CM and Relative Coordinates; Reduced Mass

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SUMMARY

The discussion focuses on the dynamics of two particles, m1 and m2, connected by a massless spring with natural length L and force constant k. The motion is analyzed using Lagrangian mechanics, yielding the equations of motion for both particles. The final position equations are derived as y1 = (L + (m1/M)*v0*t - 0.5gt^2) + (m2*v0/M*w)sin(wt) and y2 = ((m1/M)*v0*t - 0.5gt^2) - (m1*v0/M*w)sin(wt), where M is the total mass and w is the angular frequency. The analysis separates the motion into the center of mass dynamics and the relative motion of the particles.

PREREQUISITES
  • Understanding of Lagrangian mechanics
  • Familiarity with simple harmonic motion
  • Knowledge of reduced mass concept
  • Basic principles of projectile motion
NEXT STEPS
  • Study Lagrangian mechanics in detail, focusing on applications to multi-body systems
  • Explore the concept of reduced mass in oscillatory systems
  • Learn about the dynamics of coupled oscillators
  • Investigate the effects of initial conditions on the motion of connected particles
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Physics students, mechanical engineers, and anyone interested in classical mechanics and the dynamics of multi-body systems.

Midas_Touch
Two particles of mass m1 and m2 are joined by a massless spring of natural length L and force constant k. Initially, m2 is resting on a table and I am holding m2 vertically above m2 at a height L. At time t=0, I project m2 vertically upward with initial velocity v0. Find the positions of the two masses at any subsequent time (before or either mass returns to the table) and describe the motion.

okay, so i set up the equations and motions and used langrangian equations to yield y = c0+c1t-1/2gt^2

where co and c1 are constants. so for this equation, i have to take the first derivative and plug in t=0. so does that mean, y = c0 and y' = c1?

further, i let the centrifugal force equal to the spring force

F_r = mu*(r'') = -k(r-L) where mu is m1m2/M where M = m1+m2

where L is the unstretched length

anyway since the equation is a simple harmonic equation, i could find two constants b1, and b2 and i got b1 = r - L and b2 = r'/w ( where w =omega) when i plugged in t=0 for the derivatives.


however, how do these initial conditions fit in with the answer, with the position equations y1 = Y + (m2/M)*r and y2= Y - (m1/M)*r

and ultimately, the answer to the problem
which is y1 = (L + (m1/M)*v0*t - 0.5gt^2) + (m2*v0/M*w)sin(wt)
and y2 = ( (m1/M)*v0*t - 0.5gt^2) - (m1*v0/M*w)sin(wt)
 
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So you're breaking this problem into two parts: the motion of the center of mass, and the motion of the two particles relative to the center of mass.

The first one is a simple projectile motion problem, and you should use the equation you mentioned, y = c0+c1t-1/2gt^2. You need to calculate the initial position and the initial velocity of the center of mass, and plug these in for c0 and c1. The position of the center of mass should be easy, and for the velocity, use vcm = ptot/mtot.

The second part is the motion of two objects connected by a spring. Using the reduced mass equation you mentioned, you can find the distance between the objects as a function of time, and then you can use this to find the position of the objects relative to the center of mass, so that you can combine this with your previous result. The initial conditions here are the initial length of the spring and the initial rate of change of the length of the spring.
 

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