## Physics Olympiads Series Problem 15

- Category: Olympiads
- Written by fisikastudycenter

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}/_{12}ML^{2}, now we use point P as the pivot, there is ^{1}/_{2}L position shifting from the center of mass.

Using parallel axis theorema, apply to point P as a pivot of rotation, subtitute ^{1}/_{2}L 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}/_{2}L from the eart/floor, called poisition I. (KE = 0), when the bar reaches the floor, called position II (PE = 0).

PE_{1} + KE_{1} = PE_{2} + KE_{2}

Mg(^{1}/_{2}L) + 0 = 0 + ^{1}/_{2} I ω^{2}

Subtitute the inertia of the bar, finding the result

Notes:

Potential Energy PE = mgh

Kinetic Energy KE_{rotation} = ^{1}/_{2} I ω^{2}

Prepared by fisikastudycenter.com

**Literatures**

-Seri buku schaum edisi 8, Frederick J. Bueche

-Marthen Kanginan, Fisika SMA 2A, 2B