Motion of a freely falling object

AI Thread Summary
The discussion revolves around measuring the displacement of a freely falling ball bearing using strobe light intervals and a meter stick. The user calculated displacement and average velocity but initially confused total displacement with interval displacement. Clarification was provided that displacement should be measured for each time interval rather than cumulatively from the start. The user also sought advice on calculating average acceleration, which resulted in a value significantly lower than the expected gravitational acceleration. The conversation concluded with an acknowledgment of the misunderstanding and gratitude for the clarification.
chom
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Hello everyone, I'm new here and a physics noob. I'm doing Physics 024 correspondence via web and my tutor has a (if I'm lucky) a once a week schedule to talk with him. Thought I might have better luck here since I'm on a strict time frame to get this done.
Here goes...

I measured the displacement for each interval of a ball bearing free falling (in a picture in my textbook) next to a meter stick with a strobe light flashing in 0.050 s intervals. From this measurement I calculated the following data:
100mm (length of ruler on page 77)
I then factored the scale by using: 1000mm/100mm (actual length of meter stick)/(length of meter stick on print) = 10mm scale factor



Time(s) 0.050 0.100 0.150 0.200 0.250 0.300 0.350 0.400 0.450 0.500 0.550 0.600

∆d(mm)displacement
Measured for
interval 0.5 4.0 10.0 18.0 29.0 42.5 59.0 77.5 99.0 123.0 150.0 178.5

∆d (cm)actual
Displacement for
interval .50 4.0 10.0 18.0 29.0 43.0 59.0 78.0 99.0 123.0 150.0 180.0

V (cm/s)actual average velocity
For interval 10.0 80.0 200.0 360.0 580.0 860.0 1200.0 1600.0 2000.0 2500.0 3000.0 3600.0

Sorry, the forum wouldn't let me upload my word doc for word 2007 for a proper data table.

To find actual displacement the calculation I used was ∆d(mm) x 10mm(scale factor) = ∆d(mm), then I converted to cm to get the actual displacement for ∆d(cm).

To find actual average velocity I used the calculation: v=(∆d(cm))/0.050s for each interval in question.

Next, I drew a Velocity-Time graph for the motion, I calculated average acceleration using: a=∆v/∆t,
a=3600/0.600 = 6000 = 6.0x10^3cm/s^2
converted to m/s is 6.0x10^1m/s^2 which is very far from the value of g = 9.8m/s^2.

I am fairly new to all of this and any advice to where my error is would be greatly appreciated.

Chom
 
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Are you sure those displacements over each interval? And not total displacement since t=0?
 
So are you saying I should be measuring the displacement of each interval as 0 to d1, d1 to d2, d2 to d3, etc? Instead of total displacement being 0 to d1, 0 to d2, 0 to d3, etc. Maybe that is what my books means "measure the displacement for each interval of time. I was having trouble understanding exactly what it meant. Please verify. Then as far as the t=0 could you elaborate on that a bit more?
 
chom said:
So are you saying I should be measuring the displacement of each interval as 0 to d1, d1 to d2, d2 to d3, etc? Instead of total displacement being 0 to d1, 0 to d2, 0 to d3, etc. Maybe that is what my books means "measure the displacement for each interval of time. I was having trouble understanding exactly what it meant. Please verify. Then as far as the t=0 could you elaborate on that a bit more?


Yes, the displacement should be over each interval... If d1,d2,d3 are positions... that means they are the displacement since t=0...

ie I was asking if d2 is the displacement from t=0 to t2... or the displacement from t1 to t2...

You should be using displacement from t0 to t1... t1 to t2... t2 to t3... etc... because you're calculating the average velocity over that time period... so you should only use displacement over that time period...
 
Thank you for the help. That total makes sense. For some reason i was still using the method for making a position time graph. Can't believe it took me that long to figure it out. Anyways, thank you very much.

Chom
 
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