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How can I show that there is a minimum angular velocity $\omega_{min}$, different from zero, such that if we chose an $\omega$ smaller than $\omega_{min}$, then it is not possible to have a circular motion as in the picture?

figure

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  • $\begingroup$ Why does it merit a bounty? It looks straight forward. Are we missing something? Is this a quantum mechanics question, where there is a minimum energy level? $\endgroup$ – mmesser314 Dec 9 '14 at 5:35
  • $\begingroup$ They convinced me they were right!, the correct answer is that there is a minimum $\omega_{min}$, which is the one requires for an angle $\theta=0$ $\endgroup$ – Wolphram jonny Dec 9 '14 at 17:53
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If you solve the problem for the two forces the vertical and the horizontal force (which is required for the circular motion) you obtain the relation

$$\omega^2=\frac{g}{L\cos\theta}$$

Hence the minimum required $\omega$ is $\sqrt{g/l}$, the same as the angular frequency for motion of a bob in a plane.

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  • $\begingroup$ this answer is wrong, why do you think that w is a minimum one instead to that corresponding to circular moption for a given angle? remember that given an angle, there is only one possible omega, there is NOT a range of possible omegas. $\endgroup$ – Wolphram jonny Dec 9 '14 at 4:38
  • $\begingroup$ The minimum occurs when $cos \theta$ equals 1, or $\theta $ equals 0. Which means there is no circular motion. $\endgroup$ – LDC3 Dec 9 '14 at 4:39
  • $\begingroup$ OK, then you agree, there is no minimum for circular motion, just that at theta=0 there is no motion at all! $\endgroup$ – Wolphram jonny Dec 9 '14 at 4:47
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    $\begingroup$ No value smaller than w=√g/L can lead to circular motion for simple reason. $\endgroup$ – SAKhan Dec 9 '14 at 5:07
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Expanding the correct answer of @SAKhan:

Assume that the conical pendulum is rotating at an angle $\theta$ at an angular velocity $\omega$.

Note also that the radius of the circle is given by $$r=L\sin(\theta)$$

For the point mass to move in a horizontal circle, the total vertical force is zero:$$T\cos(\theta)=mg$$

The net horizontal force must supply the needed centripetal force:$$T\sin(\theta)=m\omega^2r=m\omega^2L\sin(\theta)$$Combining these two equations to eliminate $T$, we get:$$\omega^2=\frac{g}{L\cos(\theta)}$$

The maximum value of $\cos(\theta)$ is $1$, when $\theta=0$, so $$\omega_{minimum}=\sqrt{\frac{g}{L}}$$

Edited for clarity:

So, what does this math mean?

Assume for the sake of simplicity that the length, $L$ of the pendulum is $9.8$ meters, Then the equation for $\omega$ reduces to$$\omega = \sqrt{\frac{1}{\cos(\theta)}}$$ Now, we repeatedly start the pendulum into circular motion, each time at a some different angle $\theta$. (This could take some fiddling!) For each of these set-ups, once we are sure that the pendulum is moving in a circle, we measure the angle $\theta$ and the angular velocity $\omega$. This angular velocity can be measured by taking the period of the circular motion, and dividing it into $2\pi$.

If we took all the data and plot them, we would obtain graph that shows that the value of $\omega$ approaches $1$ as $\theta$ approaches $0$. Strictly speaking, $\omega=1$ is a lower limit (rather than a minimum) that is approached asymptotically as the angle $\theta$ approaches zero.

enter image description here

The vertical axis is $\omega$, in radians, and the horizontal axis is $\theta$ in degrees

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  • $\begingroup$ why you conlcude that this be the minimum omega for circular motion instead of conlcusing that it is the omega required for circular motion at angle theta? $\endgroup$ – Wolphram jonny Dec 9 '14 at 4:33
  • $\begingroup$ The minimum occurs when θ equals 0; which means the pendulum is hanging vertically. $\endgroup$ – LDC3 Dec 9 '14 at 4:42
  • $\begingroup$ The graph nicely demonstrates that there is one value of $\omega$ for each value of $\theta$, and that this value tends to a non-zero value when the circular motion becomes very small. I wish you had labeled the vertical axis in units of $\frac{L}{g}$ rather than showing that $\omega$ tends to $1$ which is slightly misleading. Perhaps define $\omega_0=\sqrt{L/g}$ and plot $\frac{\omega}{\omega_0}$ ... $\endgroup$ – Floris Dec 9 '14 at 14:15
  • $\begingroup$ Nicely explained $\endgroup$ – SAKhan Oct 1 '17 at 6:25
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There is such minimun agngular speed: You will always find an an angle that results in circular motion for any given angular speed. The angle is given by:

$$\cos \theta=\frac{g}{L \omega^2}$$

Update: I got to this expression by using the equations of motion:

$m\omega^2L \sin \theta=T\sin\theta$

and

$T\cos\theta=mg$

What this means is that the minimum speed is reached for $\theta=0$, where \omega_min=\sqrt{L/g}. Any other circular motion will require a larger angular velocity. Thus, if we give the pendulum a spees less that the minimum, it will not be able to undergo a circular motion and starts to oscilate.

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  • $\begingroup$ @Floris I updated my answer, are you sure I forgot some force? $\endgroup$ – Wolphram jonny Dec 9 '14 at 3:37
  • $\begingroup$ Have you forgotten that the bob will move like a regular pendulum below its natural frequency? There is a restoring force proportional to $\sin\theta$ when there is no rotation. I believe that sets the lower limit on $\omega$ $\endgroup$ – Floris Dec 9 '14 at 3:41
  • $\begingroup$ Yes, thanks! I missed a sin (theta) when I updated the answer, I'll correct it (but the solution is correct) $\endgroup$ – Wolphram jonny Dec 9 '14 at 3:41
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    $\begingroup$ @Wolphramjonny great! But now your equation contradicts your sentence "There is no such minimun agngular speed" (meaning, what if $g/L\omega^2>1$?) $\endgroup$ – user12029 Dec 9 '14 at 4:33
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    $\begingroup$ But the question is asking the opposite: for a given $\omega$ can you always find an angle? (or rather, the question is presenting the assertion that you cannot, and asking for a proof) $\endgroup$ – David Z Dec 9 '14 at 8:13

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