Can you find the positions of two rolling discs connected by a spring?

[touqra]

Two discs of masses m_1 and m_2, both of radius R, have centers connected by a spring so that they can roll without slipping. At the initial moment the centers are at x_1(0) = 0, x_2(0) = 2L_0 and have initial speeds -v_0 and 2v_0 respectively. Find their positions at all later times. The unstretched spring has length L_0 and a spring constant
k = \frac{9 v_0^2 m_1 m_2}{2 L_0^2 (m_1 + m_2)}

I seem to arrive at an equation which does not need the spring constant. Can you start me off ?

 

[Rainbow Child]

First of all, the amount \Delta l of the spring's deformation can be related with the positions x_1,x_2 of the disc. At any instance of time the length of the spring is L=x_2-x_1. If we assume that the spring is stretched then we have \begin{equation}\Delta l=L-L_0\Rightarrow \Delta l=x_2-x_1-L_0 \end{equation}.

Returning to the the diss, if each disc is homogenous then its moment of inertia about an axis perpendicular to the disc passing through it's center is I_C=\frac{1}{2}mR^2. When a disc is rolling without slipping we have a_{CM}=R\,\alpha.
Thus for a rolling without slipping 2nd disc, Newton's second law reads
\Sigma\tau=I_C\,\alpha\Rightarrow f\,R=\frac{1}{2}mR^2\,\alpha\Rightarrow f=\frac{1}{2}mR\,\alpha \Rightarrow f=\frac{1}{2}m\,a_{CM} and
\Sigma F=m\,a_{CM}\Rightarrow -k\,\Delta l-f=m\,a_{CM}
where f stands for the friction which is requiried for the rotation.
Adding the two equations we arrive to
-k\,\Delta l=\frac{3}{2}\,m\,\ddot{x}_2 \quad (2)
Likewise for the1st disc we arrive to
k\,\Delta l=\frac{3}{2}\,m\,\ddot{x}_1 \quad (3)

Adding equations (2),(3) we have

m_1\,a_1+m_2\,a_2=0\Rightarrow m_1\,\dot{x}_1+m_2\,\dot{x}_2=-m_1\,v_0+2\,m_2\,v_0 \,(conservation\, of\, momentum) \Rightarrow m_1\, x_1+m_2\,x_2=(m_1\,v_0+2\,m_2\,v_0 )\,t+2\,m_2\,L_0 \quad (4)

With the help of (1) and (4) we can express \Delta l with respect to x_1 or x_2, e.g.

\Delta l=(1+\frac{m_2}{m_1})\,x_2+(1-2\,\frac{m_2}{m_1})\,v_0\,t-(1+2\,\frac{m_2}{m_1})\,L_0 \quad (5)

Plugging (5) into (2) we can compute x_2(t), and from (4) we have x_1(t). The results are

x_1(t)=-\frac{m_2\,L_0}{m_1+m_2}\,(\-1 + \cos \frac{{\sqrt{3}}\,t\,{v_o}}{{L_o}} + <br /> {\sqrt{3}}\,\sin \frac{{\sqrt{3}}\,t\,{v_o}}{{L_o}})-\frac{ {m_1} - 2\,{m_2} }<br /> {{m_1} + {m_2}}\,{v_o} \,t

x_2(t)=\frac{m_1\,L_0}{m_1+m_2}\,(\-1 + \cos \frac{{\sqrt{3}}\,t\,{v_o}}{{L_o}} + <br /> {\sqrt{3}}\,\sin \frac{{\sqrt{3}}\,t\,{v_o}}{{L_o}})-\frac{ {m_1} - 2\,{m_2} }<br /> {{m_1} + {m_2}}\,{v_o} \,t +\frac{2\,m_2\,L_0}{m_1+m_2}




Hope, I helped!