# Need Help finding theoretical and actual acceleration (Atwood)

Tags:
1. Oct 14, 2014

### Student7987

So, I have an atwood machine lab. What I have been given is the masses of either end and time it takes the heavier mass to hit the ground from a distance.

M1= .2kg
M2 = .15kg
avg time = 1.12 sec

What i think I have to do: To find the actual acceleration I have to use v = vi + at and solve for acceleration.
To find theoretical acceleration, I have to use a = ( (M2-M1) (g) )/ (M1+M2).
Theoretical: 1.4m/s/s
Actual: .697m/s/s

Kindly tell me if I am doing this right or not and how should I go about doing it correctly. It seems like these values Ive obtained are incorrect and I'm missing something considering I didn't go into the topic we are studying very much. No use of free body diagrams or even sum of forces.
Thanks.

2. Oct 15, 2014

### ShayanJ

One important thing you missed is the uncertainties in your measurements. If they are high enough, you can say those accelerations are in agreement within the uncertainties!
The other point is, you missed air resistance. Also your atwood machine may not be a good one! And many other things that only you, the experimenter, can tell because you know how is the situation around the machine and with the machine itself.

3. Oct 15, 2014

### ehild

Welcome to PF!

I am not sure how you calculated the "experimental" acceleration. What was the height the bigger mass fall from in 1.12 s?

ehild

4. Oct 15, 2014

### Orodruin

Staff Emeritus
What did you use for the velocities here? If you used vi = 0 and the average velocity for v, then your result will be off by a factor of 2 (which seems deceptively similar to what you have).

5. Oct 15, 2014

### Student7987

I'm sorry I forgot to mention a few things. The greater mass was positioned .875 meters off the ground. The second mass, when the greater mass was released traveled those .875 meters up. There is no air resistance as well.
V=vi+at was supposed to be
.781 (distance/time) = 0 +a(1.12)

So what would I use for v if not the average. I don't even think I used the right formula.

6. Oct 15, 2014

### NTW

There's an equation for uniformly accelerated motion that relates acceleration, space, and time... Use it...

7. Oct 15, 2014

### Orodruin

Staff Emeritus
To add to what NTW just said: The equation you quoted is an expression for the velocity at time t while what you were using was the average velocity between 0 and 1.12 s. However, using the expression for the velocity at time t, you could relatively easy deduce the average velocity between two given times (in particular since the velocity in this case depends linearly on the time).

8. Oct 15, 2014

### Student7987

I'm still not sure what equation using space you are talking about though. But from what we were taught in AP Physics 1 (Current class) there was no equation using space. Are there other acceleration equations using distance, time, and mass.

9. Oct 15, 2014

### Student7987

Also I haven't use the sum of forces at all yet but I'm guessing it ties into my theoretical acceleration equation

10. Oct 15, 2014

### Orodruin

Staff Emeritus
Let me again suggest that you think about how the average velocity is related to the final velocity in a system with constant acceleration.