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Question-143868

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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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