What is the mathematical proportionality for centripetal acceleration?

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The discussion revolves around determining the mathematical proportionality for centripetal acceleration based on an experimental formula. The initial formula presented is UgM = 4π²rm/T², leading to observations that T is proportional to M^(-0.35), m^(0.53), and r^(0.5). A participant clarifies that T can be expressed as T = Constant * M^(-0.35) * m^(0.53) * r^(0.5), where the constant may depend on other fixed variables. The empirical exponents derived from plotted data suggest a slight discrepancy with the original formula but remain reasonably close. The discussion emphasizes the importance of logarithmic graphs in deducing the constant and understanding the relationships between the variables.
bayan
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Hi everyone!

We have been doing a n experiment about Cent acceleration.

We had a formula which was UgM= 4Pie^2rm/T^2

Then we have T being ptoportional to M, R and m

Can someone help me to figure out the proportionality?

It is to be determined using maths although I have got some result (not using maths, only from our observation of graphs and making the LOG graph to find gradiant..) I have got it to be
T proportional to M^-.35
T proportional to m^.53
T proportional to r^.5


Is there anyone that can help me to fing what the mathematical proportionality would be?

Also any work out would be apriciated to explain.
 
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Your plotted data has given you empirical exponents for the dependence of T on each of three variables. I assume your data was collected by allowing only one of the three to vary at any time. When you say "T proportional to M^-.35" You are saying T is some constant times M^-.35, but that constant may depend on the other two variables that you are holding fixed. Similar reasoning appplies to the other two variables. Putting them all together gives you

T = Constant*M^-.35*m^.53*r^.5

The value of the constant can be deduced from the intercepts of logarithmic graphs you plotted.

Your data does not quite agree with your starting formula, but it is not terribly far off.
 

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