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

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Homework Help Overview

The problem involves a physical pendulum consisting of a uniform stick with specified dimensions and mass, pivoting about a point. The objective is to determine the period of oscillation given the distance from the pivot to the center of gravity.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply a formula for the period of a pendulum but realizes it may not be suitable for a physical pendulum. They express uncertainty about how to account for the stick's dimensions in calculating the moment of inertia.
  • Some participants suggest determining the moment of inertia and the distance from the center of mass to the pivot point, indicating a need for a different approach.
  • There is a question regarding how the width of the stick affects the moment of inertia, with attempts to apply a formula that may not be appropriate for the given dimensions.

Discussion Status

The discussion is ongoing, with participants exploring different aspects of the problem. Some guidance has been provided regarding the need to calculate the moment of inertia and the correct formula for the period of a physical pendulum. However, there is no explicit consensus on the approach to take, and further clarification is needed regarding the implications of the stick's width.

Contextual Notes

Participants are navigating the complexities of applying the correct formulas for a physical pendulum, with specific attention to the dimensions of the stick and the pivot point. There is an acknowledgment of the limitations of certain formulas in this context.

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

ehild
 

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