Calculating Speed and Time for a Sliding Disk on an Elevated Air Table

AI Thread Summary
The discussion centers on calculating the speed and time for a disk sliding on an elevated air table at a 5° angle. The user seeks assistance in creating tables for position and average speed based on measurements taken every 0.1 seconds. Initial calculations for average speed between points indicate a velocity of 1.4 m/s for the first interval, leading to an average of 0.7 m/s. However, discrepancies arise when checking final velocities, suggesting that the time interval may not be consistent due to friction factors. The user ultimately expresses gratitude for the assistance received while recognizing the complexities involved in the calculations.
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Hi all!

I'd like to know fi anybody can help me out with this, I just need some info so I can do it by myself.. formulas or something.

I have a disk (looks like a hockey puck) sliding on an air table which is elevated with an angle of 5°. [weight of the disk: 0,558g] Every 0,1s, the disk makes a mark on a sheet of paper.

I need to make a table of the position (y) and time (x) and another one with average speed and time spent between each points (wich is 0,1s) .. anyone know how to do that?

Thanks for the help ;)
- Alex

ps: Sorry, I'm not too use with physics english terms.. :rolleyes:
 
Last edited:
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What is the length of one mark?
 
Last edited:
Werg22 said:
What is the length of one mark?
the length is smaller between each points, here's what it looks like:

Between Points...
[1-2] 7,0 cm
[2-3] 6,2 cm
[3-4] 5,9 cm
[4-5] 5,0 cm
[5-6] 3,9 cm
[6-7] 3,3 cm
[7-8] 2,3 cm

There we go ;) sorry I didn't mention this in the first post
 
So between point 1 and 2, v1=0 and v2=2d+v1/t. Vavg: v2-v1/2.

So I'll only do 1-2:

v1=0
v2=2(0.07)+0/0.1=1.4 m/s
vavg=1.4-0/2=0.7 m/s
 
Last edited:
Werg22 said:
So between point 1 and 2, v1=0 and v2=2d+v1/t. Vavg: v2-v1/2.

So I'll only do 1-2:

v1=0
v2=2(0.07)+0/0.1=1.4 m/s
vavg=1.4-0/2=0.7 m/s
Allright, I got it. Thanks a lot ;)
 
The problem is not true when checking:

vf^2=2(0.07)sin5*g=0.35 m/s
so vf=1.15 m/s

and vavg=0.35+0/2=0.175 m/s

The time isn't 0.1 s unless the coefficient of friction is 16.37 (wich no matter has).
 
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