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A rectangle made of two uniform bars lengths L, mass A, and two of length W, mass B, oscillates with SHM about an axis at one corner.

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SEE ATTACHMENT #1 for a diagram of the given parameters.

A rectangular frame is made from two thin, uniform bars of length L= 4 m, mass A= 6 kg, and two of length W= 3 m, mass B= 1.5 kg. The frame is placed on a pivot axis at one corner and executes SHM about that axis.
Find the period of this SHM oscillation.

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Solution Summary

The frame of a pivot axis corner is determined. In a step by step solution, the problem is solved.

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Step 1.
You may recall that for a physical pendulum, its period is given by:
(1) T= (2 PI)(sqrt[(Ip)/(M g ho)] in which Ip is the total moment of inertia of the system about the pivot axis, and ho is the translation distance, from the c.m. of the system to the pivot axis and M is the total mass of the system.

Step 2.
You may also recall ...

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