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Question Number 22370 by Tinkutara last updated on 16/Oct/17

$$\mathrm{A}\mathrm{uniform}\mathrm{flexible}\mathrm{chain}\mathrm{of}\mathrm{length}\frac{3}{2}\mathrm{m}$$$$\mathrm{rests}\mathrm{on}\mathrm{a}\mathrm{fixed}\mathrm{smooth}\mathrm{sphere}\mathrm{of}$$$$\mathrm{radius}R=\frac{2}{\pi }\mathrm{m}\mathrm{such}\mathrm{that}\mathrm{one}\mathrm{end}A\mathrm{of}$$$$\mathrm{chain}\mathrm{is}\mathrm{on}\mathrm{the}\mathrm{top}\mathrm{of}\mathrm{the}\mathrm{sphere}\mathrm{while}$$$$\mathrm{the}\mathrm{other}\mathrm{end}B\mathrm{is}\mathrm{hanging}\mathrm{freely}.\mathrm{Chain}$$$$\mathrm{is}\mathrm{held}\mathrm{stationary}\mathrm{by}\mathrm{a}\mathrm{horizontal}$$$$\mathrm{thread}PA.\mathrm{Calculate}\mathrm{the}\mathrm{acceleration}$$$$\mathrm{of}\mathrm{chain}\mathrm{when}\mathrm{the}\mathrm{horizontal}\mathrm{string}PA$$$$\mathrm{is}\mathrm{burnt}.(g=10\mathrm{m}/{\mathrm{s}}^2)$$

Commented by mrW1 last updated on 17/Oct/17

$$\stackrel{⌢}{\mathrm{AC}}=\frac{\pi \mathrm{r}}{2}=\frac{\pi }{2}\times \frac{2}{\pi }=1\mathrm{m}$$$$\mathrm{BC}=\frac{3}{2}-1=\frac{1}{2}\mathrm{m}$$$$\rho =\mathrm{mass}\mathrm{each}\mathrm{meter}$$$$\mathrm{a}=\mathrm{acceleration}$$$${\mathrm{T}}_0=\mathrm{tension}\mathrm{at}\mathrm{point}\mathrm{C},\mathrm{at}\theta =0$$$$\frac{1}{2}\rho \mathrm{g}-{\mathrm{T}}_0=\frac{1}{2}\rho \mathrm{a}$$$$\Rightarrow {\mathrm{T}}_0=\frac{1}{2}\rho (\mathrm{g}-\mathrm{a})$$$$$$$$-\mathrm{dT}+\rho \mathrm{gdscos}\theta =\mathrm{a}\rho \mathrm{ds}$$$$-\mathrm{dT}=\rho \mathrm{ds}(\mathrm{a}-\mathrm{gcos}\theta )$$$$-\mathrm{dT}=\rho \mathrm{r}(\mathrm{a}-\mathrm{gcos}\theta )\mathrm{d}\theta$$$$-{\int }_{{\mathrm{T}}_0}^0\mathrm{dT}=\rho \mathrm{r}{\int }_0^{\frac{\pi }{2}}(\mathrm{a}-\mathrm{gcos}\theta )\mathrm{d}\theta$$$${\mathrm{T}}_0=\rho \mathrm{r}{[\mathrm{a}\theta -\mathrm{gsin}\theta ]}_0^{\frac{\pi }{2}}$$$${\mathrm{T}}_0=\rho \mathrm{r}[\frac{\mathrm{a}\pi }{2}-\mathrm{g}]$$$$\frac{1}{2}\rho (\mathrm{g}-\mathrm{a})=\rho \frac{2}{\pi }(\frac{\mathrm{a}\pi }{2}-\mathrm{g})$$$$(\mathrm{g}-\mathrm{a})=\frac{4}{\pi }(\frac{\mathrm{a}\pi }{2}-\mathrm{g})$$$$\mathrm{g}-\mathrm{a}=2\mathrm{a}-\frac{4}{\pi }\mathrm{g}$$$$\Rightarrow \mathrm{a}=\frac{4+\pi }{3\pi }\mathrm{g}\approx 7.58\mathrm{m}/{\mathrm{s}}^2$$

Commented by Tinkutara last updated on 16/Oct/17

Commented by Tinkutara last updated on 17/Oct/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

Answered by revenge last updated on 18/Oct/17

Commented by revenge last updated on 18/Oct/17

$$Considerasmallelementofmassdmonthespherewhich$$$$haslengthRd\theta .Massofthissmallelementisg\sin \theta dm.$$$$dm=\lambda dxordm=\lambda Rd\theta ;where\lambda =\frac{2M}{3}kg/m.$$$$Totalmassofsuchelements=\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\left(g\sin \theta \right)\left(\lambda Rd\theta \right)$$$$=\lambda Rg\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\sin \theta d\theta =\lambda Rg=\frac{4Mg}{3\pi }$$$$andmassofhangingpartwhichhaslength\frac{1}{2}m=\frac{Mg}{3}.$$$$Accelerationofchain=\frac{Totalweight\left(force\right)}{Totalmass}=\frac{\frac{4Mg}{3\pi }+\frac{Mg}{3}}{M}$$$$=\frac{4g}{3\pi }+\frac{g}{3}=\frac{(\pi +4)g}{3\pi }$$

Commented by Tinkutara last updated on 18/Oct/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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