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The method with measuring the instantaneous weight while swinging the arm with a fixed angular velocity sounds rather impractical. You might try to combine data from multiple sources. I assumed that you are talking about the bit of limb from elbow up to and including the hand, with the hand stretched, but it turns out that you wish to exclude the hand. The ...

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There is no need to calculate the moment of inertia I, which could be quite difficult for an irregular lamina. You can find both I and g from the same experiment. The period T of small amplitude oscillations of a compound pendulum is give by (see Source below) : $(\frac{T}{2\pi})^2 = \frac{k^2+h^2}{gh}$ where k is the radius of gyration (related to ...

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Well, this will certainly depend of the shape and mass distribution of your pendulum. For instance, for a pendulum made of a rod and a thin cylindrical disk, the moment of inertia of the rod (about its own symmetry axis) would be $\frac{1}{12}M_rL^2$, and the moment of inertia of the disk $\frac{1}{2}M_sR^2$. If you have a solid sphere instead of the disk, ...

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See explanation in the following article : http://nautil.us/blog/reading-the-tea-leaves-how-particles-can-travel-upstream

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You need to define boiling more accurately. For nucleate boiling it will actually slow the process down as agitation mixes the system. For a full boil less rocking does decrease your energy loss due to sound but thats likely minimal. In reality the agitation will likely decrease the time to a full boil by both dispersing the nucleated regions increasing ...

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