Menu Close

Question-226538




Question Number 226538 by mr W last updated on 02/Dec/25
Answered by mr W last updated on 07/Dec/25
length of cylinder =L  cross−section =2RL  at speed v  Mv=(M+2RLnmvdt)(v+dv)  Mdv+2RLnmv^2 dt=0  (dv/v^2 )=−((2RLnm)/M)dt  (1/v_0 )−(1/v)=−((2RLnmt)/M)  ⇒(1/v)=(1/v_0 )+((2RLnmt)/M)  say at t=t_1 : v=(v_0 /2)  (2/v_0 )=(1/v_0 )+((2RLnmt_1 )/M)  ⇒t_1 =(M/(2RLnmv_0 ))
$${length}\:{of}\:{cylinder}\:={L} \\ $$$${cross}−{section}\:=\mathrm{2}{RL} \\ $$$${at}\:{speed}\:{v} \\ $$$${Mv}=\left({M}+\mathrm{2}{RLnmvdt}\right)\left({v}+{dv}\right) \\ $$$${Mdv}+\mathrm{2}{RLnmv}^{\mathrm{2}} {dt}=\mathrm{0} \\ $$$$\frac{{dv}}{{v}^{\mathrm{2}} }=−\frac{\mathrm{2}{RLnm}}{{M}}{dt} \\ $$$$\frac{\mathrm{1}}{{v}_{\mathrm{0}} }−\frac{\mathrm{1}}{{v}}=−\frac{\mathrm{2}{RLnmt}}{{M}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{v}}=\frac{\mathrm{1}}{{v}_{\mathrm{0}} }+\frac{\mathrm{2}{RLnmt}}{{M}} \\ $$$${say}\:{at}\:{t}={t}_{\mathrm{1}} :\:{v}=\frac{{v}_{\mathrm{0}} }{\mathrm{2}} \\ $$$$\frac{\mathrm{2}}{{v}_{\mathrm{0}} }=\frac{\mathrm{1}}{{v}_{\mathrm{0}} }+\frac{\mathrm{2}{RLnmt}_{\mathrm{1}} }{{M}} \\ $$$$\Rightarrow{t}_{\mathrm{1}} =\frac{{M}}{\mathrm{2}{RLnmv}_{\mathrm{0}} } \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *