[Math4u] Re: Re: Particle path problem

Hi Rich

let L to be :
L = MAX(ABS(d2x/dt2)), 0 <= t <= t2
find x=x(t) such that L value becomes as low as possible.

Thnaks

Rmath4u2@aol.com wrote:

maximum value of absolute of particle acceleration becomes minimum

maximum value of absolute of particle acceleration becomes minimum  What does this mean?  How can a maximum become a minimum?

Rich B. 

-----Original Message-----
From: Mojtaba Fathi <modjtabaf@yahoo.com>
To: Math4u@yahoogroups.com; moji <mojtaba.fathi@gmail.com>
Sent: Tue, 9 Oct 2007 8:13 am
Subject: [Math4u] Particle path problem

Hi all

Please help me to solve this problem. I faced this when designing a cam and follower mechanism :

Consider a particle moving along x-axis. at the beginning, the particle is in origin and hes no velocity, ie x(0)=0, v(0)=0.
Now, wa want to move particle from it's initial position to a new position so that x(t2)=x2,v(t2)=v2. (t2 > 0 , x2 > 0 and v2 > 0 are given).
Find the equation of motion x=x(t) such that
1. x(t) is continuous within range 0 <= t <= t2
2. v(t)=dx(t)/dt is continuous within range 0 <= t <= t2
3. maximum value of absolute of particle acceleration becomes minimum within range 0 <= t <= t2, i.e. L=max(abs(d2x(t)/dt2) becomes as minimum as possible.

Thanks

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