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Suppose a car moves with a constant speed of $20 \text{m/s}$ a quarter of a circle, and completes the quarter in $5$ seconds. One way to calculate the circumference is simply $20 \cdot 5 \cdot 4 = 400 \text{m}$. However, I know that $a=\frac{v^2}{R}$ and the circumference is $2 \pi R$. The acceleration is defined as $\frac{\Delta v}{\Delta t}$ and thus from Pythagorean theorem we get $a = \frac{\sqrt{20^2+20^2}}{5}=4 \sqrt{2} ~~ \text{m/s}^2$. Now to find the circumference we plug our result into $2 \pi \frac{v^2}{a}$, thus the circumference: $2 \pi \frac{20^2}{4 \sqrt{2}} \approx 444.29 \text{m}$ which is clearly wrong. So why the second way gave me a wrong result?

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Acceleration is the derivative of velocity with respect to time, which is not the same as just dividing the difference in velocity by the difference in time.

$$a=\frac{dv}{dt}\neq \frac{\Delta v}{\Delta t}$$

$dv$ and $dt$ can be thought of as infinitely small $\Delta v$ and $\Delta t$. So if your $\Delta v$ and $\Delta t$ are quite small, the result of your calculation will be pretty close to the actual acceleration. But in your case, $\Delta v$ and $\Delta t$ are pretty big. Therefore, your calculation is way of. We can see that our result becomes more accurate if we consider an eighth of a circle:

$$\frac{\Delta v}{\Delta t}=\frac{ \sqrt{(v_{x0}-v_{x1})^2 + (v_{y0}-v_{y1})^2} }{5/2}$$

Now we plug in $v_{x0}=20$, $v_{y0}=0$ and $v_{x1}=v_{y1}=20\cdot\sqrt{\frac{1}{2}}$.

$$\frac{\Delta v}{\Delta t}=2\frac{ \sqrt{20^2(1-\sqrt{\frac{1}{2}})^2 + 20^2(0-\sqrt{\frac{1}{2}})^2} }{5}=40\frac{ \sqrt{1-2\sqrt{\frac{1}{2}} + \frac{1}{2} + \frac{1}{2}} }{5}=8\sqrt{2-\sqrt{2}} \approx a$$

And we use

$$a=\frac{v^2}{R} \Rightarrow R =\frac{v^2}{a}$$

$$C=2\pi R = 2\pi \frac{v^2}{a} \approx 2\pi \frac{20^2}{8\sqrt{2-\sqrt{2}}} \approx 410.46\space m$$

As you can see, this is much closer to the real value of the circumference. This is because we chose a smaller part of the circle, and because of that a smaller $\Delta v$ and $\Delta t$. If you pick smaller and smaller parts of the circle, your value will get closer and closer to the real value. And in the limit of choosing a part of the circle with length 0, it will be the exact value. This is exactly what the '$d$' in $\frac{dv}{dt}$ means.

You can also try choosing a larger part of the circle, and you ranswer will come out a lot worse. In the worst case, if you use the entire circle, you will find that $\Delta v=0$, and therefore $a=0$, and conclude that the circle has an infinite circumference.

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    $\begingroup$ Could the downvoter comment why the downvote? This seems to be a perfectly valid answer to me. $\endgroup$ – Ruslan Oct 7 '13 at 14:39
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Why are you using Pythagoras' theorem? a triangle of two sides $R$ and one side hypotenuse $vt$ is a wrong depiction of $v$. you are essentially tracing a square path for the car with its 4 vertices on a circle of radius $R$. geometrically, $vt$ is the arc length $\theta R$, and shouldnt be chord length $R \sqrt{2-2\cos\theta}$ which is what you produce from using PT.

this perception makes the car seem faster than it actually is, by a factor $\frac{\theta R}{R \sqrt{2-2\cos\theta}}$$\approx1.11$

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  • $\begingroup$ I did it because $\vec{\Delta v} = \vec{v_2} - \vec{v_1} = \vec{v_2} + (-\vec{v_1})$ and $v_1=v_2=20 \text{m/s}$, which gives a right-angled triangle. $\endgroup$ – www Oct 7 '13 at 11:00
  • $\begingroup$ then the vector v→ is no longer tangential to the circle. $\endgroup$ – gregsan Oct 7 '13 at 15:09
  • $\begingroup$ But why should $\vec{\Delta v}$ be tangential to the circle? $\vec{\Delta v}$ is what determines the direction of the vector $\vec{a}$ - the centripetal acceleration. So it will be directed to the center of the circle. $\endgroup$ – www Oct 7 '13 at 15:19
  • $\begingroup$ v is always orthogonal to a for circular motion $\endgroup$ – gregsan Oct 7 '13 at 17:02
  • $\begingroup$ the use of pythagoras' theorem violates this. i even showed that using PT will cause your answer to be 1.11x the proper value. $\endgroup$ – gregsan Oct 7 '13 at 17:08

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