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]R[/math] and the skateboard has a mass [math]m[/math].
Note: because the skateboard is frictionless, its wheels are not going to turn.
Step 1: System
The skateboard of mass [math]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.
300 px
Step 3: Free Body Diagram
Step 4: External Force vectors
- [math]\vec{F}_g = -mg \cos \theta \hat{r} - mg \sin \theta \hat{\phi}[/math]
- [math]\vec{N} = N \hat{r}[/math]
Step 5: apply Netwon's 2nd Law
- [math]\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] r=R, \dot{r} = 0[/math]
- [math]\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]mg \cos \theta - N = -m R\dot{\phi}^2[/math]
- [math]N = m \left ( g \cos \theta + R\dot{\phi}^2 \right )[/math]
- [math]N = mg \cos \theta + ma_c[/math]
- [math]a_c = \frac{v^2}{R} = R\dot{\phi}^2 =[/math] centripetal acceleration
The phi-hat direction
- [math]-mg \sin \theta = mR \ddot \phi[/math]
- [math]\ddot \phi = - \frac{g}{R} \sin \theta[/math]
- small angle approximation
If we release the skateboard close to the bottom
Then
- [math]\sin \theta \approx \theta[/math]
Forest_UCM_NLM#Oscillatiions
