Question Number 54167 by rahul 19 last updated on 30/Jan/19
Commented by rahul 19 last updated on 30/Jan/19
$${Describe}\:{Quantitatively}\:{the}\:{motion} \\ $$$${of}\:{the}\:{skaters}\:{after}\:{they}\:{are}\:{connected} \\ $$$${by}\:{the}\:{pole}. \\ $$
Commented by rahul 19 last updated on 30/Jan/19
$${Ans}:\:{Pure}\:{rotation}\:{about}\:{C}.{O}.{M}\:{with} \\ $$$$\omega\:=\:\frac{\mathrm{20}}{\mathrm{3}}\:{rad}/{s}. \\ $$
Answered by ajfour last updated on 30/Jan/19
$${about}\:{c}.{o}.{m}.\:\:{angular}\:{momentum} \\ $$$${remains}\:{conserved};\: \\ $$$$\omega=\frac{{v}}{\left({l}/\mathrm{2}\right)}\:=\:\frac{\mathrm{10}}{\left(\mathrm{3}/\mathrm{2}\right)}\:=\:\frac{\mathrm{20}}{\mathrm{3}}\:{rad}/{s}\:. \\ $$
Commented by rahul 19 last updated on 31/Jan/19
$${thanks}\:{sir}! \\ $$$${But}\:{how}\:{do}\:{you}\:{know}\:{initially}\:{torque} \\ $$$$\left({external}\right)\:{is}\:{zero}\:? \\ $$