fit from the graph of weight difference versus acceleration is given as 5.801 g/m/s². The equation for moment of inertia is provided as I = 1/2mr², but the radius of the pulley remains unknown, complicating the calculation. Two trials were conducted with specific mass values and corresponding accelerations, highlighting the need for the radius to connect linear and angular motion.

PREREQUISITES
  • Understanding of Atwood Machine dynamics
  • Familiarity with linear regression and slope calculations
  • Knowledge of moment of inertia and its formula I = 1/2mr²
  • Basic physics concepts related to acceleration and forces
NEXT STEPS
  • Research the relationship between linear acceleration and angular acceleration in rotational systems
  • Learn how to measure the radius of a pulley accurately in experimental setups
  • Explore the implications of frictionless assumptions in physics experiments
  • Study the derivation of moment of inertia for different shapes and configurations
USEFUL FOR

Students and educators in physics, particularly those involved in mechanics experiments, as well as anyone interested in understanding the dynamics of Atwood Machines and rotational motion.

doub
Messages
15
Reaction score
0

Homework Statement



The question is from a lab. The lab used an Atwood Machine with two masses suspended string looped over a pulley. The time for one mass to reach the ground was measured and acceleration of the mass was calculated. The results were plotted on a graph of weight difference (Y) vs acceleration (x).

The question is from the linear fit, slope value, calculate the moment of inertia of the pulley.

Linear Fit equation y=mx+b

m(slope): 6.762 g/m/s^2


Homework Equations



I= 1/2mr^2

The Attempt at a Solution



Many attempts at trying to determine the moment of interia from the slope, however I get lost because the radius of the pulley was not measured in the lab. The assumption is made that pulley it self in frictionless.

Thanks in advance
 
Physics news on Phys.org
doub said:

Homework Statement



The question is from a lab. The lab used an Atwood Machine with two masses suspended string looped over a pulley. The time for one mass to reach the ground was measured and acceleration of the mass was calculated. The results were plotted on a graph of weight difference (Y) vs acceleration (x).

The question is from the linear fit, slope value, calculate the moment of inertia of the pulley.

Linear Fit equation y=mx+b

m(slope): 6.762 g/m/s^2


Homework Equations



I= 1/2mr^2

The Attempt at a Solution



Many attempts at trying to determine the moment of inertia from the slope, however I get lost because the radius of the pulley was not measured in the lab. The assumption is made that pulley it self in frictionless.

Thanks in advance
Hello doub. Welcome to PF !

Do you know the equation of motion for Atwood Machine ?

You need the two mass values as well as the radius of the pulley. The radius of the pulley gives the connection between the acceleration of the two masses and the angular acceleration of the pulley.

See the Wikipedia article for an Atwood Machine: http://en.wikipedia.org/wiki/Atwood%27s_machine
 
Last edited:
I know the two mass values, however we do not know the radius of the pulley.

In the first trial the m1 mass is 122.59 g and the m2 mass is 113.46 g with the accelerations calculated as 0.07 m/s^2

Trial 2 m1 mass 124.34 g m2 mass 111.71 g with the acceleration calculated as 0.21 m/s^2

And a correction on the slope. The slope value s 5.801 g/m/s^2

I'm not sure if that clarifies things
 
doub said:
I know the two mass values, however we do not know the radius of the pulley.

In the first trial the m1 mass is 122.59 g and the m2 mass is 113.46 g with the accelerations calculated as 0.07 m/s^2

Trial 2 m1 mass 124.34 g m2 mass 111.71 g with the acceleration calculated as 0.21 m/s^2

And a correction on the slope. The slope value s 5.801 g/m/s^2

I'm not sure if that clarifies things
The best that you can do is to fine I/r2 .

You'll have two values for that. One from each trial.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
4K
Replies
2
Views
2K
Replies
3
Views
3K
  • · Replies 13 ·
Replies
13
Views
4K
  • · Replies 28 ·
Replies
28
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
Replies
8
Views
14K
  • · Replies 30 ·
2
Replies
30
Views
4K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 40 ·
2
Replies
40
Views
6K