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physics.fisikastudycenter.com, Olympiads problem series, for beginner level. Problem number 15. Basic concepts: Inertia, circular motion, angular velocity, conservation of energy.

Problem 15
PQ is a uniform rod with mass of M and length of L at rest. PQ could freely rotate at P as a pivot.

If the rod then is released and starts to rotate, find the angular velocity of the rod when reaches the floor, show the answer in ω, g, dan L, where ω is the angular velocity of the rod, and g is the acceleration due to gravity.
Problem source: Rotasi benda tegar, Seri Buku Schaum

Disscussion
First step, find the inertia of the rod at point P as the axis. Use the parallel axis theorema: As we know, the inertia of rod with the axis through the center (perpendicular to length) of mass is I = ¹/₁₂ML², now we use point P as the pivot, there is ¹/₂L position shifting from the center of mass.
Using parallel axis theorema, apply to point P as a pivot of rotation, subtitute ¹/₂L to the x, where x is the shifting of the pivot/axis position, the previous pivot/axis is in the center of mass, then shifts to point P



Next step, use the conservation of energy law, involve the Potential energy and the kinetic energy of the bar.
The center of mass in initial position is ¹/₂L from the eart/floor, called poisition I. (KE = 0), when the bar reaches the floor, called position II (PE = 0).

PE₁ + KE₁ = PE₂ + KE₂
Mg(¹/₂L) + 0 = 0 + ¹/₂ I ω²
Subtitute the inertia of the bar, finding the result

Notes:
Potential Energy PE = mgh
Kinetic Energy KE_rotation = ¹/₂ I ω²

Prepared by fisikastudycenter.com
Literatures
-Seri buku schaum edisi 8, Frederick J. Bueche
-Marthen Kanginan, Fisika SMA 2A, 2B