Forest UCM NLM Oscilations

Skate boarder in Half pipe

Consider a frictionless skateboard released from the top of a semi-circle (half pipe) and oriented to fall directly towards the bottom. The semi-circle has a radius [math]\displaystyle{ R }[/math] and the skateboard has a mass [math]\displaystyle{ m }[/math].

Note: because the skateboard is frictionless, its wheels are not going to turn.

Determine its equation of motion

Step 1: System

The skateboard of mass [math]\displaystyle{ m }[/math] is the system.

Step 1: Coordinate system

Polar coordinate may be a good coordinate system to use since the skateboard's motion will be along the half circle.

File:Forest UCM NLM Osc Fig1.png

Step 4: External Force vectors

[math]\displaystyle{ \vec{F}_g = -mg \cos \theta \hat{r} - mg \sin \theta \hat{\phi} }[/math]

[math]\displaystyle{ \vec{N} = N \hat{r} }[/math]

Step 5: apply Netwon's 2nd Law

[math]\displaystyle{ \sum \vec{F}_{ext} = \vec{F}_g + \vec{N} = m \left ( \ddot{r} -r\dot{\phi}^2 \right) \hat{r} + \left ( 2\dot{r} \dot{\phi} +r \ddot{\phi} \right ) \hat{\phi} }[/math]

For the case of circular motion at constant [math]\displaystyle{ r=R, \dot{r} = 0 }[/math]

[math]\displaystyle{ \vec{F}_g + \vec{N} = m \left ( -R\dot{\phi}^2 \hat{r} + R \ddot{\phi} \hat{\phi} \right ) }[/math]

The r-hat direction

[math]\displaystyle{ mg \cos \theta - N = -m R\dot{\phi}^2 }[/math]

[math]\displaystyle{ N = m \left ( g \cos \theta + R\dot{\phi}^2 \right ) }[/math]

[math]\displaystyle{ N = mg \cos \theta + ma_c }[/math]

[math]\displaystyle{ a_c = \frac{v^2}{R} = R\dot{\phi}^2 = }[/math] centripetal acceleration

The phi-hat direction

[math]\displaystyle{ -mg \sin \theta = mR \ddot \phi }[/math]

[math]\displaystyle{ \ddot \phi = - \frac{g}{R} \sin \theta }[/math]

small angle approximation

If we release the skateboard close to the bottom

Then

[math]\displaystyle{ \sin \theta \approx \theta }[/math]

This approximation make the differential equation easier to solve as it looks more like the simple harmonic motion typical of a spring.

[math]\displaystyle{ \ddot \phi = - \frac{g}{R} \theta }[/math]

where

[math]\displaystyle{ \omega = \sqrt{\frac{g}{R}} = }[/math] oscillation frequency

The general solution for this second order differential equation may be written as

[math]\displaystyle{ \theta = A \cos (\omega t) + B \sin (\omega t) }[/math]

Theorem

the solutions of an nth order equation contain n independent constants

Once you have a general solution to a differential equation, your next step is to evaluate the constant A and B using constriant given in the problem such as the initial conditions ( t=0) or boundary conditions.

In this problem we can say that at time t=0

[math]\displaystyle{ \theta(t=0) = \theta_0= A \cos (0) + B \sin (0) = A }[/math]

and

[math]\displaystyle{ \dot \theta (t=0) = 0 = - A \omega \sin(0) + B \cos(0) =B }[/math]

thus the specific solution is

[math]\displaystyle{ \phi(t) = \phi_0 \cos ( \omega t) }[/math]

Forest_UCM_NLM#Oscillatiions