Uniform circular motion - frequency vs. centripetal force.

[titaniumfever]

I need to find a relationship between the centripetal force and frequency of horizontal circular motion, and then find the theoretical gradient of the graph between log(frequency) vs. log(centripetal force). The radius of the string used was constant at 60cm.

I think that f = k^Fc where k is any constant is an equation, but not completely sure of that.

It isn't helping that the equation above provides me with log(f) = Fc*log(k), when I need to find log(f) as a function of log(Fc)...

 

[Andrew Mason]

I need to find a relationship between the centripetal force and frequency of horizontal circular motion, and then find the theoretical gradient of the graph between log(frequency) vs. log(centripetal force). The radius of the string used was constant at 60cm.

I think that f = k^Fc where k is any constant is an equation, but not completely sure of that.

What is the relationship between centripetal force and tangential speed? How is tangential speed related to angular speed or the frequency of circular motion?

AM

 

[HallsofIvy]

I think that f = k^Fc where k is any constant is an equation, but not completely sure of that.

Well, that certainly is an equation, but it doesn't have anything to do with circular motion!

If an object moves on a circle of radius R, we can write the "position vector" as $Rcos(\omega t)\vec{i}+ Rsin(\omega t)\vec{j}$. Since sine and cosine have period $2\pi$, that will have period T where $\omega T= 2\pi$ or $T= 2\pi/\omega$ and $\omega= 2\pi/T$. The frequency, in radians per second, is $T/2\pi$. With $\vec{r}= R cos(2\pi t/T)\vec{i}+ R sin(2\pi t/T)\vec{j}$, the velocity vector is the derivative, $\vec{v}= -2\pi R/T sin(2\pi t/T)\vec{i}+ 2\pi R/T cos(2\pi t/T)\vec{j}$, and the acceleration is the second deriative, $\vec{a}= -4\pi^2 R/T^2 cos(2\pi t/T)\vec{i}- 4\pi^2 R/T^2 sin(2\pi t/T)\vec{j}$. Since "Force = mass *acceleration", the strength of the force holding the mass in circular motion must be $F= 4m\pi^2 R/T^2$ and, since $f= 2\pi/T$, $F= mRf^2$

 

[Andrew Mason]

Well, that certainly is an equation, but it doesn't have anything to do with circular motion!

If an object moves on a circle of radius R, we can write the "position vector" as $Rcos(\omega t)\vec{i}+ Rsin(\omega t)\vec{j}$. Since sine and cosine have period $2\pi$, that will have period T where $\omega T= 2\pi$ or $T= 2\pi/\omega$ and $\omega= 2\pi/T$. The frequency, in radians per second, is $T/2\pi$. With $\vec{r}= R cos(2\pi t/T)\vec{i}+ R sin(2\pi t/T)\vec{j}$, the velocity vector is the derivative, $\vec{v}= -2\pi R/T sin(2\pi t/T)\vec{i}+ 2\pi R/T cos(2\pi t/T)\vec{j}$, and the acceleration is the second deriative, $\vec{a}= -4\pi^2 R/T^2 cos(2\pi t/T)\vec{i}- 4\pi^2 R/T^2 sin(2\pi t/T)\vec{j}$. Since "Force = mass *acceleration", the strength of the force holding the mass in circular motion must be $F= 4m\pi^2 R/T^2$ and, since $f= 2\pi/T$, $F= mRf^2$

I think it is simpler to use $F = ma = mv^2/r$ where $v = 2\pi r/T = 2\pi r\nu$. It is pretty easy to work out F as a function of frequency $\nu$ from that.

AM