Period of Rotating Pendulum: 2.2m, 9.0cm, 4.5kg

[juggalomike]

Homework Statement

A uniform stick of length L = 2.2 m, width W = 9.0 cm, and mass M = 4.5 kg oscillates as a physical pendulum and pivots about point O as shown in the Figure. What is the period of the pendulum if x, the distance from the pivot point to the center of gravity of the pendulum, is equal to 0.68 m?

Homework Equations

T=2*pi*sqrt(L/g)

The Attempt at a Solution

I am stuck, i know 2*pi*sqrt((1.1+.68)/9.81) is not going to be right because the section of the stick above the pivot needs to be taken into account, I am just not sure how.

 

[ehild]

It is a physical pendulum. The formula for T you cited is not valid for it.

Determine I, the moment of inertia with respet to the pivot point. Find the distance d between the centre of mas and the pivot.

T = 2 pi sqrt[I/(mgd)]

ehild

 

[juggalomike]

It is a physical pendulum. The formula for T you cited is not valid for it.

Determine I, the moment of inertia with respet to the pivot point. Find the distance d between the centre of mas and the pivot.

T = 2 pi sqrt[I/(mgd)]

ehild

ah ok, how does width factor into the inertia? i was going to do
(4.5*(1.1+.68)^2/(2.2*3))+(4.5*(2.2-1.10.68)^2/(2.2*3)) but 1/3mL^2 is for a stick that is infinitly thin, not sure how to compensate for width.

 

[ehild]

It is not infinitely thin. Find a formula somewhere which fits to the picture of this pendulum. (You did not show it)

ehild