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I have a pendulum with the following properties:

  1. The diameter of the sphere is 1 meter.
  2. The sphere of the pendulum is made of pure titanium.
  3. The pendulum is connected to the roof with a rigid string 12 meters long. It does not have mass. The union between the string and the sphere is rigid, so it cannot rotate.
  4. The oscillation angle is 60 degrees from the vertical.

We suppose there's no air, so there's no friction.

With this data, I need to know the period of this pendulum. I have not worked with pendulums so I do not know how to start. Any hint/guide?

I know that with 1. and 2. we can obtain the mass of the pendulum, using the density of the Titanium, $d_{titanium}=4506\;kg/m^3$. The volume of the sphere is $V=\frac{4}{3}\pi$, so the mass is $m=d\cdot V=4506\cdot\frac{4}{3}\pi=6803\pi$.

Antonio Gamiz Delgado
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  • The period of the pendulum is independent of its mass. – David G. Stork Nov 11 '20 at 17:18
  • Oh, you're right. Thanks, I will edit the question! – Antonio Gamiz Delgado Nov 11 '20 at 17:19
  • The period is also independent of the initial angle. And independent of any "push" (which is equivalent to a larger starting angle). And your phrase "infinite resistance" makes no sense. – David G. Stork Nov 11 '20 at 17:21
  • Really? Then the period does not depend on any initial condition? Maybe the initial "push"? – Antonio Gamiz Delgado Nov 11 '20 at 17:23
  • The motion may be independent of mass, but without mass, there will be no movement. Nor will there be movement without gravitation, And it will depend on just how much gravitation there is. So the problem is missing a lot of essential data. –  Nov 11 '20 at 17:31
  • @ProfessorVector. The problem provides all relevant information (assuming you know the gravitational constant), and realize you're given that there is some mass— any mass—in the pendulum. Specifically, what do you imagine has been left out???? – David G. Stork Nov 11 '20 at 17:34
  • I have readded the previous data, do you need more information? Gravity is $g=9.807\frac{m}{s^2}$. – Antonio Gamiz Delgado Nov 11 '20 at 17:36
  • And you agree that the period is independent of mass and yet leave all the irrelevant discussion of material, density, volume, and mass!!! And you even leave in the non-sensical phrase "infinite resistance." What does that even mean?!? – David G. Stork Nov 11 '20 at 17:45
  • @David G. Stork LOL. The gravitational constant may be a constant (there are a few differing opinions among physicists), the gravitational acceleration on earth isn't, and we aren't told the pendulum is on earth. So stop trolling, please, would you? –  Nov 11 '20 at 17:49
  • So a newbie who states "I have not worked with pendulums," doesn't know that a pendulum's period is independent of mass and initial angle, and wrote $g = 9.807 \frac{s^2}{m}$ should be concerned that some elite physicists are debating whether $g$ may vary over eons by minuscule amounts, or by $0.7%$ on the earth? This isn't "trolling." It is expert instruction by a PhD physicist who has held faculty positions in elite colleges and universities and written leading textbooks stating facts clearly and directly. Talking of "differing opinions among physicists" about $g$ here seems trolling. – David G. Stork Nov 11 '20 at 18:01
  • @ProfessorVector: what are you trying to do ? Just making fun ? Also review your statements. A photon has no mass but is in motion. In zero-gravity, an pendulum will circle, and this is a movement. –  Nov 11 '20 at 21:54
  • The period does depend on the angle. See https://math.stackexchange.com/questions/2257095/exact-period-of-simple-pendulum?rq=1 For "small" initial angle (amplitude) period is approximately independent of amplitude. – sammy gerbil Nov 11 '20 at 22:12
  • https://en.wikipedia.org/wiki/Gravitational_constant (please note the disclaimer: "Not to be confused with g, the gravity of Earth.") –  Nov 11 '20 at 22:22

1 Answers1

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Let's assume that the gravitational acceleration $g$ is known. You did not specify where the string is connected to the sphere (center or a point on the surface). I will use the center. If it's at the surface, all you need to do is change the distance to the center by adding half the diameter. The moment of inertia of the sphere with respect to the center of the mass is $$I_{CM}=\frac 25 MR^2$$ With respect to the point on the roof around which it oscillates it is $$I=I_{CM}+ML^2=M\left(\frac25 R^2+L^2\right)$$ If you want, you can say that $R^2\ll L^2$. In that case the problem will not depend on the radius of the sphere.

Then using the zero of potential energy at $\theta=0$, where $\theta$ is measured from the vertical axis, you have $$MgL(1-\cos\theta)+\frac12I\dot\theta^2=MgL(1-\cos\theta)+\frac12M\left(\frac25 R^2+L^2\right)\dot\theta^2=MgL(1-\cos\theta_0)$$ $\theta_0$ is the initial angle of $60^\circ$. Calculating $\dot\theta$ from here, you get $$\dot\theta=\sqrt{\frac{2gL}{\frac25 R^2+L^2}(\cos\theta-\cos\theta_0)}$$ Using $\dot\theta=\frac{d\theta}{dt}$, you get $dt=\frac{d\theta}{\dot\theta}$. To get the period$$T=2\int_{-\theta_0}^{\theta_0}\frac{d\theta}{\dot\theta}=2\sqrt{\frac{L^2+\frac25R^2}{2Lg}}\int_{-\theta_0}^{\theta_0}\frac{d\theta}{\sqrt{\cos\theta-\cos\theta_0}}$$

Andrei
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