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Question-143868




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. (Assume sufficient  friction at the table corner )
$${Find}\:{minimum}\:{h}\:{such}\:{that} \\ $$$${stick}\:{just}\:{loses}\:{contact}\:{with} \\ $$$${table}\:{upon}\:{being}\:{hit}\:{by}\:{a}\:{small} \\ $$$${ball}\:{of}\:{mass}\:{m}\:{released}\:{a} \\ $$$${height}\:{h}\:{above}\:{the}\:{table}\:{top} \\ $$$${level}.\:\left({Assume}\:{sufficient}\right. \\ $$$$\left.{friction}\:{at}\:{the}\:{table}\:{corner}\:\right) \\ $$
Answered by ajfour last updated on 21/Jun/21
Angular momentum  conservation during elastic  collision of ball with stick:  2m(√(2gh))(kL)=Iω_0   After collision stick rotates;  now say at some point of  time thereafter, say t=t,  angle of stick from vertical  is θ; then  rise in potential energy   = fall in kinetic energy  Mg((L/2)−kL)cos θ=(1/2)I(ω_0 ^2 −ω^2 )  And let Normal reaction   then vanishes ⇒  say  (L/2)−kL=r  Mgrsin θ=−Iα  ⇒  Mgrsin θ=−I(((ωdω)/dθ))  &  ω^2 rsin θ=−(((ωdω)/dθ))rcos θ      &  ω^2 rcos θ−(((ωdω)/dθ))rsin θ=g  ⇒  ω^2 r=((Mgr^2 cos θ)/I)         ω^2 rcos θ+((Mgr^2 sin^2 θ)/I)=g     ........
$${Angular}\:{momentum} \\ $$$${conservation}\:{during}\:{elastic} \\ $$$${collision}\:{of}\:{ball}\:{with}\:{stick}: \\ $$$$\mathrm{2}{m}\sqrt{\mathrm{2}{gh}}\left({kL}\right)={I}\omega_{\mathrm{0}} \\ $$$${After}\:{collision}\:{stick}\:{rotates}; \\ $$$${now}\:{say}\:{at}\:{some}\:{point}\:{of} \\ $$$${time}\:{thereafter},\:{say}\:{t}={t}, \\ $$$${angle}\:{of}\:{stick}\:{from}\:{vertical} \\ $$$${is}\:\theta;\:{then} \\ $$$${rise}\:{in}\:{potential}\:{energy} \\ $$$$\:=\:{fall}\:{in}\:{kinetic}\:{energy} \\ $$$${Mg}\left(\frac{{L}}{\mathrm{2}}−{kL}\right)\mathrm{cos}\:\theta=\frac{\mathrm{1}}{\mathrm{2}}{I}\left(\omega_{\mathrm{0}} ^{\mathrm{2}} −\omega^{\mathrm{2}} \right) \\ $$$${And}\:{let}\:{Normal}\:{reaction}\: \\ $$$${then}\:{vanishes}\:\Rightarrow \\ $$$${say}\:\:\frac{{L}}{\mathrm{2}}−{kL}={r} \\ $$$${Mgr}\mathrm{sin}\:\theta=−{I}\alpha \\ $$$$\Rightarrow\:\:{Mgr}\mathrm{sin}\:\theta=−{I}\left(\frac{\omega{d}\omega}{{d}\theta}\right)\:\:\& \\ $$$$\omega^{\mathrm{2}} {r}\mathrm{sin}\:\theta=−\left(\frac{\omega{d}\omega}{{d}\theta}\right){r}\mathrm{cos}\:\theta\:\:\:\:\:\:\& \\ $$$$\omega^{\mathrm{2}} {r}\mathrm{cos}\:\theta−\left(\frac{\omega{d}\omega}{{d}\theta}\right){r}\mathrm{sin}\:\theta={g} \\ $$$$\Rightarrow\:\:\omega^{\mathrm{2}} {r}=\frac{{Mgr}^{\mathrm{2}} \mathrm{cos}\:\theta}{{I}} \\ $$$$\:\:\:\:\:\:\:\omega^{\mathrm{2}} {r}\mathrm{cos}\:\theta+\frac{{Mgr}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}{{I}}={g} \\ $$$$\:\:\:…….. \\ $$
Answered by TheHoneyCat last updated on 21/Jun/21
Let′s assume that the bar of mass M will not slide  To fall, it therefore needs to end up vertical    The ball will fall on the bar and transfer all of its kinetic energy.  that energy beeing its potential energy at the beginning because, at that time, it had no speed, and since then, it has been in free fall.  Choosing z=0 for the bar′s height.    E_m_(initially)  = E_m_(ball initially)  + E_m_(bar initially)    =mgh+(Mg).0  =mgh    E_m_(after contact)   = E_m_(ball left)  +E_m_(bar)    =^1 /_2 mv_(ball left) ^2  + E_p_(vertical bar)  +E_c_(bar)    =^1 /_2 mv_(ball left) ^2  +∫_(bar) ((δMv^2 )/2) + ∫_(bar) δMgz  supposing that the mass is uniformaly distributed  =^1 /_2 mv_(ball left) ^2  +∫_(bar) ((δmv^2 )/2) +(M/L)g ∫_(bar) z dz  =^1 /_2 mv_(ball left) ^2  +∫_(bar) ((δmv^2 )/2) +(M/(2L))g ((1−k)^2 −k^2 )  =^1 /_2 mv_(ball left) ^2  +∫_(bar) ((δmv^2 )/2) +(M/(2L))g (1−2k)  ≥(M/(2L))g (1−2k)    by the conservation of energy  E_(initially) =E_(after)   so: mgh≥(M/(2L))g(1−2k)  ⇔2L mh≥M(1−2k)_■
$$\mathrm{Let}'\mathrm{s}\:\mathrm{assume}\:\mathrm{that}\:\mathrm{the}\:\mathrm{bar}\:\mathrm{of}\:\mathrm{mass}\:{M}\:\mathrm{will}\:\mathrm{not}\:\mathrm{slide} \\ $$$$\mathrm{To}\:\mathrm{fall},\:\mathrm{it}\:\mathrm{therefore}\:\mathrm{needs}\:\mathrm{to}\:\mathrm{end}\:\mathrm{up}\:\mathrm{vertical} \\ $$$$ \\ $$$$\mathrm{The}\:\mathrm{ball}\:\mathrm{will}\:\mathrm{fall}\:\mathrm{on}\:\mathrm{the}\:\mathrm{bar}\:\mathrm{and}\:\mathrm{transfer}\:\mathrm{all}\:\mathrm{of}\:\mathrm{its}\:\mathrm{kinetic}\:\mathrm{energy}. \\ $$$$\mathrm{that}\:\mathrm{energy}\:\mathrm{beeing}\:\mathrm{its}\:\mathrm{potential}\:\mathrm{energy}\:\mathrm{at}\:\mathrm{the}\:\mathrm{beginning}\:\mathrm{because},\:\mathrm{at}\:\mathrm{that}\:\mathrm{time},\:\mathrm{it}\:\mathrm{had}\:\mathrm{no}\:\mathrm{speed},\:\mathrm{and}\:\mathrm{since}\:\mathrm{then},\:\mathrm{it}\:\mathrm{has}\:\mathrm{been}\:\mathrm{in}\:\mathrm{free}\:\mathrm{fall}. \\ $$$$\mathrm{Choosing}\:{z}=\mathrm{0}\:\mathrm{for}\:\mathrm{the}\:\mathrm{bar}'\mathrm{s}\:\mathrm{height}. \\ $$$$ \\ $$$${E}_{{m}_{{initially}} } =\:{E}_{{m}_{{ball}\:{initially}} } +\:{E}_{{m}_{{bar}\:{initially}} } \\ $$$$={mgh}+\left({Mg}\right).\mathrm{0} \\ $$$$={mgh} \\ $$$$ \\ $$$${E}_{{m}_{{after}\:{contact}} } \:=\:{E}_{{m}_{{ball}\:{left}} } +{E}_{{m}_{{bar}} } \\ $$$$=^{\mathrm{1}} /_{\mathrm{2}} {mv}_{{ball}\:{left}} ^{\mathrm{2}} \:+\:{E}_{{p}_{{vertical}\:{bar}} } +{E}_{{c}_{{bar}} } \\ $$$$=^{\mathrm{1}} /_{\mathrm{2}} {mv}_{{ball}\:{left}} ^{\mathrm{2}} \:+\underset{{bar}} {\int}\frac{\delta{Mv}^{\mathrm{2}} }{\mathrm{2}}\:+\:\underset{{bar}} {\int}\delta{Mgz} \\ $$$${supposing}\:{that}\:{the}\:{mass}\:{is}\:{uniformaly}\:{distributed} \\ $$$$=^{\mathrm{1}} /_{\mathrm{2}} {mv}_{{ball}\:{left}} ^{\mathrm{2}} \:+\underset{{bar}} {\int}\frac{\delta{mv}^{\mathrm{2}} }{\mathrm{2}}\:+\frac{{M}}{{L}}{g}\:\underset{{bar}} {\int}{z}\:{dz} \\ $$$$=^{\mathrm{1}} /_{\mathrm{2}} {mv}_{{ball}\:{left}} ^{\mathrm{2}} \:+\underset{{bar}} {\int}\frac{\delta{mv}^{\mathrm{2}} }{\mathrm{2}}\:+\frac{{M}}{\mathrm{2}{L}}{g}\:\left(\left(\mathrm{1}−{k}\right)^{\mathrm{2}} −{k}^{\mathrm{2}} \right) \\ $$$$=^{\mathrm{1}} /_{\mathrm{2}} {mv}_{{ball}\:{left}} ^{\mathrm{2}} \:+\underset{{bar}} {\int}\frac{\delta{mv}^{\mathrm{2}} }{\mathrm{2}}\:+\frac{{M}}{\mathrm{2}{L}}{g}\:\left(\mathrm{1}−\mathrm{2}{k}\right) \\ $$$$\geqslant\frac{{M}}{\mathrm{2}{L}}{g}\:\left(\mathrm{1}−\mathrm{2}{k}\right) \\ $$$$ \\ $$$$\mathrm{by}\:\mathrm{the}\:\mathrm{conservation}\:\mathrm{of}\:\mathrm{energy} \\ $$$$\mathrm{E}_{{initially}} ={E}_{{after}} \\ $$$$\mathrm{so}:\:{mgh}\geqslant\frac{{M}}{\mathrm{2}{L}}{g}\left(\mathrm{1}−\mathrm{2}{k}\right) \\ $$$$\Leftrightarrow\mathrm{2}{L}\:{mh}\geqslant{M}\left(\mathrm{1}−\mathrm{2}{k}\right)_{\blacksquare} \\ $$
Commented by TheHoneyCat last updated on 21/Jun/21
so obviously the minimum value of h is:  ((M(1−2k))/(2Lm))
$${so}\:{obviously}\:{the}\:{minimum}\:{value}\:{of}\:{h}\:{is}: \\ $$$$\frac{{M}\left(\mathrm{1}−\mathrm{2}{k}\right)}{\mathrm{2}{Lm}} \\ $$
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.
$${The}\:{stick}\:{doesn}'{t}\:{slide}\:{granted}, \\ $$$${but}\:{stick}\:{may}\:{jump}\:{during} \\ $$$${collision}\:{and}\:{leave}\:{contact} \\ $$$${at}\:{once};\:{i}\:{see}\:{a}\:{fair}\:{possibility} \\ $$$${of}\:{that}. \\ $$

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