0
$\begingroup$

Using the formula $F=mv^2/r$ I have rearranged the formula and found that the radius is proprtional to the velocity squared. I then found a website that affirms the relationship that I found. On the website, it says "As you increase the radius you square the speed." I just don't understand how that works though. As in, if you increase the radius from 1m to 1.1m, how are you squaring the velocity (if the other factors are kept constant). Could someone plase show me an example where as the radius increases, the velocity squares?

Question: Is it true that as the radius increses, the velocity increases (as r∝v^2)?

Improve this question
$\endgroup$
6
  • 2
    $\begingroup$ This got a close vote for "unclear what you are asking". The thing that is unclear is the sentence that confused the OP. His question is quite clear: he can't make any sense out of a sentence that is senseless to start with. $\endgroup$
    – garyp
    Commented Oct 12, 2015 at 14:24
  • $\begingroup$ You are misunderstanding the background of the statement, taking it out of context. That statement is in response to a question about a controlled experiment. Go back and spend some time reading the experiment and I believe your confusion will go away. It is dangerous to take statements out of context. $\endgroup$
    – Bill N
    Commented Oct 12, 2015 at 17:50
  • $\begingroup$ I'm voting to close this question as off-topic because the OP has taken the statement out of context. This shows a lack of research. $\endgroup$
    – Bill N
    Commented Oct 12, 2015 at 17:51
  • $\begingroup$ Seems to me that this is the sort of question that we want here: OP is clearly confused about relationships between variables and asked for our help. Voting to reopen. $\endgroup$
    – Kyle Kanos
    Commented Oct 13, 2015 at 10:31
  • $\begingroup$ @BillN I looked at the context again. Context doesn't help. The phrase is simply poorly worded. As it stands it is a statement about increases, which to me suggests a relationship between $\mathrm{d}r$ and $\mathrm{d}v$. $\endgroup$
    – garyp
    Commented Oct 13, 2015 at 15:02

2 Answers 2

Reset to default
2
$\begingroup$

$$F=\frac{mv^2}r\Leftrightarrow r=\frac mF v^2\quad\text{so}\quad r \propto v^2$$

As you say, the proportionality relationship is clear. You can read it like this:

  • If $r$ is doubled, then $v^2$ is doubled (so $v$ is multiplied by $\sqrt{2}$)
  • If $r$ is tripled, then $v^2$ is tripled (so $v$ is multiplied by $\sqrt{3}$)
  • $\cdots$
  • If $r$ is multiplied by any number $k$, then $v^2$ is multiplied by $k$ (so $v$ is multiplied by $\sqrt{k}$).

You can convince yourself of the $\sqrt k$ by testing it - let's multiply $r$ with $k$:

$$k\,r=k\frac mF v^2=\sqrt k^2\frac mF v^2=\frac mF(\sqrt k\,v)^2$$

So if the new radius is $k\,r$, then the new speed is $\sqrt k\,v$.

What your sentence fails to tell you is simply the same thing said in reverse. Instead of considering what happens to $v$ when $r$ is changed, we can consider $r$ when $v$ is changed:

  • If $v$ is doubled, then $r$ is multiplied with 4
  • If $v$ is tripled, then $r$ is multiplied with 9
  • $\cdots$
  • If $v$ is multiplied by any number $k$, then $r$ is multiplied by $k^2$

Same check as before: $$\frac mF (kv)^2=\frac mF k^2v^2=k^2\,r$$

So, $r$ increases quadrativally with $v$. That is what the sentence should have said.

Improve this answer
$\endgroup$
9
  • 1
    $\begingroup$ Thanks for that lovely answer. I have one more question: As when r∝v^2, you say r increases quadrativally with v. What if (example, I know it isn't) r∝1/v^2? How would you decribe it? $\endgroup$
    – J. Doe
    Commented Oct 12, 2015 at 20:50
  • $\begingroup$ @J.Doe Then $r$ is inversely proportional to $v^2$. Then $r$ increases not quadratically but "reversly" quadratically with $v^2$, so to say. So when you multiply $v$ with $k$, $r$ is multiplied with $1/k^2$. $\endgroup$
    – Steeven
    Commented Oct 12, 2015 at 21:32
  • $\begingroup$ Some of my friends have been telling me that as you increase the radius, the velocity actually decreases. But, from my deirvation, I found that r∝v^2, which means that as the radius increases, so does the velocity right? $\endgroup$
    – J. Doe
    Commented Oct 13, 2015 at 9:44
  • $\begingroup$ For $r\propto v^2$, an increase in one means an increase in the other. For $r\propto 1/v^2$ an increase in one means a decrease in the other. What exactly are you confused about? $\endgroup$
    – Steeven
    Commented Oct 13, 2015 at 10:25
  • 1
    $\begingroup$ When you increase the radius the angular velocity does decrease. $\endgroup$
    – garyp
    Commented Oct 13, 2015 at 14:19
1
$\begingroup$

That is a really poorly worded sentence. It should say something like "Radius increases quadratically with velocity, if force and mass are kept constant". In fact, I can't make any sense of that sentence at all.

This would put me on alert that there might be other poorly worded phrases in the document.

update after comment

I can't show you the example the you ask for, because the sentence in question doesn't make any sense.

I think you (the Original Poster) understand correctly what centripetal force is all about: $r =\frac{mv^2}{F}$. The sentence that confuses you (and me) seems to imply that if your initial radius is 1 m, and your initial speed is 2 m/s, and you change the radius to 1.1 m then you should expect the velocity to become 4 m/s.

The wording can also be interpreted as an answer to the question: how does velocity change if I change the radius? Now that's even more confusing since $v = \sqrt\frac{rF}{m}$

Improve this answer
$\endgroup$
4
  • $\begingroup$ I think it's an answer. But I'll reword it to make that clear. $\endgroup$
    – garyp
    Commented Oct 12, 2015 at 14:26
  • $\begingroup$ The sentence "Velocity increases quadratically with radius should be the other way round: "Radius increases quadratically with velocity". $\endgroup$
    – Steeven
    Commented Oct 12, 2015 at 15:00
  • $\begingroup$ Thanks for that help very much. I have one more question: As when r∝v^2, you say r increases quadrativally with v. What if (example, I know it isn't) r∝1/v^2? How would you decribe it? $\endgroup$
    – J. Doe
    Commented Oct 12, 2015 at 20:51
  • $\begingroup$ "$r$ varies as the inverse of $v^2$, or "inversely with $v^2$" $\endgroup$
    – garyp
    Commented Oct 12, 2015 at 21:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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