Why is the velocity squared in the formula for centripetal force?

[Oilaye]

The formula for centripetal force is

The force required to keep an object of a constant mass in circular motion is equal to the mass of the object * the velocity of the object squared, then divided by the radius of the circular 'path'. My question is, why would the velocity be squared? I'm having a tough time understanding this. I can work with it and solve problems, but I can't say I truly understand it. I'm not sure how to graph this sort of thing just yet either, as we haven't dealt much in the way of curves yet.

If someone could explain this I would really appreciate it.

Thank you.

 

[wjm11]

The formula for centripetal force is

The force required to keep an object of a constant mass in circular motion is equal to the mass of the object * the velocity of the object squared, then divided by the radius of the circular 'path'. My question is, why would the velocity be squared? I'm having a tough time understanding this. I can work with it and solve problems, but I can't say I truly understand it. I'm not sure how to graph this sort of thing just yet either, as we haven't dealt much in the way of curves yet.

If someone could explain this I would really appreciate it.

Thank you.

I'm not sure what to tell you. It depends on what you already know and accept as true. Do you accept Newton's law that F = ma?

Do you accept that centripetal acceleration is equal to velocity squared divided by the radius of curvature, a = (v^2)/r ?

If so, when you combine those two, you get F = m[(v^2)/r].

Hope that helps.

 

[Oilaye]

I'm not sure what to tell you. It depends on what you already know and accept as true. Do you accept Newton's law that F = ma?

Do you accept that centripetal acceleration is equal to velocity squared divided by the radius of curvature, a = (v^2)/r ?

If so, when you combine those two, you get F = m[(v^2)/r].

Hope that helps.

We never went over those formulas. This one was given as an introduction to combined variation. Maybe I should just go with it for now and question it later. I'm sure I'll see it again, more detailed in the future.

Not understanding something bugs me, but I suppose this was only trying to introduce the idea of the different types of variation; there was no need to understand the logic behind the particular formula. I was just really worried at first is all.

 

[HallsofIvy]

Consider an object moving in a circle of radius R with angular speed $\displaystyle \omega$. Setting up a coordinate system with (0, 0) at the center of the circle, we can write $\displaystyle x= Rcos(\omega t)$, $\displaystyle y= Rsin(\omega t)$. The "position vector" is given by $\displaystyle <R cos(\omega t), R sin(\omega t)>$ . The "velocity vector" is its derivative, $\displaystyle <-R\omega sin(\omega t), R\omega cos(\omega t)>$. The "acceleration vector" is its derivative, \(\displaystyle <-R\omega^2 cos(t), -R\omega^2 sin(t)>= -\omega^2<R cos(\omega t), R sin(\omega t)>, which, because of the negative sign, points in the opposite direction of the position vector, back toward the center of the circle. But the crucial point is that the length of the acceleration vector, the "scalar acceleration", is \(\displaystyle R\omega^2\). From "force equals mass times acceleration" the force is equal to $\displaystyle Rm\omega^2$. Your "v" is the linear speed of the object around the circle- $\displaystyle v= R\omega$ ($\displaystyle \omega$ is the speed in radians per second, say. Then with an angular speed of $\displaystyle \omega$ radians per second, since there are $\displaystyle 2\pi$ radians in a circle, we would go once around the circle in $\displaystyle 2\pi/\omega$ seconds. The circumference of the circle is $\displaystyle 2\pi R$ so we would have gone a distance $\displaystyle 2\pi R$ in $\displaystyle 2\pi/\omega$ seconds for a linear speed of $\displaystyle v= (2\pi R/(2\pi/\omega)= R\omega$. So $\displaystyle \omega= \frac{v}{R}$, $\displaystyle \omega^2= \frac{v^2}{R^2}$ and then $\displaystyle F= Rm\omega^2= Rm\frac{v^2}{R^2}= \frac{mv^2}{R}$.\)

 

[Oilaye]

Consider an object moving in a circle of radius R with angular speed $\displaystyle \omega$. Setting up a coordinate system with (0, 0) at the center of the circle, we can write $\displaystyle x= Rcos(\omega t)$, $\displaystyle y= Rsin(\omega t)$. The "position vector" is given by $\displaystyle <R cos(\omega t), R sin(\omega t)>$ . The "velocity vector" is its derivative, $\displaystyle <-R\omega sin(\omega t), R\omega cos(\omega t)>$. The "acceleration vector" is its derivative, \(\displaystyle <-R\omega^2 cos(t), -R\omega^2 sin(t)>= -\omega^2<R cos(\omega t), R sin(\omega t)>, which, because of the negative sign, points in the opposite direction of the position vector, back toward the center of the circle. But the crucial point is that the length of the acceleration vector, the "scalar acceleration", is \(\displaystyle R\omega^2\). From "force equals mass times acceleration" the force is equal to $\displaystyle Rm\omega^2$. Your "v" is the linear speed of the object around the circle- $\displaystyle v= R\omega$ ($\displaystyle \omega$ is the speed in radians per second, say. Then with an angular speed of $\displaystyle \omega$ radians per second, since there are $\displaystyle 2\pi$ radians in a circle, we would go once around the circle in $\displaystyle 2\pi/\omega$ seconds. The circumference of the circle is $\displaystyle 2\pi R$ so we would have gone a distance $\displaystyle 2\pi R$ in $\displaystyle 2\pi/\omega$ seconds for a linear speed of $\displaystyle v= (2\pi R/(2\pi/\omega)= R\omega$. So $\displaystyle \omega= \frac{v}{R}$, $\displaystyle \omega^2= \frac{v^2}{R^2}$ and then $\displaystyle F= Rm\omega^2= Rm\frac{v^2}{R^2}= \frac{mv^2}{R}$.\)

\(\displaystyle


Yep, this is most definitely beyond the level I'm at right now, haha. I talked with a friend last night. She told me this sort of thing becomes easier to understand once you learn about derivatives and how to 'take apart' the variables. I'm sure I will revisit this in the future; I will learn it then. I do appreciate you taking the time to help though. Thank you. \)