physical pendulum physics question?

Question

A rod suspended on its end and acting as a physical pendulum swings with a period of 2.26 s. What is its length? (g = 9.80 m/s^2)

Answer 1

time period of physical pendulum = 2pi x square root (effective length / g ),, where effective length =( K^2 + L^2 ) / L
K is radius of gyration of rod for given point of rotation.
let the length of the pendulum be L, therefore moment of inertia of rod about point of suspension is = (mL^2 )/3, therefor radius of gyration K = L/ square root of 3.
on putting these values we get effective length = 4L/3,
T = 2pi squareroot( 4L/3g).
2.26 = 2pi squareroot ( 4l/3g),
on solving we get L = 0.970 meters.

Answer 2

Good question!


I think the period is
T=2pi sqrt(I/(mgR))

I - moment of inertia = (1/12) mL^2
L - length of the rod
R - distance from pivot to center of mass
m - mass of the pendulum
g- acceleration due to gravity= 9.80 m/s^2
pi = 3.14...
R=L/2

we have

T= 2 pi sqrt ( mL^2/( 12 m g L/2))
T=2 pi sqrt(L/(6g))

L= 6g(2T/2pi)^2
L=(3/2)g(T/ pi)^2
L=(3/2)x 9.80 x(2.26 / 3.14...)^2=
L=7.6 m