Solving Atwood Machine: m1 & m2 Reaching Floor & Ascending Height

[DmytriE]

Homework Statement

One mass, m₁ = 215g, of an ideal Atwood machine rests on the floor 1.10m below the other mass, m₂ = 255g. (a) If the masses are released from rest, how long does it take m₂ to reach the floor? (b) How high will mass m₁ ascend from the floor?

Known variables:
m₁ = 215g = 0.215kg = F₁
m₂ = 255g = 0.255kg = F₂
y_d = 1.10m

Homework Equations

F₂ - F₁ = (m₁ + m₂)a
y_d = v₀t + 1/2at²

The Attempt at a Solution

A.
2.5N - 2.1N = (0.255kg + 0.215kg)*a
0.4N = 0.47kg*a
a = 0.851m/s²

1.10m = 1/2*(0.851m/s²)*t²
t = 1.61s

B. I think I have to use equation 'b' in the relevant equations section but I do not know how to use it properly. For this part I have the masses or each object, the acceleration of the entire apparatus, and the different forces. Any insight on this problem would be greatly appreciated!

 

[gneill]

Hi DmytriE.

You might want to keep a few more decimal places in your intermediate values so that rounding errors don't creep into your results.

You may need to think about recruiting some additional equations for part (B). When m2 hits the floor at the time t that you've calculated, what happens to the forces acting on m1? What's m1's state of motion at that instant?

 

[DmytriE]

That's the problem. I don't know what other equations I need to use to figure out this problem. This is what I have:

a = 0.834 m/s²
v₀ = 0 m/s
t = 1.62s
x = 1.10 m

As m₂ is accelerating downward toward the ground m₁ is accelerating upward. It will continue to accelerate for 1.10m (the point where m₂ hits the ground) and then m₁ begins to slow down.

x_m1 = 1.10 + the distance it takes for a = 0.

Maybe I should be looking at a different acceleration for m₁ since it's lighter than m₂? m₁ should accelerate quicker than m₂.


NOTE: I have just graduated college with a BA in biochemistry. This is no longer homework for me but an interest and a learning experience. I will not be getting homework points.

 

[gneill]

That's the problem. I don't know what other equations I need to use to figure out this problem. This is what I have:

a = 0.834 m/s²
v₀ = 0 m/s
t = 1.62s
x = 1.10 m

As m₂ is accelerating downward toward the ground m₁ is accelerating upward. It will continue to accelerate for 1.10m (the point where m₂ hits the ground) and then m₁ begins to slow down.

x_m1 = 1.10 + the distance it takes for a = 0.

Maybe I should be looking at a different acceleration for m₁ since it's lighter than m₂? m₁ should accelerate quicker than m₂.

As long as the Atwood machine's parts are free from external interference they will move together -- the same velocity and the same acceleration. The string will remain tensioned.

When m2 hits the floor, m1 continues its upward journey and the string (presumably the much vaunted "light inextensible string" of physics folklore) will go slack. At this point m1 is behaving as a projectile fired upwards with whatever velocity it had attained immediately before m2 hit the floor. So to go further you need:

1. The height when m1 is "launched" (you have this)
2. The velocity of m1 when it is "launched"
3. The new situation for forces acting on m1 after launch
4. Appropriate projectile motion equations (Hint: conservation of energy would help here)

NOTE: I have just graduated college with a BA in biochemistry. This is no longer homework for me but an interest and a learning experience. I will not be getting homework points.

That's most commendable. Won't change what happens here though!

 

[DmytriE]

I understand gneill. Your post had me look at the problem in an entirely different way. Thanks!

Sometimes a day or two off helps reset the brain.