Thread: atwood machine, physics

Imagine a pulley rigged with a separate mass on each side connected by a string.
m1 = 2.00 kg and m2 = 8.00 kg
The masses of the pulley and string are negligible by comparison. The pulley turns without friction and the string does not stretch. The lighter object is released with a sharp push that sets it into motion at vi = 2.80 m/s downward.

How far will m1 descend below its initial level?

I am not quite sure how to conclude solving this.

I have the tension of the string = (2*16*9.8)/(10) = 31.36
I have the acceleration = (+/-) (8-2)/(2+8)*g = 3/5*g = (+/-) 5.88

I just do not know how to go from here.

I know Vf^2=Vi^2+2*a*dy = Vf^2=2.8^2+2*(-5.88)*dy

dy is my answer, but then how do I get Vf^2 from what I know?

the velocity function is 5.88t-2.8
the position function is 2.94t^2-2.8t

K. I got it. VF^2=0

0-2.8^2=2*5.88*dy

-7.84=11.76dy

dy=-7.84/11.76

dy= -2/3

Originally Posted by thedoge

Imagine a pulley rigged with a separate mass on each side connected by a string.
m1 = 2.00 kg and m2 = 8.00 kg
The masses of the pulley and string are negligible by comparison. The pulley turns without friction and the string does not stretch. The lighter object is released with a sharp push that sets it into motion at vi = 2.80 m/s downward.

How far will m1 descend below its initial level?

I am not quite sure how to conclude solving this.

I have the tension of the string = (2*16*9.8)/(10) = 31.36
I have the acceleration = (+/-) (8-2)/(2+8)*g = 3/5*g = (+/-) 5.88

I just do not know how to go from here.

I know Vf^2=Vi^2+2*a*dy = Vf^2=2.8^2+2*(-5.88)*dy

dy is my answer, but then how do I get Vf^2 from what I know?

the velocity function is 5.88t-2.8
the position function is 2.94t^2-2.8t

K. I got it. VF^2=0

0-2.8^2=2*5.88*dy

-7.84=11.76dy

dy=-7.84/11.76

dy= -2/3

You are correct in your answers except for two things: significant digits (2/3 is NOT the same as 0.667) and:

Pet Peeve Warning!! Danger! Danger!

What the heck happened to the all the units in your answers??

(Pet peeve not over, to be continued indefinitely.)

I'll start this from the beginning, since it looks like you are applying final formulas to get your answer.

I have a diagram that puts the 2 kg mass on the left side of the pulley and the 8 kg mass on the right. To analyze this note that the pulley is massless and frictionless and we have an ideal string, so neither really influences the problem except that the pulley changes the direction of the tension in the string. (And the tension along an ideal string is the same at all points.)

So. Free-Body Diagram on the 2 kg mass (m1). We have a weight (w1) acting downward and a tension (T) acting upward. Since we know that the other mass is larger than this one we can state that the acceleration of this mass is upward. So I am going to go with my usual convention and put the +y direction in upward: in the direction of the acceleration of the object. So Newton's 2nd says:


or


For the FBD on the 8 kg mass (m2), we have a weight (w2) acting downward and a tension (T) acting upward. Since the two masses are connected by an ideal string both tensions will be equal and so will the accelerations. BUT the acceleration on this mass is now downward. To be consistent with the first FBD I am going to define +y to be downward here, else we have to play with an extra negative sign. So Newton's 2nd says:
<-- Note the effect of the change in positive direction here.

or


So we have a system of equations for a and T:



We may solve these in any way we wish, substitution being the method most likely to be used by a Physicist, and obtain:


(The appropriate directions of these are given by the FBDs.)

So I get that the magnitude of the acceleration is:


The magnitude of the tension is:
(To the correct number of sig. digs.)
Now, we know the acceleration of the 2.00 kg mass is upward and constant, and the initial velocity of it is 2.80 m/s downward. In the coordinate system of the FBD for the 2.00 kg mass (+y is upward), thus we use a = 5.88 m/s^2 upward (ie. use a as a positive number) and v0 = 2.80 m/s downward (ie. use v0 as a negative number.)

So we want the lowest distance this mass goes. At this point v = 0 m/s. Thus:
v^2 = v_0^2 + 2a(y - y_0) <-- LaTeX won't do this for some reason.
where y represents the lowest position of the mass and v = 0 m/s. I will set the 0 point of the coordinate system to be where the mass starts at t = 0 s.







So this is a very long message stating your answers are correct, except for ignoring the units on your answers.

-Dan

Wow, thank you for the input Dan. You put a lot of work into it and only because I was too lazy to carry my units through and use sig figs.

As this is web based work AND a one dimensional problem, I tend to disregard extensive carrying of units until the end. There was no conversion of units that I needed to track. The final answer indeed was -2/3, but you were right that the correct answer set to sig figs was -.667, which I input as my final answer. I merely adjusted for the correct number of sig figs once I had found my answer.

The reason I do this is because if I convert things such as tension or acceleration to 3 sig figs and use that number to continue my calculations, the answer is often WRONG when inputed into the homework input box online. This is why I do not convert until the end.

Though I should get in the habit of expressing units and numbers to the correct extent. Thank you

I admit my comments were a bit peevish , but I have also found that the best way to ensure you put the units on the final answer is to always make yourself do it. As for the sig digs, yes, I too usually keep all the digits until the final answer (or at least as many digits as my calculator is willing to carry.) This cuts down on rounding errors.

-Dan