Question Number 5357 by Junaid Mirza last updated on 11/May/16
Commented by Yozzii last updated on 11/May/16
$${Let}\:{the}\:{lowest}\:{position}\:{of}\:{the}\:{bob}\:{be} \\ $$$${that}\:{where}\:{its}\:{gravitational}\:{potential} \\ $$$${energy}\:{is}\:{zero}.\:{Then},\:{by}\:{the}\:{law}\:{of} \\ $$$${conservation}\:{of}\:{energy}\:{for}\:{the}\:{bob}, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Delta{K}.{E}=\Delta{P}.{E}. \\ $$$$\Delta{K}.{E}=\frac{{m}}{\mathrm{2}}\left({v}^{\mathrm{2}} −{u}^{\mathrm{2}} \right)\:{and}\:\Delta{P}.{E}={mgl}\left({cos}\alpha−{cos}\theta\right) \\ $$$${where}\:\theta=\:{initial}\:{angular}\:{displacement}\:{of}\:{the} \\ $$$${pendulum}\:{from}\:{the}\:{centre}\:{position}, \\ $$$$\alpha={final}\:{angular}\:{displacement}\:{from}\:{centre}\:{position} \\ $$$${u}={initial}\:{speed},\:{v}={final}\:{speed},\:{m}={mass}\:{of}\:{bob} \\ $$$${g}={acceleration}\:{due}\:{to}\:{gravity},\:{l}={length}\:{of}\:{pendulum}. \\ $$$$ \\ $$$$\therefore\:\frac{{m}}{\mathrm{2}}\left({v}^{\mathrm{2}} −{u}^{\mathrm{2}} \right)={mgl}\left({cos}\alpha−{cos}\theta\right) \\ $$$${m}>\mathrm{0}\Rightarrow{v}^{\mathrm{2}} ={u}^{\mathrm{2}} +\mathrm{2}{gl}\left({cos}\alpha−{cos}\theta\right) \\ $$$$\Rightarrow{v}=\sqrt{{u}^{\mathrm{2}} +\mathrm{2}{gl}\left({cos}\alpha−{cos}\theta\right)} \\ $$$${Initially},\:{u}=\mathrm{0},\:\theta=\mathrm{60}°.\:{l}=\mathrm{2}\:,\:{g}\approx\mathrm{9}.\mathrm{8}{ms}^{−\mathrm{2}} . \\ $$$$\alpha=\mathrm{0}°\:{at}\:{centre}\:{position}. \\ $$$$\therefore{v}\approx\sqrt{\mathrm{0}+\mathrm{2}×\mathrm{2}×\mathrm{9}.\mathrm{8}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$${v}\approx\sqrt{\mathrm{19}.\mathrm{6}}\:{ms}^{−\mathrm{1}} \approx\mathrm{4}.\mathrm{43}\:{ms}^{−\mathrm{1}} \\ $$
Commented by Junaid Mirza last updated on 12/May/16
$$\mathrm{Thanks}\:\mathrm{you}\:\mathrm{are}\:\mathrm{correct} \\ $$