Question

This can be done by taking the weight difference (m1−m2)g as the y variable and the acceleration a as the x variable. What is the slope and the y-intercept of the resulting linear equation? what do they represent? and if the y intercept is nonzero why? and what is the inertia calculated from the slope value?

Answer 1

Atwood's Machine

Frictionless case, neglecting pulley mass Equation of Free-body motion for diagram Free-body diagram for m2 Equation of motion for for mi = m-8- =m,a Ca a acceleration For this idealized case the tension T is the same on both sides of the pulley. The acceleration a is the same for both masses. Solving for T gives Substituting T into the equation for m2 gives The equation of motion for the two-mass system is then m m2

If the acceleration is towards m1, then (m1-m2)g = (m1+m2)a

comparing with y= mx + c,

slope, m = (m1+m2)

If you plotted y = ( m 2 − m 1 ) g vs x = a and you get line y = b x + c, you compare that to the relation you are testing which was (assuming typo): ( m 1 − m 2 ) g = ( m 1 + m 2 ) a + I a / R ^2

This is y = ( m 1 + m 2 ) x + I a / R^ 2

y=(m1+m2)x+Ia/R^2 so compare with your y = m x + c

See why the slope was expected to be m = m 1 + m 2

The actual theoretical line should be: y = m x where m = m 1 + m 2 + I / R ^2

moment of inertia I = (m-m1-m2)*R^2