Physics Asked by Aderam on August 18, 2021
I am looking into pendulum physics for a real life application. In short I need to have a reasonable estimate of the horizontal stress applied by a swing mechanism supported by a wooden structure.
I started with an ideal pendulum representation (weightless string + bob of mass $m$, no damping or further complications), although I cannot use a small angle approximation.
From energy conservation I found the following expression for the angular velocity $omega$ as a function of the pendulum angle $theta$, provided that the pendulum starts at rest at angle $theta_0$ and has length $l$:
$$ omega = sqrt{frac{2g}{l}(cos(theta)-cos(theta_0))} $$
So the tension $T$ sustained at any time (i.e. angle) by the structure should be:
$$T = mgcos(theta)+mlomega^2 = mg(3cos(theta)-2cos(theta_0))$$
It boggles me that the tension does not depend on the pendulum length, as the centripetal component is related to the bob tangential velocity while the velocity will clearly be affected by the pendulum length. Something really seems wrong here.
Is the tension as seen by the structure exactly radial ? In which case I could simply take the sine of it to get an estimate of the horizontal stress?
Provided that the derivation is correct, is it consistent with a rigid rod setup rather than a string?
Thanks
If you increase the length and keep $theta_0$ constant, the motion is exactly the same except everything happens more slowly. So why is it surprising that the tension only depends on the angle? In a sense, $l$ and $g$ just define a natural "unit of time" - i.e. the period of the oscillations.
The tension is exactly radial so long as your assumption are exactly correct. Of course none of your assumption are exactly correct in real life.
It depends whether you assume the rigid rod is massless or not. A rigid heavy rod will have both radial and shear forces which vary along its length.
Correct answer by alephzero on August 18, 2021
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