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Homework Help: Find positions of spring

  1. Dec 23, 2007 #1
    Two discs of masses [tex] m_1 [/tex] and [tex] m_2 [/tex], 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 [tex] x_1(0) = 0, x_2(0) = 2L_0 [/tex] and have initial speeds [tex] -v_0 [/tex] and [tex] 2v_0 [/tex] respectively. Find their positions at all later times. The unstretched spring has length [tex] L_0 [/tex] and a spring constant
    [tex] k = \frac{9 v_0^2 m_1 m_2}{2 L_0^2 (m_1 + m_2)} [/tex]

    I seem to arrive at an equation which does not need the spring constant. Can you start me off ?
     
  2. jcsd
  3. Dec 23, 2007 #2
    First of all, the amount [tex] \Delta l [/tex] of the spring's deformation can be related with the positions [tex] x_1,x_2 [/tex] of the disc. At any instance of time the length of the spring is [tex] L=x_2-x_1 [/tex]. If we assume that the spring is stretched then we have [tex] \begin{equation}\Delta l=L-L_0\Rightarrow \Delta l=x_2-x_1-L_0 \end{equation}[/tex].

    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 [tex] I_C=\frac{1}{2}mR^2 [/tex]. When a disc is rolling without slipping we have [tex] a_{CM}=R\,\alpha [/tex].
    Thus for a rolling without slipping 2nd disc, Newton's second law reads
    [tex] \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}[/tex] and
    [tex] \Sigma F=m\,a_{CM}\Rightarrow -k\,\Delta l-f=m\,a_{CM} [/tex]
    where [tex] f [/tex] stands for the friction which is requiried for the rotation.
    Adding the two equations we arrive to
    [tex] -k\,\Delta l=\frac{3}{2}\,m\,\ddot{x}_2 \quad (2)[/tex]
    Likewise for the1st disc we arrive to
    [tex] k\,\Delta l=\frac{3}{2}\,m\,\ddot{x}_1 \quad (3)[/tex]

    Adding equations (2),(3) we have

    [tex] 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)[/tex]

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

    [tex] \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)[/tex]

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

    [tex] 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 [/tex]

    [tex] 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}[/tex]




    Hope, I helped!
     
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