Deriving a mathematical model for a stick falling over

  • #1
janneman
2
0
TL;DR Summary
build a mathematical model that describes the motion of a stick falling over
this is how far i have come with my model, i am trying to first the most simple model, meaning no friction involved and then testing that against an actual stick falling by using tracking software. I am currently stuck as my model still has an acceleration in the y direction that i cannot seem to get rid off. i am trying to model it only in terms of mass gravitational acceleration length and the angular acceleration and velocity. Could one use relative motion analyses to get rid of the acceleration in the y?
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  • #2
The rotation and the motion in the y direction are dependent on each other. If you can specify the constraint, then you can eliminate one of them.
 
  • #3
what constraint would let me elimate the acceleration in the y?
 
  • #4
janneman said:
what constraint would let me elimate the acceleration in the y?
The dependence of ##y## as a function of ##\theta##.
 

FAQ: Deriving a mathematical model for a stick falling over

What are the basic assumptions for deriving a mathematical model for a stick falling over?

The basic assumptions typically include that the stick is a rigid body with uniform mass distribution, it starts from a vertical position, and there is no slipping at the pivot point. Additionally, air resistance is often neglected, and the stick is assumed to fall under the influence of gravity only.

What equations govern the motion of the falling stick?

The motion of the falling stick can be described using the equations of rotational dynamics. The primary equation is the rotational analog of Newton's second law: \( \tau = I \alpha \), where \( \tau \) is the torque, \( I \) is the moment of inertia of the stick about the pivot point, and \( \alpha \) is the angular acceleration. For a stick of length \( L \) and mass \( m \), the moment of inertia about the pivot point is \( I = \frac{1}{3}mL^2 \).

How do you determine the torque acting on the stick?

The torque \( \tau \) acting on the stick is due to the gravitational force acting at the center of mass of the stick. If the stick makes an angle \( \theta \) with the vertical, the torque is given by \( \tau = mg \frac{L}{2} \sin(\theta) \), where \( g \) is the acceleration due to gravity.

What is the differential equation for the angle of the stick as it falls?

Combining the expressions for torque and the moment of inertia, the differential equation governing the angle \( \theta \) of the stick as it falls is \( \frac{d^2 \theta}{dt^2} = \frac{3g}{2L} \sin(\theta) \). This is a nonlinear second-order differential equation that describes the angular motion of the falling stick.

How can the differential equation be solved?

The differential equation \( \frac{d^2 \theta}{dt^2} = \frac{3g}{2L} \sin(\theta) \) is generally solved using numerical methods due to its nonlinearity. Analytical solutions are possible for small angles using the small-angle approximation \( \sin(\theta) \approx \theta \), which simplifies the equation to a linear form \( \frac{d^2 \theta}{dt^2} = \frac{3g}{2L} \theta \). However, for larger angles, numerical integration techniques like the Runge-Kutta method are typically employed.

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