Circular Motion rubber stopper lab

[g4orce]

I have a circular motion lab, in which we spin a rubber stopper attached at one end and a mass at the other end of a string. We calculate the period by the recording the time it takes for 20 revolutions. And we figure out the force Fc.

So we do three different graphs:

one for Fc vs Period
2nd for Radius vs Period
3rd for Mass vs Period


And we get the following eqn's of the line once, notice I have subsituted for x and y from the above variables.

Fc = 3.0381/T^2 - 2.6566
R=1.5962T^2 - 0.0778
m= 0.1031T^2 - 0.0024

The problems is that I have to Write each eqn in terms of T (period) and then write One final eqn that relates Fc, m, r and T. How do i do this?

 

[HallsofIvy]

The first one, "Write each eqn in terms of T" you've already done. The equations you give ARE in terms of T.

Tp do the second one, recognize that what you are saying is that the constants in each equation must depend on the other variables (you don't say it but I assume that each graph was gotten by holding the other variables constant). The crucial point is that R (minus a constant that MIGHT depend on m) and m (minus a constant that MIGHT depend on R) are both proportional to T^2 while F is proportional to 1/T^2.

 

[M98Ranger]

To write each equation in terms of T, you can simply substitute the value of T from the second equation into the first and third equations. This will give you:

Fc = 3.0381/(1.5962T^2 - 0.0778)^2 - 2.6566
m = 0.1031/(1.5962T^2 - 0.0778)^2 - 0.0024

To write one final equation that relates Fc, m, r, and T, you can use the equation for centripetal force, Fc = mω^2r, where ω is the angular velocity. Since we know that ω = 2π/T, we can substitute this into the equation to get:

Fc = (m(2π/T)^2)r

This equation relates all the variables in terms of T, as requested.