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Question Number 214132 by ajfour last updated on 29/Nov/24
Answered by MathematicalUser2357 last updated on 29/Nov/24
b^2 =a^2 +c^2 −2ac cos θ cos θ=((a^2 −b^2 +c^2 )/(2ac)) Apply the formula cos θ=(√(1−sin^2 θ)) sin θ=(√(1−(((a^2 −b^2 +c^2 )/(2ac)))^2 )) Summarize height h=a sin θ+r h=a sin θ+r=a(√(1−(((a^2 −b^2 +c^2 )/(2ac)))^2 ))+r Calculate time with equation Since the height equation is −t^2 +h So let f(t)=−t^2 +h f(t)=−t^2 +h=0 t=(√h) ⇒ So, time=t=(√(a(√(1−(((a^2 −b^2 +c^2 )/(2ac)))^2 ))+r)) Calculate speed with formula s=(d/dt)f(t) (d/dt)f(t)=(d/dt)(−t^2 +h)=−2t ⇒ So, speed=−2t=−2(√(a(√(1−(((a^2 −b^2 +c^2 )/(2ac)))^2 ))+r)) Please correct me if i′m wrong
$${b}^{\mathrm{2}} ={a}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{ac}\:\mathrm{cos}\:\theta \\ $$$$\mathrm{cos}\:\theta=\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{2}{ac}} \\ $$$$\mathrm{Apply}\:\mathrm{the}\:\mathrm{formula}\:\mathrm{cos}\:\theta=\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \theta} \\ $$$$\mathrm{sin}\:\theta=\sqrt{\mathrm{1}−\left(\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{2}{ac}}\right)^{\mathrm{2}} } \\ $$$$\mathrm{Summarize}\:\mathrm{height}\:{h}={a}\:\mathrm{sin}\:\theta+{r} \\ $$$${h}={a}\:\mathrm{sin}\:\theta+{r}={a}\sqrt{\mathrm{1}−\left(\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{2}{ac}}\right)^{\mathrm{2}} }+{r} \\ $$$$ \\ $$$$\mathrm{Calculate}\:\mathrm{time}\:\mathrm{with}\:\mathrm{equation} \\ $$$$\mathrm{Since}\:\mathrm{the}\:\mathrm{height}\:\mathrm{equation}\:\mathrm{is}\:−{t}^{\mathrm{2}} +{h} \\ $$$$\mathrm{So}\:\mathrm{let}\:{f}\left({t}\right)=−{t}^{\mathrm{2}} +{h} \\ $$$${f}\left({t}\right)=−{t}^{\mathrm{2}} +{h}=\mathrm{0} \\ $$$${t}=\sqrt{{h}}\:\Rightarrow\:\mathrm{So},\:\mathrm{time}={t}=\sqrt{{a}\sqrt{\mathrm{1}−\left(\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{2}{ac}}\right)^{\mathrm{2}} }+{r}} \\ $$$$ \\ $$$$\mathrm{Calculate}\:\mathrm{speed}\:\mathrm{with}\:\mathrm{formula}\:{s}=\frac{{d}}{{dt}}{f}\left({t}\right) \\ $$$$\frac{{d}}{{dt}}{f}\left({t}\right)=\frac{{d}}{{dt}}\left(−{t}^{\mathrm{2}} +{h}\right)=−\mathrm{2}{t}\:\Rightarrow\:\mathrm{So},\:\mathrm{speed}=−\mathrm{2}{t}=−\mathrm{2}\sqrt{{a}\sqrt{\mathrm{1}−\left(\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{2}{ac}}\right)^{\mathrm{2}} }+{r}} \\ $$$$ \\ $$$$\mathrm{Please}\:\mathrm{correct}\:\mathrm{me}\:\mathrm{if}\:\mathrm{i}'\mathrm{m}\:\mathrm{wrong} \\ $$
Commented by MathematicalUser2357 last updated on 29/Nov/24
Commented by ajfour last updated on 29/Nov/24
−t^2 +h=0 how?
$$−{t}^{\mathrm{2}} +{h}=\mathrm{0} \\ $$$${how}? \\ $$$$ \\ $$
Commented by MathematicalUser2357 last updated on 29/Nov/24
Since the height equation is −t^2 +h, So let f(t)=−t^2 +h Plus, to the gravity law, the height equation is y=−t^2 +h
$$\mathrm{Since}\:\mathrm{the}\:\mathrm{height}\:\mathrm{equation}\:\mathrm{is}\:−{t}^{\mathrm{2}} +{h},\:\mathrm{So}\:\mathrm{let}\:{f}\left({t}\right)=−{t}^{\mathrm{2}} +{h} \\ $$$$\mathrm{Plus},\:\mathrm{to}\:\mathrm{the}\:\mathrm{gravity}\:\mathrm{law},\:\mathrm{the}\:\mathrm{height}\:\mathrm{equation}\:\mathrm{is}\:{y}=−{t}^{\mathrm{2}} +{h} \\ $$
Commented by ajfour last updated on 29/Nov/24
Its not a free fall..
$${Its}\:{not}\:{a}\:{free}\:{fall}.. \\ $$
Answered by mr W last updated on 29/Nov/24
Commented by mr W last updated on 30/Nov/24
M=m_1 +m_2 at t=0: θ_0 =cos^(−1) ((a^2 +b^2 −c^2 )/(2ab)) h_(C0) c=Δ_(ABP) =((√((a+b+c)(−a+b+c)(a−b+c)(a+b−c)))/4) ⇒h_(C0) =((√((a+b+c)(−a+b+c)(a−b+c)(a+b−c)))/(4c)) at t=T: θ=π=180° at t: let ω=(dθ/dt) S, T are mass centers of rods SP=(a/2), TP=(b/2) C=mass center of both rods AB=l=(√(a^2 +b^2 −2ab cos θ)) (dl/dθ)=((ab sin θ)/( (√(a^2 +b^2 −2ab cos θ))))=((ab sin θ)/l) ST=(l/2) SC=(m_2 /(m_1 +m_2 ))×ST=((m_2 l)/(2M)) TC=(m_1 /(m_1 +m_2 ))×ST=((m_1 l)/(2M)) point C moves only in vertical direction. ω_1 =(dϕ_1 /dt)=ω(dϕ_1 /dθ) ω_2 =(dϕ_2 /dt)=ω(dϕ_2 /dθ) ((sin ϕ_1 )/b)=((sin ϕ_2 )/a)=((sin θ)/( l)) ((cos ϕ_1 )/b)×(dϕ_1 /dθ)=((cos ϕ_2 )/a)×(dϕ_2 /dθ)=((cos θ)/l)−((sin θ)/l^2 )×(dl/dθ) ((cos ϕ_1 )/b)×ω_1 =((cos ϕ_2 )/a)×ω_2 =(cos θ−((ab sin^2 θ)/l^2 ))(ω/l) ω_1 =(cos θ−((ab sin^2 θ)/l^2 ))((ωb)/(l cos ϕ_1 )) ⇒ω_1 =(cos θ−((ab sin^2 θ)/l^2 ))(ω/( (√((l^2 /b^2 )− sin^2 θ)))) similarly ⇒ω_2 =(cos θ−((ab sin^2 θ)/l^2 ))(ω/( (√((l^2 /a^2 )− sin^2 θ)))) lh_C =((ab sin θ)/2)=Δ_(ABC) ⇒h_C =((ab sin θ)/(2l)) u_(Sx) =((d(SC))/dt)=((m_2 ω)/(2M))×(dl/dθ)=((m_2 ωab sin θ)/(2Ml)) u_(Tx) =((d(TC))/dt)=((m_1 ω)/(2M))×(dl/dθ)=((m_1 ωab sin θ)/(2Ml)) u_(Sy) =u_(Ty) =−(dh_C /dt)=−ω(dh_C /dθ)=−((ωab)/2)(((cos θ)/l)−((sin θ)/l^2 )×((ab sin θ)/l)) u_(Sy) =u_(Ty) =−((ωab)/(2l))(cos θ−((ab sin^2 θ)/l^2 )) Mg(h_(C0) −h_C )=(1/2)m_1 (u_(Sx) ^2 +u_(Sy) ^2 )+(1/2)×((m_1 a^2 ω_1 ^2 )/(12)) +(1/2)m_2 (u_(Tx) ^2 +u_(Ty) ^2 )+(1/2)×((m_2 b^2 ω_2 ^2 )/(12)) ......
$${M}={m}_{\mathrm{1}} +{m}_{\mathrm{2}} \\ $$$${at}\:{t}=\mathrm{0}: \\ $$$$\theta_{\mathrm{0}} =\mathrm{cos}^{−\mathrm{1}} \frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}} \\ $$$${h}_{{C}\mathrm{0}} {c}=\Delta_{{ABP}} =\frac{\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)}}{\mathrm{4}} \\ $$$$\Rightarrow{h}_{{C}\mathrm{0}} =\frac{\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)}}{\mathrm{4}{c}} \\ $$$${at}\:{t}={T}: \\ $$$$\theta=\pi=\mathrm{180}° \\ $$$${at}\:{t}: \\ $$$${let}\:\omega=\frac{{d}\theta}{{dt}} \\ $$$${S},\:{T}\:{are}\:{mass}\:{centers}\:{of}\:{rods} \\ $$$${SP}=\frac{{a}}{\mathrm{2}},\:{TP}=\frac{{b}}{\mathrm{2}} \\ $$$${C}={mass}\:{center}\:{of}\:{both}\:{rods} \\ $$$${AB}={l}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\:\mathrm{cos}\:\theta} \\ $$$$\frac{{dl}}{{d}\theta}=\frac{{ab}\:\mathrm{sin}\:\theta}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\:\mathrm{cos}\:\theta}}=\frac{{ab}\:\mathrm{sin}\:\theta}{{l}} \\ $$$${ST}=\frac{{l}}{\mathrm{2}} \\ $$$${SC}=\frac{{m}_{\mathrm{2}} }{{m}_{\mathrm{1}} +{m}_{\mathrm{2}} }×{ST}=\frac{{m}_{\mathrm{2}} {l}}{\mathrm{2}{M}} \\ $$$${TC}=\frac{{m}_{\mathrm{1}} }{{m}_{\mathrm{1}} +{m}_{\mathrm{2}} }×{ST}=\frac{{m}_{\mathrm{1}} {l}}{\mathrm{2}{M}} \\ $$$${point}\:{C}\:{moves}\:{only}\:{in}\:{vertical} \\ $$$${direction}. \\ $$$$\omega_{\mathrm{1}} =\frac{{d}\varphi_{\mathrm{1}} }{{dt}}=\omega\frac{{d}\varphi_{\mathrm{1}} }{{d}\theta} \\ $$$$\omega_{\mathrm{2}} =\frac{{d}\varphi_{\mathrm{2}} }{{dt}}=\omega\frac{{d}\varphi_{\mathrm{2}} }{{d}\theta} \\ $$$$ \\ $$$$\frac{\mathrm{sin}\:\varphi_{\mathrm{1}} }{{b}}=\frac{\mathrm{sin}\:\varphi_{\mathrm{2}} }{{a}}=\frac{\mathrm{sin}\:\theta}{\:{l}} \\ $$$$\frac{\mathrm{cos}\:\varphi_{\mathrm{1}} }{{b}}×\frac{{d}\varphi_{\mathrm{1}} }{{d}\theta}=\frac{\mathrm{cos}\:\varphi_{\mathrm{2}} }{{a}}×\frac{{d}\varphi_{\mathrm{2}} }{{d}\theta}=\frac{\mathrm{cos}\:\theta}{{l}}−\frac{\mathrm{sin}\:\theta}{{l}^{\mathrm{2}} }×\frac{{dl}}{{d}\theta} \\ $$$$\frac{\mathrm{cos}\:\varphi_{\mathrm{1}} }{{b}}×\omega_{\mathrm{1}} =\frac{\mathrm{cos}\:\varphi_{\mathrm{2}} }{{a}}×\omega_{\mathrm{2}} =\left(\mathrm{cos}\:\theta−\frac{{ab}\:\mathrm{sin}^{\mathrm{2}} \:\theta}{{l}^{\mathrm{2}} }\right)\frac{\omega}{{l}} \\ $$$$\omega_{\mathrm{1}} =\left(\mathrm{cos}\:\theta−\frac{{ab}\:\mathrm{sin}^{\mathrm{2}} \:\theta}{{l}^{\mathrm{2}} }\right)\frac{\omega{b}}{{l}\:\mathrm{cos}\:\varphi_{\mathrm{1}} } \\ $$$$\Rightarrow\omega_{\mathrm{1}} =\left(\mathrm{cos}\:\theta−\frac{{ab}\:\mathrm{sin}^{\mathrm{2}} \:\theta}{{l}^{\mathrm{2}} }\right)\frac{\omega}{\:\sqrt{\frac{{l}^{\mathrm{2}} }{{b}^{\mathrm{2}} }−\:\mathrm{sin}^{\mathrm{2}} \:\theta}} \\ $$$${similarly} \\ $$$$\Rightarrow\omega_{\mathrm{2}} =\left(\mathrm{cos}\:\theta−\frac{{ab}\:\mathrm{sin}^{\mathrm{2}} \:\theta}{{l}^{\mathrm{2}} }\right)\frac{\omega}{\:\sqrt{\frac{{l}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\:\mathrm{sin}^{\mathrm{2}} \:\theta}} \\ $$$${lh}_{{C}} =\frac{{ab}\:\mathrm{sin}\:\theta}{\mathrm{2}}=\Delta_{{ABC}} \\ $$$$\Rightarrow{h}_{{C}} =\frac{{ab}\:\mathrm{sin}\:\theta}{\mathrm{2}{l}} \\ $$$${u}_{{Sx}} =\frac{{d}\left({SC}\right)}{{dt}}=\frac{{m}_{\mathrm{2}} \omega}{\mathrm{2}{M}}×\frac{{dl}}{{d}\theta}=\frac{{m}_{\mathrm{2}} \omega{ab}\:\mathrm{sin}\:\theta}{\mathrm{2}{Ml}} \\ $$$${u}_{{Tx}} =\frac{{d}\left({TC}\right)}{{dt}}=\frac{{m}_{\mathrm{1}} \omega}{\mathrm{2}{M}}×\frac{{dl}}{{d}\theta}=\frac{{m}_{\mathrm{1}} \omega{ab}\:\mathrm{sin}\:\theta}{\mathrm{2}{Ml}} \\ $$$${u}_{{Sy}} ={u}_{{Ty}} =−\frac{{dh}_{{C}} }{{dt}}=−\omega\frac{{dh}_{{C}} }{{d}\theta}=−\frac{\omega{ab}}{\mathrm{2}}\left(\frac{\mathrm{cos}\:\theta}{{l}}−\frac{\mathrm{sin}\:\theta}{{l}^{\mathrm{2}} }×\frac{{ab}\:\mathrm{sin}\:\theta}{{l}}\right) \\ $$$${u}_{{Sy}} ={u}_{{Ty}} =−\frac{\omega{ab}}{\mathrm{2}{l}}\left(\mathrm{cos}\:\theta−\frac{{ab}\:\mathrm{sin}^{\mathrm{2}} \:\theta}{{l}^{\mathrm{2}} }\right) \\ $$$${Mg}\left({h}_{{C}\mathrm{0}} −{h}_{{C}} \right)=\frac{\mathrm{1}}{\mathrm{2}}{m}_{\mathrm{1}} \left({u}_{{Sx}} ^{\mathrm{2}} +{u}_{{Sy}} ^{\mathrm{2}} \right)+\frac{\mathrm{1}}{\mathrm{2}}×\frac{{m}_{\mathrm{1}} {a}^{\mathrm{2}} \omega_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{12}} \\ $$$$\:\:\:\:\:\:\:\:\:+\frac{\mathrm{1}}{\mathrm{2}}{m}_{\mathrm{2}} \left({u}_{{Tx}} ^{\mathrm{2}} +{u}_{{Ty}} ^{\mathrm{2}} \right)+\frac{\mathrm{1}}{\mathrm{2}}×\frac{{m}_{\mathrm{2}} {b}^{\mathrm{2}} \omega_{\mathrm{2}} ^{\mathrm{2}} }{\mathrm{12}} \\ $$$$…… \\ $$

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