Question on pendulum

Question

2-1. A 1-meter-long non-massless rod is suspended about a pivot at one end and has a small sphere attached at the other, forming a physical pendulum. The rod and sphere have the equal masses. What is the angular frequency of oscillation?

(a) 2.37 rad/s     (b) 3.16 rad/s       (c) 3.35 rad/s       (d) 3.87 rad/s     (e) 4.00 rad/s

Do you treat the mass as a point particle? Is it a sphere? If it is a sphere do we neglect the radius? How do you solve?

Answer 1

Hi Yogita S.,

Two questions in one day! You treat the sphere as a point mass. You should be able to correctly average the center of mass from that. This is not a complicated case, the center of mass determines the moment of inertia, since everything is in the same direction -- unlike cases with counterweighted pendulums, i.e. a musical metronome.

And you don't have to worry about your initial coefficient (e.g. the (1/3) in (1/3)mL^2 [integrated as (L^2 * dm from 0 to L] when considering a rod rotating at one end, or the (1/12) when you chop your tibia in two for rotation in the middle), since it's just a point mass swinging at the "end" of the appropriate rod..

-- Cheers, -- Mr. d.