Falling pencil (inverted pendulum)

In summary, the problem involves a pencil of length 0.2 m being balanced on its point, with all of its mass at the tip. The goal is to find the time it takes for the pencil to fall, assuming the small angle approximation. After attempting a solution using the inverted pendulum and integrating, it is concluded that the small angle approximation may not hold in this case. However, it is still used as an approximation to solve the problem. The final answer is attached for reference.
  • #1
rogeralms
19
0

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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  • #2
if the whole pencil can fall then θ will be π/2 how can the 'small angle approximation' hold?
 
  • #3
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...
 
  • #4
The small angle approximation helps you to get started. ##\theta(0)=0## doesn't. :smile:
 
  • #5
Answer to pencil problem

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

Attachments

  • ans_pencil.docx
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Related to Falling pencil (inverted pendulum)

What is a falling pencil (inverted pendulum)?

A falling pencil, also known as an inverted pendulum, is a physical system that consists of a pencil or stick attached to a pivot point or string. When released from an upright position, the pencil will fall and swing back and forth like a pendulum.

What factors affect the motion of a falling pencil (inverted pendulum)?

The motion of a falling pencil is affected by various factors, including the length and weight of the pencil, the angle it is released from, the force of gravity, and any external forces acting on it, such as air resistance.

What is the equation of motion for a falling pencil (inverted pendulum)?

The equation of motion for a falling pencil is given by the simple pendulum equation: θ''(t) + (g/L)sin(θ(t)) = 0, where θ is the angle of the pencil from the vertical, g is the acceleration due to gravity, and L is the length of the pencil.

How can the motion of a falling pencil (inverted pendulum) be controlled?

The motion of a falling pencil can be controlled by adjusting the initial conditions, such as the angle of release or the length of the pencil. External forces, such as damping or feedback control, can also be applied to control the motion.

What are the real-world applications of a falling pencil (inverted pendulum)?

The inverted pendulum is a classic example in dynamics and control systems and has various real-world applications, including balancing robots, self-balancing scooters, and stabilizing systems such as segways and drones.

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