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Question-187216

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Question Number 187216 by ajfour last updated on 14/Feb/23
Commented by mr W last updated on 15/Feb/23
till the rod striks the ground, the ball moves upwards and the string is always under tension. at the moment as the rod strikes the groud the ball has a certain velocity upwards and the tension in string gets zero. then the ball moves as free body upwards with this initial velocity. all in all the ball moves only in vertical direction, so i don′t understand your diagram which shows as if the ball also moves sidewards.
$${till}\:{the}\:{rod}\:{striks}\:{the}\:{ground},\:{the}\:{ball} \\ $$$${moves}\:{upwards}\:{and}\:{the}\:{string}\:{is} \\ $$$${always}\:{under}\:{tension}.\:\:{at}\:{the}\:{moment} \\ $$$${as}\:{the}\:{rod}\:{strikes}\:{the}\:{groud}\:{the}\:{ball} \\ $$$${has}\:{a}\:{certain}\:{velocity}\:{upwards}\:{and} \\ $$$${the}\:{tension}\:{in}\:{string}\:{gets}\:{zero}.\:{then} \\ $$$${the}\:{ball}\:{moves}\:{as}\:{free}\:{body}\:{upwards} \\ $$$${with}\:{this}\:{initial}\:{velocity}. \\ $$$${all}\:{in}\:{all}\:{the}\:{ball}\:{moves}\:{only}\:{in}\: \\ $$$${vertical}\:{direction},\:{so}\:{i}\:{don}'{t}\: \\ $$$${understand}\:{your}\:{diagram}\:{which} \\ $$$${shows}\:{as}\:{if}\:{the}\:{ball}\:{also}\:{moves} \\ $$$${sidewards}. \\ $$
Answered by mr W last updated on 15/Feb/23
Commented by mr W last updated on 15/Feb/23
Commented by mr W last updated on 15/Feb/23
when the rod falls, the tension T changes and the inclination of left string changes, but the right string remains always in vertical position, i think.
$${when}\:{the}\:{rod}\:{falls},\:{the}\:{tension}\:{T} \\ $$$${changes}\:{and}\:{the}\:{inclination}\:{of}\:{left} \\ $$$${string}\:{changes},\:{but}\:{the}\:{right}\:{string} \\ $$$${remains}\:{always}\:{in}\:{vertical}\:{position}, \\ $$$${i}\:{think}. \\ $$
Commented by mr W last updated on 15/Feb/23
θ=30° tan ϕ=((L(1−((√3)/2)))/((L/2)+a))=((2−(√3))/(1+((2a)/L)))=k, say v cos ϕ=ωL ⇒ω=((v cos ϕ)/L)=(v/(L(√(1+k^2 )))) h+((L/2)+a)(1/(cos ϕ))=2a h=2a−((L/2)+a)(√(1+k^2 )) ((Iω^2 )/2)+((mv^2 )/2)=((MgL)/2) sin θ−mg(a−h) ((L^2 ω^2 )/3)+((mv^2 )/M)=((gL)/2)−((2mg)/M)[((L/2)+a)(√(1+k^2 ))−a] [(1/(3(1+k^2 )))+(m/M)]v^2 =((gL)/2){1−((4m)/M)[((1/2)+(a/L))(√(1+k^2 ))−(a/L)]} ⇒v=(√((((gL)/2){1−((4m)/M)[((1/2)+(a/L))(√(1+k^2 ))−(a/L)]})/((1/(3(1+k^2 )))+(m/M))))
$$\theta=\mathrm{30}° \\ $$$$\mathrm{tan}\:\varphi=\frac{{L}\left(\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)}{\frac{{L}}{\mathrm{2}}+{a}}=\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{\mathrm{1}+\frac{\mathrm{2}{a}}{{L}}}={k},\:{say} \\ $$$${v}\:\mathrm{cos}\:\varphi=\omega{L} \\ $$$$\Rightarrow\omega=\frac{{v}\:\mathrm{cos}\:\varphi}{{L}}=\frac{{v}}{{L}\sqrt{\mathrm{1}+{k}^{\mathrm{2}} }} \\ $$$${h}+\left(\frac{{L}}{\mathrm{2}}+{a}\right)\frac{\mathrm{1}}{\mathrm{cos}\:\varphi}=\mathrm{2}{a} \\ $$$${h}=\mathrm{2}{a}−\left(\frac{{L}}{\mathrm{2}}+{a}\right)\sqrt{\mathrm{1}+{k}^{\mathrm{2}} } \\ $$$$\frac{{I}\omega^{\mathrm{2}} }{\mathrm{2}}+\frac{{mv}^{\mathrm{2}} }{\mathrm{2}}=\frac{{MgL}}{\mathrm{2}}\:\mathrm{sin}\:\theta−{mg}\left({a}−{h}\right) \\ $$$$\frac{{L}^{\mathrm{2}} \omega^{\mathrm{2}} }{\mathrm{3}}+\frac{{mv}^{\mathrm{2}} }{{M}}=\frac{{gL}}{\mathrm{2}}−\frac{\mathrm{2}{mg}}{{M}}\left[\left(\frac{{L}}{\mathrm{2}}+{a}\right)\sqrt{\mathrm{1}+{k}^{\mathrm{2}} }−{a}\right] \\ $$$$\left[\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{1}+{k}^{\mathrm{2}} \right)}+\frac{{m}}{{M}}\right]{v}^{\mathrm{2}} =\frac{{gL}}{\mathrm{2}}\left\{\mathrm{1}−\frac{\mathrm{4}{m}}{{M}}\left[\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{{a}}{{L}}\right)\sqrt{\mathrm{1}+{k}^{\mathrm{2}} }−\frac{{a}}{{L}}\right]\right\} \\ $$$$\Rightarrow{v}=\sqrt{\frac{\frac{{gL}}{\mathrm{2}}\left\{\mathrm{1}−\frac{\mathrm{4}{m}}{{M}}\left[\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{{a}}{{L}}\right)\sqrt{\mathrm{1}+{k}^{\mathrm{2}} }−\frac{{a}}{{L}}\right]\right\}}{\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{1}+{k}^{\mathrm{2}} \right)}+\frac{{m}}{{M}}}} \\ $$
Commented by ajfour last updated on 15/Feb/23
I should have mentioned the radius of pulley say R. [ but as soon as rod strikes ground, string gets horizontal, so thats an additional condition given that might help find R =f(L,a)] Under varying tension i think angle of contact of rope with pulley circumference should change.. thanks for solution of this simpler case, Sir.
$${I}\:{should}\:{have}\:\:{mentioned}\:{the}\: \\ $$$${radius}\:{of}\:{pulley}\:{say}\:{R}.\: \\ $$$$\left[\:{but}\:{as}\:{soon}\:{as}\:{rod}\:{strikes}\right. \\ $$$${ground},\:{string}\:{gets}\:{horizontal},\:{so} \\ $$$${thats}\:{an}\:{additional}\:{condition}\:{given} \\ $$$$\left.{that}\:{might}\:{help}\:{find}\:{R}\:={f}\left({L},{a}\right)\right] \\ $$$$\:{Under}\:{varying}\:{tension}\:{i}\:{think} \\ $$$$\:{angle}\:{of}\:{contact}\:{of}\:{rope}\:{with}\:{pulley} \\ $$$${circumference}\:{should}\:{change}.. \\ $$$${thanks}\:{for}\:{solution}\:{of}\:{this}\:{simpler} \\ $$$${case},\:{Sir}. \\ $$
Commented by mr W last updated on 15/Feb/23
why should the string get horizontal when the rod strikes the ground?
$${why}\:{should}\:{the}\:{string}\:{get}\:{horizontal} \\ $$$${when}\:{the}\:{rod}\:{strikes}\:{the}\:{ground}? \\ $$
Commented by ajfour last updated on 15/Feb/23
Thank you sir, you are right, i got misguided by the real mass string and centrifygal force on it as it turns round the pulley..
$${Thank}\:{you}\:{sir},\:{you}\:{are}\:{right},\:{i}\: \\ $$$${got}\:{misguided}\:{by}\:{the}\:{real}\:{mass} \\ $$$${string}\:{and}\:{centrifygal}\:{force}\:{on} \\ $$$${it}\:{as}\:{it}\:{turns}\:{round}\:{the}\:{pulley}.. \\ $$
Commented by ajfour last updated on 16/Feb/23
τ=TL+T(((L(√3))/2)−2R)−mg((L/2))cos θ T=mg+m(dv/dt) v=−L(dθ/dt) (((ML^2 )/3))(dθ/dt)+mv(((L(√3))/2)−2R) ={mg+m(dv/dθ)((dθ/dt))}(L+((L(√3))/2)−2R) −mg((L/2))cos θ say M=pm, R=qL −(p/3)ω+ω(((√3)/2)−2q)+(g/(2L))cos θ =((g/L)+((ωdω)/dθ))(1+((√3)/2)−2q) say 1+((√3)/2)−2q=k (k−1−(p/3))ω+(g/(2L))(cos θ−2k) =((kωdω)/dθ) ....
$$\tau={TL}+{T}\left(\frac{{L}\sqrt{\mathrm{3}}}{\mathrm{2}}−\mathrm{2}{R}\right)−{mg}\left(\frac{{L}}{\mathrm{2}}\right)\mathrm{cos}\:\theta \\ $$$${T}={mg}+{m}\frac{{dv}}{{dt}} \\ $$$${v}=−{L}\frac{{d}\theta}{{dt}} \\ $$$$\left(\frac{{ML}^{\mathrm{2}} }{\mathrm{3}}\right)\frac{{d}\theta}{{dt}}+{mv}\left(\frac{{L}\sqrt{\mathrm{3}}}{\mathrm{2}}−\mathrm{2}{R}\right) \\ $$$$\:\:\:\:=\left\{{mg}+{m}\frac{{dv}}{{d}\theta}\left(\frac{{d}\theta}{{dt}}\right)\right\}\left({L}+\frac{{L}\sqrt{\mathrm{3}}}{\mathrm{2}}−\mathrm{2}{R}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:−{mg}\left(\frac{{L}}{\mathrm{2}}\right)\mathrm{cos}\:\theta \\ $$$${say}\:\:\:{M}={pm},\:\:{R}={qL} \\ $$$$−\frac{{p}}{\mathrm{3}}\omega+\omega\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\mathrm{2}{q}\right)+\frac{{g}}{\mathrm{2}{L}}\mathrm{cos}\:\theta \\ $$$$\:\:\:\:\:\:\:\:\:=\left(\frac{{g}}{{L}}+\frac{\omega{d}\omega}{{d}\theta}\right)\left(\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\mathrm{2}{q}\right) \\ $$$${say}\:\:\:\:\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\mathrm{2}{q}={k} \\ $$$$\left({k}−\mathrm{1}−\frac{{p}}{\mathrm{3}}\right)\omega+\frac{{g}}{\mathrm{2}{L}}\left(\mathrm{cos}\:\theta−\mathrm{2}{k}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{{k}\omega{d}\omega}{{d}\theta} \\ $$$$…. \\ $$

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