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Question Number 23346 by ajfour last updated on 29/Oct/17

Commented by Physics lover last updated on 29/Oct/17

$$Nicetricksir.$$

Commented by ajfour last updated on 29/Oct/17

$$Thecentreofmassvelocityof$$$$theconveyorbeltvehicleof$$$$massMrollingalongalevel$$$$roadis{\mathit{v}}_0.\mathit{F}inditskineticenergy.$$

Commented by Physics lover last updated on 29/Oct/17

Commented by Physics lover last updated on 29/Oct/17

$$Sorry,MrAjfour,Iflaggedthis$$$$questionbymistake.$$$$Didnotintendtodosoatall.$$

Commented by mrW1 last updated on 29/Oct/17

$$\mathrm{M}={\mathrm{M}}_{\mathrm{R}}+{2\mathrm{M}}_{\mathrm{L}}$$$${\mathrm{M}}_{\mathrm{R}}=\mathrm{mass}\mathrm{of}\mathrm{circular}\mathrm{parts}(=\stackrel{⌢}{\mathrm{DA}}+\stackrel{⌢}{\mathrm{BC}})$$$${\mathrm{M}}_{\mathrm{L}}=\mathrm{mass}\mathrm{of}\mathrm{parallel}\mathrm{part}(=\mathrm{AB}=\mathrm{DC})$$$$$$$${\mathrm{KE}}_{\mathrm{R}}=\frac{1}{2}{\mathrm{M}}_{\mathrm{R}}{\mathrm{v}}_0^2+\frac{1}{2}\mathrm{I}{\omega }^2$$$$=\frac{1}{2}{\mathrm{M}}_{\mathrm{R}}{\mathrm{v}}_0^2+\frac{1}{2}{\mathrm{M}}_{\mathrm{R}}{\mathrm{R}}^2{\left(\frac{{\mathrm{v}}_0}{\mathrm{R}}\right)}^2$$$$={\mathrm{M}}_{\mathrm{R}}{\mathrm{v}}_0^2$$$$$$$${\mathrm{KE}}_{\mathrm{L}}=\frac{1}{2}{\mathrm{M}}_{\mathrm{L}}{\left({2\mathrm{v}}_0\right)}^2+\frac{1}{2}{\mathrm{M}}_{\mathrm{L}}{0}^2$$$$={2\mathrm{M}}_{\mathrm{L}}{\mathrm{v}}_0^2$$$$$$$$\mathrm{total}\mathrm{KE}={\mathrm{KE}}_{\mathrm{R}}+{\mathrm{KE}}_{\mathrm{L}}$$$$={\mathrm{M}}_{\mathrm{R}}{\mathrm{v}}_0^2+{2\mathrm{M}}_{\mathrm{L}}{\mathrm{v}}_0^2$$$$=({\mathrm{M}}_{\mathrm{R}}+{2\mathrm{M}}_{\mathrm{L}}){\mathrm{v}}_0^2$$$$={\mathrm{Mv}}_0^2$$

Commented by ajfour last updated on 29/Oct/17

$$thanksagain,Sir.$$

Commented by ajfour last updated on 29/Oct/17

$$Ifweassumesmallradius,or$$$$height,then$$$$K=\frac{1}{2}\left(\frac{M}{2}\right){\left(2v_0\right)}^2+0+0+0$$$$\mathit{K}={\mathit{M}\mathit{v}}_0^2.$$

Answered by Physics lover last updated on 29/Oct/17

$$theparticlesonABardpurely$$$$translatingwithvelocity2v_0$$$$AndthesemicurcularringsAD$$$$andBCarepurerollingwithvelocity$$$$v_{0}and\omega =\frac{v_0}{R}$$$$let\lambda belinearmassdensity$$$$\Rightarrow \lambda =\frac{M}{2l+2\pi R}$$$$TotK.E.=K.E{.}_{rotational}+K.E{.}_{translational}$$$$=\left\{\frac{1}{2}\left(2\pi R\lambda \right)R^2\left(\frac{v_0^2}{R^2}\right)\right\}+\{\frac{1}{2}\left(\lambda l\right){\left(2v_0\right)}^2+\frac{1}{2}\left(2\pi R\lambda \right)v_0^2\}$$$$$$$$=\left\{\pi R\lambda v_0^2\right\}+\{2l\lambda v_0^2+\pi R\lambda v_0^2\}$$$$=2\pi {Rv}_0^2+2l\lambda v_0^2$$$$=2\lambda v_0^2(\pi R+l)$$$$=2\left(\frac{M}{2l+2\pi R}\right)v_0^2(\pi R+l)$$$$K.E.={Mv}_0^2$$$$$$$$$$$$$$

Commented by ajfour last updated on 29/Oct/17

$$Ans.isK.E.={Mv}_0^2.$$

Commented by ajfour last updated on 29/Oct/17

$$ThankyouPhysicslover.$$

Commented by Physics lover last updated on 29/Oct/17

$$Youarewelcomesir.Butiguess$$$$ididitinthemostlengthyway.lol$$

Commented by ajfour last updated on 29/Oct/17

$$Yourproofisquitegeneralone$$$$\left(answertoasubjectivequestion\right).$$

Commented by Physics lover last updated on 29/Oct/17

$$Ohthankyousir.:)$$

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