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Question-26864




Question Number 26864 by Tinkutara last updated on 30/Dec/17
Answered by ajfour last updated on 31/Dec/17
i thought one third overhanged.  Now since one third part lies on  table initially, we can think this  transmitted to join the lower  end of the hanging part by the  time it slips of the table.  (1/2)Mv^2 =((M/3))gh  h is shift in centre of mass  of the part that was on table.  (1/2)(4)v^2 =((4/3))(10)((4/3)+(1/3))  ⇒   v^2 =((100)/9)  or   v=((10)/3) (m/s) .
$${i}\:{thought}\:{one}\:{third}\:{overhanged}. \\ $$$${Now}\:{since}\:{one}\:{third}\:{part}\:{lies}\:{on} \\ $$$${table}\:{initially},\:{we}\:{can}\:{think}\:{this} \\ $$$${transmitted}\:{to}\:{join}\:{the}\:{lower} \\ $$$${end}\:{of}\:{the}\:{hanging}\:{part}\:{by}\:{the} \\ $$$${time}\:{it}\:{slips}\:{of}\:{the}\:{table}. \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{Mv}^{\mathrm{2}} =\left(\frac{{M}}{\mathrm{3}}\right){gh} \\ $$$${h}\:{is}\:{shift}\:{in}\:{centre}\:{of}\:{mass} \\ $$$${of}\:{the}\:{part}\:{that}\:{was}\:{on}\:{table}. \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{4}\right){v}^{\mathrm{2}} =\left(\frac{\mathrm{4}}{\mathrm{3}}\right)\left(\mathrm{10}\right)\left(\frac{\mathrm{4}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$$\Rightarrow\:\:\:{v}^{\mathrm{2}} =\frac{\mathrm{100}}{\mathrm{9}}\:\:{or}\:\:\:{v}=\frac{\mathrm{10}}{\mathrm{3}}\:\frac{{m}}{{s}}\:. \\ $$
Commented by Tinkutara last updated on 31/Dec/17
Thank you very much Sir! I got the answer.

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