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Question Number 16497 by Tinkutara last updated on 23/Jun/17

Two particles are revolving on two  coplanar circles with constant angular  velocities ω_1  and ω_2  respectively. Their  time periods are T_1  and T_2  then prove  that the time taken by second particle  to complete one revolution more than  the first particle, T, is given by  T = ((T_1 T_2 )/(T_1  − T_2 ))

$$\mathrm{Two}\:\mathrm{particles}\:\mathrm{are}\:\mathrm{revolving}\:\mathrm{on}\:\mathrm{two} \\ $$$$\mathrm{coplanar}\:\mathrm{circles}\:\mathrm{with}\:\mathrm{constant}\:\mathrm{angular} \\ $$$$\mathrm{velocities}\:\omega_{\mathrm{1}} \:\mathrm{and}\:\omega_{\mathrm{2}} \:\mathrm{respectively}.\:\mathrm{Their} \\ $$$$\mathrm{time}\:\mathrm{periods}\:\mathrm{are}\:{T}_{\mathrm{1}} \:\mathrm{and}\:{T}_{\mathrm{2}} \:\mathrm{then}\:\mathrm{prove} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{time}\:\mathrm{taken}\:\mathrm{by}\:\mathrm{second}\:\mathrm{particle} \\ $$$$\mathrm{to}\:\mathrm{complete}\:\mathrm{one}\:\mathrm{revolution}\:\mathrm{more}\:\mathrm{than} \\ $$$$\mathrm{the}\:\mathrm{first}\:\mathrm{particle},\:{T},\:\mathrm{is}\:\mathrm{given}\:\mathrm{by} \\ $$$${T}\:=\:\frac{{T}_{\mathrm{1}} {T}_{\mathrm{2}} }{{T}_{\mathrm{1}} \:−\:{T}_{\mathrm{2}} } \\ $$

Answered by ajfour last updated on 23/Jun/17

 △t = ((2π)/(△ω)) = ((2π)/((((2π)/T_2 )−((2π)/T_1 )))) = ((T_1 T_2 )/(T_1 −T_2 )) .

$$\:\bigtriangleup{t}\:=\:\frac{\mathrm{2}\pi}{\bigtriangleup\omega}\:=\:\frac{\mathrm{2}\pi}{\left(\frac{\mathrm{2}\pi}{{T}_{\mathrm{2}} }−\frac{\mathrm{2}\pi}{{T}_{\mathrm{1}} }\right)}\:=\:\frac{{T}_{\mathrm{1}} {T}_{\mathrm{2}} }{{T}_{\mathrm{1}} −{T}_{\mathrm{2}} }\:. \\ $$

Commented by Tinkutara last updated on 23/Jun/17

Thanks Sir!

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

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