Question Number 143868 by ajfour last updated on 19/Jun/21
Commented by ajfour last updated on 19/Jun/21
$${Find}\:{minimum}\:{h}\:{such}\:{that} \\ $$$${stick}\:{just}\:{loses}\:{contact}\:{with} \\ $$$${table}\:{upon}\:{being}\:{hit}\:{by}\:{a}\:{small} \\ $$$${ball}\:{of}\:{mass}\:{m}\:{released}\:{a} \\ $$$${height}\:{h}\:{above}\:{the}\:{table}\:{top} \\ $$$${level}.\:\left({Assume}\:{sufficient}\right. \\ $$$$\left.{friction}\:{at}\:{the}\:{table}\:{corner}\:\right) \\ $$
Answered by ajfour last updated on 21/Jun/21
$${Angular}\:{momentum} \\ $$$${conservation}\:{during}\:{elastic} \\ $$$${collision}\:{of}\:{ball}\:{with}\:{stick}: \\ $$$$\mathrm{2}{m}\sqrt{\mathrm{2}{gh}}\left({kL}\right)={I}\omega_{\mathrm{0}} \\ $$$${After}\:{collision}\:{stick}\:{rotates}; \\ $$$${now}\:{say}\:{at}\:{some}\:{point}\:{of} \\ $$$${time}\:{thereafter},\:{say}\:{t}={t}, \\ $$$${angle}\:{of}\:{stick}\:{from}\:{vertical} \\ $$$${is}\:\theta;\:{then} \\ $$$${rise}\:{in}\:{potential}\:{energy} \\ $$$$\:=\:{fall}\:{in}\:{kinetic}\:{energy} \\ $$$${Mg}\left(\frac{{L}}{\mathrm{2}}−{kL}\right)\mathrm{cos}\:\theta=\frac{\mathrm{1}}{\mathrm{2}}{I}\left(\omega_{\mathrm{0}} ^{\mathrm{2}} −\omega^{\mathrm{2}} \right) \\ $$$${And}\:{let}\:{Normal}\:{reaction}\: \\ $$$${then}\:{vanishes}\:\Rightarrow \\ $$$${say}\:\:\frac{{L}}{\mathrm{2}}−{kL}={r} \\ $$$${Mgr}\mathrm{sin}\:\theta=−{I}\alpha \\ $$$$\Rightarrow\:\:{Mgr}\mathrm{sin}\:\theta=−{I}\left(\frac{\omega{d}\omega}{{d}\theta}\right)\:\:\& \\ $$$$\omega^{\mathrm{2}} {r}\mathrm{sin}\:\theta=−\left(\frac{\omega{d}\omega}{{d}\theta}\right){r}\mathrm{cos}\:\theta\:\:\:\:\:\:\& \\ $$$$\omega^{\mathrm{2}} {r}\mathrm{cos}\:\theta−\left(\frac{\omega{d}\omega}{{d}\theta}\right){r}\mathrm{sin}\:\theta={g} \\ $$$$\Rightarrow\:\:\omega^{\mathrm{2}} {r}=\frac{{Mgr}^{\mathrm{2}} \mathrm{cos}\:\theta}{{I}} \\ $$$$\:\:\:\:\:\:\:\omega^{\mathrm{2}} {r}\mathrm{cos}\:\theta+\frac{{Mgr}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}{{I}}={g} \\ $$$$\:\:\:........ \\ $$
Answered by TheHoneyCat last updated on 21/Jun/21
$$\mathrm{Let}'\mathrm{s}\:\mathrm{assume}\:\mathrm{that}\:\mathrm{the}\:\mathrm{bar}\:\mathrm{of}\:\mathrm{mass}\:{M}\:\mathrm{will}\:\mathrm{not}\:\mathrm{slide} \\ $$$$\mathrm{To}\:\mathrm{fall},\:\mathrm{it}\:\mathrm{therefore}\:\mathrm{needs}\:\mathrm{to}\:\mathrm{end}\:\mathrm{up}\:\mathrm{vertical} \\ $$$$ \\ $$$$\mathrm{The}\:\mathrm{ball}\:\mathrm{will}\:\mathrm{fall}\:\mathrm{on}\:\mathrm{the}\:\mathrm{bar}\:\mathrm{and}\:\mathrm{transfer}\:\mathrm{all}\:\mathrm{of}\:\mathrm{its}\:\mathrm{kinetic}\:\mathrm{energy}. \\ $$$$\mathrm{that}\:\mathrm{energy}\:\mathrm{beeing}\:\mathrm{its}\:\mathrm{potential}\:\mathrm{energy}\:\mathrm{at}\:\mathrm{the}\:\mathrm{beginning}\:\mathrm{because},\:\mathrm{at}\:\mathrm{that}\:\mathrm{time},\:\mathrm{it}\:\mathrm{had}\:\mathrm{no}\:\mathrm{speed},\:\mathrm{and}\:\mathrm{since}\:\mathrm{then},\:\mathrm{it}\:\mathrm{has}\:\mathrm{been}\:\mathrm{in}\:\mathrm{free}\:\mathrm{fall}. \\ $$$$\mathrm{Choosing}\:{z}=\mathrm{0}\:\mathrm{for}\:\mathrm{the}\:\mathrm{bar}'\mathrm{s}\:\mathrm{height}. \\ $$$$ \\ $$$${E}_{{m}_{{initially}} } =\:{E}_{{m}_{{ball}\:{initially}} } +\:{E}_{{m}_{{bar}\:{initially}} } \\ $$$$={mgh}+\left({Mg}\right).\mathrm{0} \\ $$$$={mgh} \\ $$$$ \\ $$$${E}_{{m}_{{after}\:{contact}} } \:=\:{E}_{{m}_{{ball}\:{left}} } +{E}_{{m}_{{bar}} } \\ $$$$=^{\mathrm{1}} /_{\mathrm{2}} {mv}_{{ball}\:{left}} ^{\mathrm{2}} \:+\:{E}_{{p}_{{vertical}\:{bar}} } +{E}_{{c}_{{bar}} } \\ $$$$=^{\mathrm{1}} /_{\mathrm{2}} {mv}_{{ball}\:{left}} ^{\mathrm{2}} \:+\underset{{bar}} {\int}\frac{\delta{Mv}^{\mathrm{2}} }{\mathrm{2}}\:+\:\underset{{bar}} {\int}\delta{Mgz} \\ $$$${supposing}\:{that}\:{the}\:{mass}\:{is}\:{uniformaly}\:{distributed} \\ $$$$=^{\mathrm{1}} /_{\mathrm{2}} {mv}_{{ball}\:{left}} ^{\mathrm{2}} \:+\underset{{bar}} {\int}\frac{\delta{mv}^{\mathrm{2}} }{\mathrm{2}}\:+\frac{{M}}{{L}}{g}\:\underset{{bar}} {\int}{z}\:{dz} \\ $$$$=^{\mathrm{1}} /_{\mathrm{2}} {mv}_{{ball}\:{left}} ^{\mathrm{2}} \:+\underset{{bar}} {\int}\frac{\delta{mv}^{\mathrm{2}} }{\mathrm{2}}\:+\frac{{M}}{\mathrm{2}{L}}{g}\:\left(\left(\mathrm{1}−{k}\right)^{\mathrm{2}} −{k}^{\mathrm{2}} \right) \\ $$$$=^{\mathrm{1}} /_{\mathrm{2}} {mv}_{{ball}\:{left}} ^{\mathrm{2}} \:+\underset{{bar}} {\int}\frac{\delta{mv}^{\mathrm{2}} }{\mathrm{2}}\:+\frac{{M}}{\mathrm{2}{L}}{g}\:\left(\mathrm{1}−\mathrm{2}{k}\right) \\ $$$$\geqslant\frac{{M}}{\mathrm{2}{L}}{g}\:\left(\mathrm{1}−\mathrm{2}{k}\right) \\ $$$$ \\ $$$$\mathrm{by}\:\mathrm{the}\:\mathrm{conservation}\:\mathrm{of}\:\mathrm{energy} \\ $$$$\mathrm{E}_{{initially}} ={E}_{{after}} \\ $$$$\mathrm{so}:\:{mgh}\geqslant\frac{{M}}{\mathrm{2}{L}}{g}\left(\mathrm{1}−\mathrm{2}{k}\right) \\ $$$$\Leftrightarrow\mathrm{2}{L}\:{mh}\geqslant{M}\left(\mathrm{1}−\mathrm{2}{k}\right)_{\blacksquare} \\ $$
Commented by TheHoneyCat last updated on 21/Jun/21
$${so}\:{obviously}\:{the}\:{minimum}\:{value}\:{of}\:{h}\:{is}: \\ $$$$\frac{{M}\left(\mathrm{1}−\mathrm{2}{k}\right)}{\mathrm{2}{Lm}} \\ $$
Commented by ajfour last updated on 21/Jun/21
$${The}\:{stick}\:{doesn}'{t}\:{slide}\:{granted}, \\ $$$${but}\:{stick}\:{may}\:{jump}\:{during} \\ $$$${collision}\:{and}\:{leave}\:{contact} \\ $$$${at}\:{once};\:{i}\:{see}\:{a}\:{fair}\:{possibility} \\ $$$${of}\:{that}. \\ $$