Difference between revisions of "Forest UCM NLM Oscilations"

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Note: because the skateboard is frictionless, its wheels are not going to turn.
 
Note: because the skateboard is frictionless, its wheels are not going to turn.
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Determine its equation of motion
  
 
==Step 1: System==
 
==Step 1: System==

Revision as of 12:34, 31 August 2014

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 R and the skateboard has a mass m.

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 m 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

Fg=mgcosθˆrmgsinθˆϕ
N=Nˆr

Step 5: apply Netwon's 2nd Law

Fext=Fg+N=m(¨rr˙ϕ2)ˆr+(2˙r˙ϕ+r¨ϕ)ˆϕ

For the case of circular motion at constant r=R,˙r=0

Fg+N=m(R˙ϕ2ˆr+R¨ϕˆϕ)

The r-hat direction

mgcosθN=mR˙ϕ2
N=m(gcosθ+R˙ϕ2)
N=mgcosθ+mac
ac=v2R=R˙ϕ2= centripetal acceleration

The phi-hat direction

mgsinθ=mR¨ϕ
¨ϕ=gRsinθ
small angle approximation

If we release the skateboard close to the bottom

Then

sinθθ

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

¨ϕ=gRθ

where

ω=gR= oscillation frequency

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

θ=Acos(ωt)+Bsin(ωt)
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

θ(t=0)=θ0=Acos(0)+Bsin(0)=A

and

˙θ(t=0)=0=Aωsin(0)+Bcos(0)=B

thus the specific solution is

ϕ(t)=ϕ0cos(ωt)

Forest_UCM_NLM#Oscillatiions