Question Number 224859 by fantastic last updated on 08/Oct/25
$${A}\:{homogeneous}\:{rod}\:{AB}\:{of}\:{length} \\ $$$${L}=\mathrm{1}.\mathrm{8}{m}\:{and}\:{mass}\:{M}\:{is}\:{pivoted} \\ $$$${at}\:{the}\:{centre}\:{O}\:{in}\:{such}\:{a}\:{way}\:{that} \\ $$$${it}\:{can}\:{rotate}\:{freely}\:{in}\:{the}\:{vertical}\:{plane}. \\ $$$$ \\ $$$${The}\:{rod}\:{is}\:{initially}\:{in}\:{the}\:{horizontal} \\ $$$${position}.{An}\:{insect}\:{S}\:\:{of}\:{the}\: \\ $$$${same}\:{mass}\:{M}\:{falls}\:{vertically} \\ $$$${with}\:{speed}\:{V}\:{on}\:{point}\:{C},\:{midway} \\ $$$${between}\:{the}\:{points}\:{O}\:{and}\:{B}. \\ $$$${Immediaty}\:{after}\:{falling},\:{the} \\ $$$${insect}\:{moves}\:{towards}\:{the}\:{end}\:{B} \\ $$$${such}\:{that}\:{the}\:{rod}\:{rotates} \\ $$$${with}\:{a}\:{constant}\:{angular}\:{velocity}\:\omega. \\ $$$$\left.{a}\right){determine}\:{the}\:{angular}\:{velocity}\:\omega \\ $$$${in}\:{terms}\:{of}\:{V}\:{and}\:{L}. \\ $$$$\left.{b}\right){If}\:{the}\:{imsect}\:{reaches}\:{the}\:{end}\:{B} \\ $$$${when}\:{the}\:{rod}\:{turned}\:{through} \\ $$$${an}\:{angle}\:\mathrm{90}^{\mathrm{0}} ,\:{determine}\:{V} \\ $$
Commented by fantastic last updated on 08/Oct/25
Answered by mr W last updated on 11/Oct/25
![a) [((ML^2 )/(12))+M×((L/4))^2 ]ω=MV×(L/4) ⇒ω=((12V)/(7L)) b) let r=distance of insect to point O I=((ML^2 )/(12))+Mr^2 ((d(Iω))/dt)=Mgr cos θ 2ωMr(dr/dt)=Mgr cos θ 2ωMrω(dr/dθ)=Mgr cos θ 2ω^2 (dr/dθ)=g cos θ (dr/dθ)=((g cos θ)/(2ω^2 )) ∫_(L/4) ^r dr=(g/(2ω^2 ))∫_0 ^θ cos θ dθ ⇒r=(L/4)+((g sin θ)/(2ω^2 )) at θ=(π/2): r=(L/2) (L/2)=(L/4)+(g/(2ω^2 )) ⇒ω^2 =((2g)/L) (((12V)/(7L)))^2 =((2g)/L) ⇒V=(7/(12))(√(2gL)) note: u=(dr/dt)=ω(dr/dθ)=((g cos θ)/(2ω))≠constant, that means the insect doesn′t move with constant speed towards point B, i.e. the relative speed of the insect to rod isn′t constant.](https://www.tinkutara.com/question/Q224869.png)
$$\left.{a}\right) \\ $$$$\left[\frac{{ML}^{\mathrm{2}} }{\mathrm{12}}+{M}×\left(\frac{{L}}{\mathrm{4}}\right)^{\mathrm{2}} \right]\omega={MV}×\frac{{L}}{\mathrm{4}} \\ $$$$\Rightarrow\omega=\frac{\mathrm{12}{V}}{\mathrm{7}{L}} \\ $$$$\left.{b}\right) \\ $$$${let}\:{r}={distance}\:{of}\:{insect}\:{to}\:{point}\:{O} \\ $$$${I}=\frac{{ML}^{\mathrm{2}} }{\mathrm{12}}+{Mr}^{\mathrm{2}} \\ $$$$\frac{{d}\left({I}\omega\right)}{{dt}}={Mgr}\:\mathrm{cos}\:\theta \\ $$$$\mathrm{2}\omega{Mr}\frac{{dr}}{{dt}}={Mgr}\:\mathrm{cos}\:\theta \\ $$$$\mathrm{2}\omega{Mr}\omega\frac{{dr}}{{d}\theta}={Mgr}\:\mathrm{cos}\:\theta \\ $$$$\mathrm{2}\omega^{\mathrm{2}} \frac{{dr}}{{d}\theta}={g}\:\mathrm{cos}\:\theta \\ $$$$\frac{{dr}}{{d}\theta}=\frac{{g}\:\mathrm{cos}\:\theta}{\mathrm{2}\omega^{\mathrm{2}} } \\ $$$$\int_{\frac{{L}}{\mathrm{4}}} ^{{r}} {dr}=\frac{{g}}{\mathrm{2}\omega^{\mathrm{2}} }\int_{\mathrm{0}} ^{\theta} \mathrm{cos}\:\theta\:{d}\theta \\ $$$$\Rightarrow{r}=\frac{{L}}{\mathrm{4}}+\frac{{g}\:\mathrm{sin}\:\theta}{\mathrm{2}\omega^{\mathrm{2}} } \\ $$$${at}\:\theta=\frac{\pi}{\mathrm{2}}:\:{r}=\frac{{L}}{\mathrm{2}} \\ $$$$\frac{{L}}{\mathrm{2}}=\frac{{L}}{\mathrm{4}}+\frac{{g}}{\mathrm{2}\omega^{\mathrm{2}} } \\ $$$$\Rightarrow\omega^{\mathrm{2}} =\frac{\mathrm{2}{g}}{{L}} \\ $$$$\left(\frac{\mathrm{12}{V}}{\mathrm{7}{L}}\right)^{\mathrm{2}} =\frac{\mathrm{2}{g}}{{L}} \\ $$$$\Rightarrow{V}=\frac{\mathrm{7}}{\mathrm{12}}\sqrt{\mathrm{2}{gL}} \\ $$$${note}: \\ $$$${u}=\frac{{dr}}{{dt}}=\omega\frac{{dr}}{{d}\theta}=\frac{{g}\:\mathrm{cos}\:\theta}{\mathrm{2}\omega}\neq{constant},\:{that} \\ $$$${means}\:{the}\:{insect}\:{doesn}'{t}\:{move}\:{with} \\ $$$${constant}\:{speed}\:{towards}\:{point}\:{B},\:{i}.{e}.\: \\ $$$${the}\:{relative}\:{speed}\:{of}\:{the}\:{insect}\:{to}\: \\ $$$${rod}\:{isn}'{t}\:{constant}. \\ $$
Commented by mr W last updated on 09/Oct/25
Commented by fantastic last updated on 09/Oct/25
$${sir}\:{the}\:{answers}\:{are} \\ $$$$\omega=\frac{\mathrm{12}{V}}{\mathrm{7}{L}}\:{and}\:{V}=\frac{\mathrm{7}}{\mathrm{12}}\sqrt{\mathrm{2}{gL}} \\ $$