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

## 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

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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?

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haruspex
Homework Helper
Gold Member
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.

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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haruspex
$x_{com}(t) = \frac{m_2l}{m_1+m_2} (1+(t-t_m)\frac{\omega}{2}) = \frac{m_2l}{m_1+m_2}(1-\frac{\pi}{4}+\frac{t\omega}{2})$
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).