How Long Does It Take for a Pencil to Fall?

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    Falling Pendulum
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Homework Help Overview

The problem involves a pencil balanced on its point, with the objective of determining the time it takes for it to fall. The scenario assumes the pencil behaves as a massless rod with all mass concentrated at the tip, and the small angle approximation is applied.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the dynamics of the pencil as an inverted pendulum and explore the implications of the small angle approximation. There are attempts to derive the time of fall using equations of motion and integration, but some express confusion regarding the validity of the assumptions made, particularly when considering larger angles.

Discussion Status

The discussion is ongoing, with participants sharing their reasoning and questioning the assumptions of the problem. Some have provided hints and insights into the nature of the small angle approximation, while others express uncertainty about its applicability in this context.

Contextual Notes

Participants note the challenge of reconciling the small angle approximation with the requirement for the pencil to fall to an angle of π/2. There is also mention of the problem being presented in a homework context, which may impose certain constraints on the assumptions used.

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Homework Statement


A pencil of length l = 0.2 m is balanced on its point. How much time does it take to fall? Assume that the pencil is a massless rod and all of its mass is at the tip. To make the math easier, assume the small angle approximation.


Homework Equations



theta (double dot) = g/l sin theta

T = 2 pi (l/g) ^ (1/2)


The Attempt at a Solution



Since the pencil seems to be an inverted pendulum and the fall would be only 1/4 of the period, I first tried T/4 = pi/2 (.2/9.8) ^ (1/2). But that is not a correct solution.

Next I tried to rearrange and integrate

1/sin theta d2theta = g/l dt2

1/sin theta = sin theta/ sin2theta= sin theta/(1 - cos2theta)

let u = cos theta so du=- sin theta d theta

Integral( 1/ u2 - 1) du = 1/2 ln(u - 1) - 1/2 ln (u + 1)

[ 1/2 ln (cos theta -1) - 1/2 ln (cos theta + 1)] d theta = gt/l dt

I get lost here because the cos pi/2 is zero which makes the above meaningless.

The way I interpret the problem, all the mass is at the balancing point. So how can you use energy to calculate potential difference or KE?

If someone could just give a hint as to the approach, I would be greatful.

Thank you.
 
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if the whole pencil can fall then θ will be π/2 how can the 'small angle approximation' hold?
 
Arkavo said:
if the whole pencil can fall then θ will be π/2 how can the 'small angle approximation' hold?

Sorry, I agree, but that was the problem that I was given.

I think as in a lot of physics problems the assumption is made to make a good approximation.

As in how to design an automatic chicken plucker: First you assume a perfectly symmetrical, spherical chicken...
 
The small angle approximation helps you to get started. ##\theta(0)=0## doesn't. :smile:
 
Answer to pencil problem

Attached is the answer for your falling pencil pleasure. Hope this helps someone.
 

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