# Find positions of spring

1. Dec 23, 2007

### 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 ?

2. Dec 23, 2007

### 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 $$$$\Delta l=L-L_0\Rightarrow \Delta l=x_2-x_1-L_0$$$$.

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
$$k\,\Delta l=\frac{3}{2}\,m\,\ddot{x}_1 \quad (3)$$

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

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

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

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

Hope, I helped!