3
$\begingroup$

In one formula given below, if the given radius $r$ is larger, $F$ would be lesser. In another formula, if the given radius $r$ is larger, $F$ would be greater.

How is this possible? Intuitively, if we increase the radius of the circular path, we would need lesser centripetal force to keep it on the path.

$$F=\frac{mv^2}{r}=m\omega^2r$$

$\endgroup$
6
$\begingroup$

The angular speed $\omega$ tells you how often you complete a trip around the circle. If $\omega = 2\pi / \text{second}$, then this means that you complete one trip around the circle ($2\pi$ radians) each second. Now imagine the circle has a very small radius. Then on each trip around the circle I don't travel very far, so I'm not going very fast. On the other hand, if the radius is large, then for each trip around the circle I have gone a bigger distance so I'm going fast. Therefore, for a given angular speed $\omega$, my actual speed $v$ increases as the radius of the circle increases.

Linear speed is distance/time, so we can write

$$v = \frac{\text{distance}}{\text{time}} = \frac{\text{circumference of circle}}{\text{time to make one trip around circle}} = \frac{2\pi r}{2\pi /\omega} = \omega r .$$

Now we see that your equations say the same thing: $$\frac{m v^2}{r} = \frac{m (\omega r)^2}{r} = m \omega^2 r .$$

Now let's understand this intuitively. Consider the equation $F = m\omega^2 r$. You said it's confusing that the force gets larger as $r$ increases because bigger $r$ means less tight of a curve. That's true. However, we need to realize that when we talk about increasing $r$ we're implicitly assuming that we're keeping $\omega$ constant. If $r$ is going up but $\omega$ is staying the same, then we are increasing our speed because the length of the path around the circle is getting bigger but we're taking the same amount of time to complete the trip. So by increasing $r$ we are simultaneously lessening the tightness of the curve and increasing our speed. Which factor dominates?

Start with $F = mv^2 / r$. If I increase $r$ by a factor of $n$ but keep the angular speed the same, then I must have increased $v$ by a factor of $n$. However, since $v$ is squared the net effect is to raise the force by a factor of $n$, as shown here:

$$F = \frac{mv^2}{r} \rightarrow \frac{m (nv)^2}{nr} = nF$$

This shows that, for constant angular speed, the force increases linearly with $r$, which is compatible with the equation $F=m\omega^2 r$.

$\endgroup$
  • $\begingroup$ thanks.. so a variable's placement in the formula may not always reflect its actual relation to the function? $\endgroup$ – KawaiKx Sep 24 '14 at 8:58
  • $\begingroup$ @Saurabh: I'm not sure what you mean by that. The placement of $r$ in $F = m\omega^2 r$ definitely affects how it relates to $F$. That formula says "if you increase $r$ while keeping $m$ and $\omega$ fixed, then you increase $F$". The other formula, $F=mv^2 / r$, says "If you increase $r$ while keeping $m$ and $v$ fixed, you decrease $F$". Both of those statements are true. $\endgroup$ – DanielSank Sep 24 '14 at 17:03
1
$\begingroup$

You are right. But, please consider that v = ωr, meaning that for a constant linear velocity v, if r increases ω drops, and in the equation ω is squared.

$\endgroup$
  • $\begingroup$ I did a sum and found it to be true.. it is counter intuitive though. $\endgroup$ – KawaiKx Sep 24 '14 at 8:56
1
$\begingroup$

Both arguments are correct. It all depends on the variables involved. For one to keep the angular speed constant, the linear speed must increase which must then require a greater centripetal force to sustain. If on the other hand the linear speed is to be kept constant, it implies the increase in r is accompanied by a decrease in v hence a lesser centripetal force to sustain the motion

$\endgroup$

protected by Community Mar 7 '18 at 14:34

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.