Rank Velocity of m2 in an Atwood Machine

[kelslee28]

Homework Statement

An Atwood Machine consists of two masses m1 and m2 attached to each other by a single massless string passing over a cylindrical pulley of mass M without slipping. The masses are released from rest, and allowed to accelerate until each has moved a distance h = 1.8 m. Assuming that upwards is positive, rank the velocity of m2 (from most negative to most positive) at this moment for the following cases:
Case A: m1 = 21 kg, m2 = 16 kg, M = 28 kg

Case B: m1 = 13 kg, m2 = 16 kg, M = 31 kg

Case C: m1 = 25 kg, m2 = 16 kg, M = 26 kg

Case D: m1 = 18 kg, m2 = 16 kg, M = 34 kg

Case E: m1 = 17 kg, m2 = 16 kg, M = 33 kg

Homework Equations

sum of the torques = I*alpha
vf^2 = vi^2 + 2a (yf - yi)

The Attempt at a Solution

I did a free body diagram of the cylinder. I did sum of the Torques = I(alpha). I found that the acceleration was equal to
a = [2g(m1+m2)]/M

Then I used a constant acceleration equation Vf^2 = Vi^2 + 2a(yf-yi). I said the initial velocity is 0, I substituted [2g(m1+m2)]/M in for a, I said that yf is the height change (1.8 m) and yi is 0.
I simplified this to say Vf = the square root of [4gh(m1+m2)]/M

I plugged in the values for the examples A through E. I made B a negative velocity because mass 2 is heavier so it would go down.

My ranking was B<D=E<A<C
Thanks for helping :)

 

[anathema]

Thank you for your detailed post and for providing all the necessary information to solve this problem. Your approach using the free body diagram and equations for torque and acceleration is correct. Your ranking of the velocities is also correct. However, I would like to offer some additional explanation for those who may be struggling with this problem.

In an Atwood Machine, the masses are connected by a single string that passes over a pulley without slipping. This means that the string is always taut and both masses move with the same acceleration. In this case, the acceleration of the system can be found by considering the net force on the system, which is equal to the sum of the forces on each mass:

Fnet = T - m1g + m2g

Since the masses are released from rest, the initial velocity is 0 and the final velocity can be found using the equation vf^2 = vi^2 + 2a(yf-yi). Plugging in the value of acceleration found above and the distance traveled (1.8 m), we get:

vf = sqrt[(2g(m1+m2)) / M]

From this, we can see that the final velocity is directly proportional to the square root of the total mass (m1 + m2) and inversely proportional to the mass of the pulley (M). This explains why in cases A, C, D, and E, where the total mass is increasing, the final velocity also increases. In case B, where the total mass is decreasing, the final velocity also decreases.

To rank the velocities from most negative to most positive, we can use the fact that the initial velocity is 0 and the acceleration is always positive. This means that the final velocity will be negative if the mass is heavier (and therefore moves down) and positive if the mass is lighter (and moves up). This explains why case B has the most negative velocity and case E has the most positive velocity.

I hope this additional explanation helps in your understanding of the Atwood Machine. Keep up the good work in your scientific studies!

 

[kerimek]

I would like to confirm that your approach and solution are correct. Your use of the equations for torque and constant acceleration are appropriate for this problem. Your ranking of the velocities is also correct, with the exception of case B, where the velocity of m2 should be more negative than case D. This is because in case B, m2 is the heavier mass and therefore would have a greater acceleration and velocity. Overall, your understanding and application of the concepts are accurate. Great job!