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Question Number 24877 by Tinkutara last updated on 28/Nov/17

Commented by Tinkutara last updated on 28/Nov/17

Commented by mrW1 last updated on 29/Nov/17

mg−T=ma  Tr=((Mr^2 )/2)×(a/r)  ⇒T=((Ma)/2)  ⇒mg−((Ma)/2)=ma  ⇒a=(m/(m+(M/2)))×g=((0.1)/(0.1+(2/2)))×g=(g/(11))  h=((at^2 )/2)  ⇒t=(√((2h)/a))  ⇒v=at=(√(2ha))=(√((2g)/(11)))≈1.35 m/s

$${mg}−{T}={ma} \\ $$$${Tr}=\frac{{Mr}^{\mathrm{2}} }{\mathrm{2}}×\frac{{a}}{{r}} \\ $$$$\Rightarrow{T}=\frac{{Ma}}{\mathrm{2}} \\ $$$$\Rightarrow{mg}−\frac{{Ma}}{\mathrm{2}}={ma} \\ $$$$\Rightarrow{a}=\frac{{m}}{{m}+\frac{{M}}{\mathrm{2}}}×{g}=\frac{\mathrm{0}.\mathrm{1}}{\mathrm{0}.\mathrm{1}+\frac{\mathrm{2}}{\mathrm{2}}}×{g}=\frac{{g}}{\mathrm{11}} \\ $$$${h}=\frac{{at}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow{t}=\sqrt{\frac{\mathrm{2}{h}}{{a}}} \\ $$$$\Rightarrow{v}={at}=\sqrt{\mathrm{2}{ha}}=\sqrt{\frac{\mathrm{2}{g}}{\mathrm{11}}}\approx\mathrm{1}.\mathrm{35}\:{m}/{s} \\ $$

Commented by ajfour last updated on 28/Nov/17

please explain.

$${please}\:{explain}. \\ $$

Commented by ajfour last updated on 28/Nov/17

If the string is simply wrapped  and pulley frictionless, then  mass falls freely.  v^2 =2gh =2(9.8m/s^2 )(1m)  v=(√(19.6)) m/s .

$${If}\:{the}\:{string}\:{is}\:{simply}\:{wrapped} \\ $$$${and}\:{pulley}\:{frictionless},\:{then} \\ $$$${mass}\:{falls}\:{freely}. \\ $$$${v}^{\mathrm{2}} =\mathrm{2}{gh}\:=\mathrm{2}\left(\mathrm{9}.\mathrm{8}{m}/{s}^{\mathrm{2}} \right)\left(\mathrm{1}{m}\right) \\ $$$${v}=\sqrt{\mathrm{19}.\mathrm{6}}\:{m}/{s}\:. \\ $$

Commented by Tinkutara last updated on 29/Nov/17

Thank you Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir}! \\ $$

Commented by jota+ last updated on 28/Nov/17

El eje no hace trabajo sobre la polea.  Tr−(Rρ=0)=Iα  R=rozamiento  ρ=radio del  eje

$${El}\:{eje}\:{no}\:{hace}\:{trabajo}\:{sobre}\:{la}\:{polea}. \\ $$$${Tr}−\left({R}\rho=\mathrm{0}\right)={I}\alpha \\ $$$${R}={rozamiento} \\ $$$$\rho={radio}\:{del}\:\:{eje} \\ $$

Commented by mrW1 last updated on 29/Nov/17

To Mr jota+:   thank you for your wonderful   contribution!  As the most users of this forum  use English, can you please kindly  also use English so that more users  like me can understand your posts?   Thank you very much!

$${To}\:{Mr}\:{jota}+:\: \\ $$$${thank}\:{you}\:{for}\:{your}\:{wonderful}\: \\ $$$${contribution}! \\ $$$${As}\:{the}\:{most}\:{users}\:{of}\:{this}\:{forum} \\ $$$${use}\:{English},\:{can}\:{you}\:{please}\:{kindly} \\ $$$${also}\:{use}\:{English}\:{so}\:{that}\:{more}\:{users} \\ $$$${like}\:{me}\:{can}\:{understand}\:{your}\:{posts}?\: \\ $$$${Thank}\:{you}\:{very}\:{much}! \\ $$

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