Question and Answers Forum

All Questions      Topic List

Mechanics Questions

Previous in All Question      Next in All Question      

Previous in Mechanics      Next in Mechanics      

Question Number 34781 by Tinkutara last updated on 11/May/18

Answered by ajfour last updated on 11/May/18

$$6{mgx}_1-3{mgx}_2=\frac{1}{2}\left(18m\right)v^2+\frac{1}{2}{kx}^2$$$$.....\left(i\right)$$$$x_1-x_2=x....\left(ii\right)$$$$v^2=2\left(\frac{3mg}{18m}\right)\left(\frac{2x_1+x_2}{3}\right)....\left(iii\right)$$$$\Rightarrow \frac{1}{2}\left(18m\right)v^2=mg(2x_1+x_2)$$$$usingin\left(i\right):$$$$3mg(2x_1-x_2)=mg(2x_1+x_2)+\frac{{kx}^2}{2}$$$$or{x}_1-x_2=\frac{{kx}^2}{8mg}$$$$using\left(ii\right):$$$$x=\frac{8mg}{k}.\left(3\right).$$

Commented by Tinkutara last updated on 11/May/18

Please explain (iii) equation.

Commented by ajfour last updated on 11/May/18

$$whenspringismaximumstretched$$$$v_{cm}^2=v_{6m}^2=v_{3m}^2=2a_{cm}{s}_{cm}$$$$a_{cm}=\frac{F_{ext}}{18m};s_{cm}=\frac{12{mx}_1+6{mx}_2}{18m}.$$

Commented by Tinkutara last updated on 11/May/18

$$Why{s}_{cm}=\frac{12{mx}_1+6{mx}_2}{18m}?{Wouldn}^{\prime }t$$$$itbealongy-axisalso?$$

Commented by ajfour last updated on 11/May/18

$$yes{y}_{cm}\ne 0,but{we}^{\prime }llhaveto$$$$gointothenormalreactionseven.$$$$Andasfarasthedisplacement$$$$ofcentreofmassisconcerned,$$$$wecanassumethemasses{\left(6m\right)}_s$$$$and{\left(3m\right)}_{s}havemovedby{x}_{1}and$$$$x_{2}horizontally.situationsare$$$$equivalentforobtaining{s}_{cm}.$$

Commented by Tinkutara last updated on 11/May/18

$$Then{s}_{cm}=\frac{12{mx}_1-6{mx}_2}{18m}because$$$$3mggoesleftifallmotionis$$$$consideredleft?$$

Commented by ajfour last updated on 11/May/18

$$x_{2}beit+veor-ve$$$$s_{cm}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}.$$

Commented by Tinkutara last updated on 12/May/18

Thank you very much Sir! I got the answer. ��������

Terms of Service

Privacy Policy

Contact: info@tinkutara.com