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Motion of c.o.m. (kleppner 3.7)

  • #1
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Homework Statement


A system composed of two blocks of mass ##m_1## and ##m_2## connected by a massless spring with spring constant ##k##. The blocks slide on a frictionless plane. The unstretched length of the spring is ##l##. Initially, ##m_2## is held so that the spring is compressed at ##\frac{l}{2}## and ##m_1## is forced against a stop. ##m_2## is released at ##t=0##.
Find the motion of the center of mass.

Homework Equations


Spring restoration force, center of mass.

The Attempt at a Solution


[/B]
First part: mass 1 does not move
Their is a time ##0\le t\le t_m## where the mass ##m_1## will be stuck to the wall, and the external force from the wall on the system will be the spring restoration force, so ##f_{ext} = k(l-x_2)##.
Since ## x_1(t) = 0 ##, the center of mass is ## x_{com}(t) = \frac{m_2}{m_1+m_2} x_2(t) ##.
We must therefore solve the differential equation
## (m_1 + m_2) \ddot x_{com} = f_{ext} \iff m_2 \ddot x_2 = k(l-x_2) ##,
which is not homogenous, so we put ## u = x_2 - l ## and solve ## \ddot u + \frac{k}{m_2} u = \ddot u + \omega^2 u = 0##.
With initial conditions ##u(0) = -\frac{l}{2}## and ##\dot u(0) = 0 ##, we find that:
## x_{com}(t) = \frac{m_2l}{m_1+m_2}(1-\frac{\cos(\omega t)}{2})##
Now the maximum stretch before mass 1 moves is attained when ## \omega t_m = \frac{\pi}{2} ## so ## t_m = \frac{\pi}{2} \sqrt{\frac{m_2}{k}}##

Second part: mass 1 moves but there are no more external forces.
Therefore the center of mass must have constant speed for ## t\ge t_m##, so its motion must be :
##x_{com}(t) = x_{com}(t_m) + (t-t_m) v(t_m) = \frac{m_2l}{m_1+m_2}(1-\frac{\pi}{4}+\frac{t\omega}{2})##

Do you think it is the right solution?
 
Last edited:

Answers and Replies

  • #2
haruspex
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maximum stretch before mass 1 moves is attained at ##t_m = \frac{\pi}{2} \sqrt{\frac{m_2}{k}}##
You can get to this a bit faster. Up to here, we could treat m1 as fixed to the stop. Going from max compression to zero will be one quarter period.
##x_{com}(t) = \frac{m_2l}{m_1+m_2}(1-\frac{\pi}{4}+\frac{t\omega}{2})##
Doesn't seem right. Please post the individual steps.
 
  • #3
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72
Hello, thanks for answering.
Ok so in details, I get :
##x_{com}(t_m) = \frac{m_2l}{m_1+m_2} ##
##v(t_m) = \dot x_{com}(t_m) = \frac{m_2l}{m_1+m_2} \frac{\omega}{2}##

So in the end, for ##t\ge t_m## :
## x_{com}(t) = \frac{m_2l}{m_1+m_2} (1+(t-t_m)\frac{\omega}{2}) = \frac{m_2l}{m_1+m_2}(1-\frac{\pi}{4}+\frac{t\omega}{2})##
 
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  • #4
haruspex
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## x_{com}(t) = \frac{m_2l}{m_1+m_2} (1+(t-t_m)\frac{\omega}{2}) = \frac{m_2l}{m_1+m_2}(1-\frac{\pi}{4}+\frac{t\omega}{2})##
Yes, that looks right, except that you probably should not leave an ##\omega## in the answer (which is why I made the mistake of thinking we had different answers).
 

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