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Question Number 214132 by ajfour last updated on 29/Nov/24

Answered by MathematicalUser2357 last updated on 29/Nov/24

$$b^2=a^2+c^2-2ac\cos \theta$$$$\cos \theta =\frac{a^2-b^2+c^2}{2ac}$$$$\mathrm{Apply}\mathrm{the}\mathrm{formula}\cos \theta =\sqrt{1-{\sin }^2\theta }$$$$\sin \theta =\sqrt{1-{\left(\frac{a^2-b^2+c^2}{2ac}\right)}^2}$$$$\mathrm{Summarize}\mathrm{height}h=a\sin \theta +r$$$$h=a\sin \theta +r=a\sqrt{1-{\left(\frac{a^2-b^2+c^2}{2ac}\right)}^2}+r$$$$$$$$\mathrm{Calculate}\mathrm{time}\mathrm{with}\mathrm{equation}$$$$\mathrm{Since}\mathrm{the}\mathrm{height}\mathrm{equation}\mathrm{is}-t^2+h$$$$\mathrm{So}\mathrm{let}f\left(t\right)=-t^2+h$$$$f\left(t\right)=-t^2+h=0$$$$t=\sqrt{h}\Rightarrow \mathrm{So},\mathrm{time}=t=\sqrt{a\sqrt{1-{\left(\frac{a^2-b^2+c^2}{2ac}\right)}^2}+r}$$$$$$$$\mathrm{Calculate}\mathrm{speed}\mathrm{with}\mathrm{formula}s=\frac{d}{dt}f\left(t\right)$$$$\frac{d}{dt}f\left(t\right)=\frac{d}{dt}(-t^2+h)=-2t\Rightarrow \mathrm{So},\mathrm{speed}=-2t=-2\sqrt{a\sqrt{1-{\left(\frac{a^2-b^2+c^2}{2ac}\right)}^2}+r}$$$$$$$$\mathrm{Please}\mathrm{correct}\mathrm{me}\mathrm{if}{\mathrm{i}}^{\prime }\mathrm{m}\mathrm{wrong}$$

Commented by MathematicalUser2357 last updated on 29/Nov/24

Commented by ajfour last updated on 29/Nov/24

$$-t^2+h=0$$$$how?$$$$$$

Commented by MathematicalUser2357 last updated on 29/Nov/24

$$\mathrm{Since}\mathrm{the}\mathrm{height}\mathrm{equation}\mathrm{is}-t^2+h,\mathrm{So}\mathrm{let}f\left(t\right)=-t^2+h$$$$\mathrm{Plus},\mathrm{to}\mathrm{the}\mathrm{gravity}\mathrm{law},\mathrm{the}\mathrm{height}\mathrm{equation}\mathrm{is}y=-t^2+h$$

Commented by ajfour last updated on 29/Nov/24

$$Itsnotafreefall..$$

Commented by MathematicalUser2357 last updated on 01/Dec/24

$$\mathrm{Then}\mathrm{linear}\mathrm{motion}?$$$$\{\begin{array}{l}{x}_0=a\sqrt{1-{\left(\frac{a^2-b^2+c^2}{2ac}\right)}^2}+r\\ x=0\\ v_0=0\\ v=\alpha t\end{array}$$$$\{\begin{array}{l}x=a\sqrt{1-{\left(\frac{a^2-b^2+c^2}{2ac}\right)}^2}+r+\frac{1}{2}\alpha t^2\\ x=a\sqrt{1-{\left(\frac{a^2-b^2+c^2}{2ac}\right)}^2}+r+vt-\frac{1}{2}\alpha t^2\\ x=a\sqrt{1-{\left(\frac{a^2-b^2+c^2}{2ac}\right)}^2}+r+\frac{1}{2}vt\end{array}$$$$uh-$$

Commented by mr W last updated on 01/Dec/24

Commented by mr W last updated on 03/Dec/24

$$thispuctureshowsthe$$$$motionsdescribedinthisquestion.$$$$forexampleyouarestandingon$$$$thetopofsuchatwinladder.$$$$sudentlythespreadersconnecting$$$$theladdersarebrocken...$$$$thequestioniswithwhatspeedyou$$$$hitthefloorandwhattimedoyou$$$$take.$$$$thefloorisslippery,nofriction.$$$$ihopeyou{didn}^{\prime }tgetserioulyinjured.$$

Answered by mr W last updated on 01/Dec/24

Commented by mr W last updated on 01/Dec/24

$$rodshaveuniformmassdistribution.$$$$rodAP:lengtha,mass{m}_1$$$$rodBP:lengthb,mass{m}_2$$$$M=m_1+m_2$$$$say{\mu }_1=\frac{m_1}{M},{\mu }_2=\frac{m_2}{M}$$$$C=centerofmassofbothrodstogether$$$$h_C=heightofCaboveground=\frac{h}{2}$$$$$$$$att=0:$$$$\theta ={\theta }_0={\cos }^{-1}\frac{a^2+b^2-c^2}{2ab}$$$$\frac{h_{0}c}{2}={\mathrm{\Delta }}_{ABP}=\frac{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}{4}$$$$\Rightarrow h_0=\frac{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}{2c}$$$$h_{C0}=\frac{h_0}{2}$$$$$$$$att=T:$$$$\theta =\pi =180°$$$$$$$$att:$$$$let\omega =\frac{d\theta }{dt}$$$$E,Faremasscentersofrods$$$$EP=\frac{a}{2},FP=\frac{b}{2}$$$$AB=l=\sqrt{a^2+b^2-2ab\cos \theta }$$$$\frac{dl}{d\theta }=\frac{ab\sin \theta }{\sqrt{a^2+b^2-2ab\cos \theta }}=\frac{ab\sin \theta }{l}$$$$EF=\frac{l}{2}$$$$EC=l_1=\frac{m_2}{m_1+m_2}\times EF=\frac{m_{2}l}{2M}$$$$FC=l_2=\frac{m_1}{m_1+m_2}\times EF=\frac{m_{1}l}{2M}$$$$pointCmovesonlyinvertical$$$$direction,becausenoforceisacting$$$$inhorizontaldirection.$$$${\omega }_1=\frac{d{\phi }_1}{dt}=\frac{d{\phi }_1}{d\theta }\times \frac{d\theta }{dt}=\omega \frac{d{\phi }_1}{d\theta }$$$${\omega }_2=\frac{d{\phi }_2}{dt}=\frac{d{\phi }_2}{d\theta }\times \frac{d\theta }{dt}=\omega \frac{d{\phi }_2}{d\theta }$$$$\frac{\sin {\phi }_1}{b}=\frac{\sin {\phi }_2}{a}=\frac{\sin \theta }{l}$$$$\frac{\cos {\phi }_1}{b}\times \frac{d{\phi }_1}{d\theta }=\frac{\cos {\phi }_2}{a}\times \frac{d{\phi }_2}{d\theta }=\frac{\cos \theta }{l}-\frac{\sin \theta }{l^2}\times \frac{dl}{d\theta }$$$$\frac{\cos {\phi }_1}{b}\times \frac{{\omega }_1}{\omega }=\frac{\cos {\phi }_2}{a}\times \frac{{\omega }_2}{\omega }=\frac{\cos \theta }{l}-\frac{ab{\sin }^2\theta }{l^3}$$$${\omega }_1=(\cos \theta -\frac{ab{\sin }^2\theta }{l^2})\frac{\omega b}{l\cos {\phi }_1}$$$${\omega }_1=(\cos \theta -\frac{ab{\sin }^2\theta }{l^2})\frac{\omega b}{l\sqrt{1-{\sin }^2{\phi }_1}}$$$$\Rightarrow {\omega }_1=(\cos \theta -\frac{ab{\sin }^2\theta }{l^2})\frac{\omega }{\sqrt{\frac{l^2}{b^2}-{\sin }^2\theta }}$$$$similarly$$$$\Rightarrow {\omega }_2=(\cos \theta -\frac{ab{\sin }^2\theta }{l^2})\frac{\omega }{\sqrt{\frac{l^2}{a^2}-{\sin }^2\theta }}$$$$\frac{2h_{C}l}{2}=\frac{ab\sin \theta }{2}={\mathrm{\Delta }}_{ABP}$$$$\Rightarrow h_C=\frac{ab\sin \theta }{2l}$$$$\frac{{dh}_C}{d\theta }=\frac{ab}{2}(\frac{\cos \theta }{l}-\frac{\sin \theta }{l^2}\times \frac{dl}{d\theta })=\frac{ab}{2l}(\cos \theta -\frac{ab{\sin }^2\theta }{l^2})$$$$u_{1x}=\frac{{dl}_1}{dt}=\frac{m_2\omega }{2M}\times \frac{dl}{d\theta }=\frac{m_2\omega ab\sin \theta }{2Ml}$$$$u_{2x}=\frac{{dl}_2}{dt}=\frac{m_1\omega }{2M}\times \frac{dl}{d\theta }=\frac{m_1\omega ab\sin \theta }{2Ml}$$$$u_{1y}=u_{2y}=-\frac{{dh}_C}{dt}=-\omega \frac{{dh}_C}{d\theta }=-\frac{\omega ab}{2l}(\cos \theta -\frac{ab{\sin }^2\theta }{l^2})$$$$lossofP.E.=gainofK.E.$$$$Mg(h_{C0}-h_C)=\frac{1}{2}{m}_1(u_{1x}^2+u_{1y}^2)+\frac{1}{2}\times \frac{m_1{a}^2{\omega }_1^2}{12}$$$$+\frac{1}{2}{m}_2(u_{2x}^2+u_{2y}^2)+\frac{1}{2}\times \frac{m_2{b}^2{\omega }_2^2}{12}$$$$2Mg(h_{C0}-h_C)={Mu}_{1y}^2+m_1{u}_{1x}^2+m_2{u}_{2x}^2+\frac{m_1{a}^2{\omega }_1^2}{12}+\frac{m_2{b}^2{\omega }_2^2}{12}$$$$2Mg(h_{C0}-h_C)=M{\left(\frac{\omega ab}{2l}\right)}^2{(\cos \theta -\frac{ab{\sin }^2\theta }{l^2})}^2+(m_1{m}_2^2+m_2{m}_1^2){\left(\frac{\omega ab\sin \theta }{2Ml}\right)}^2$$$$+\frac{m_1{a}^2}{12}\times {(\cos \theta -\frac{ab{\sin }^2\theta }{l^2})}^2\left(\frac{{\omega }^2}{\frac{l^2}{b^2}-{\sin }^2\theta }\right)+\frac{m_2{b}^2}{12}\times {(\cos \theta -\frac{ab{\sin }^2\theta }{l^2})}^2\left(\frac{{\omega }^2}{\frac{l^2}{b^2}-{\sin }^2\theta }\right)$$$$8Mg(h_{C0}-\frac{ab\sin \theta }{2l})={(\cos \theta -\frac{ab{\sin }^2\theta }{l^2})}^2\times \frac{{Ma}^2{b}^2{\omega }^2}{l^2}+\frac{m_1{m}_2{a}^2{b}^2{\omega }^2{\sin }^2\theta }{{Ml}^2}$$$$+\frac{1}{3}{(\cos \theta -\frac{ab{\sin }^2\theta }{l^2})}^2(\frac{m_1{a}^2}{\frac{l^2}{b^2}-{\sin }^2\theta }+\frac{m_2{b}^2}{\frac{l^2}{b^2}-{\sin }^2\theta }){\omega }^2$$$$4Mg(h_0-\frac{ab\sin \theta }{l})=\frac{m_1{m}_2{a}^2{b}^2{\omega }^2{\sin }^2\theta }{{Ml}^2}+\frac{1}{3}{(\cos \theta -\frac{ab{\sin }^2\theta }{l^2})}^2(\frac{m_1{a}^2}{\frac{l^2}{b^2}-{\sin }^2\theta }+\frac{m_2{b}^2}{\frac{l^2}{a^2}-{\sin }^2\theta }+\frac{3{Ma}^2{b}^2}{l^2}){\omega }^2$$$${\omega }^2=\frac{4Mg(h_0-\frac{ab\sin \theta }{l})}{\frac{m_1{m}_2{a}^2{b}^2{\sin }^2\theta }{{Ml}^2}+\frac{1}{3}{(\cos \theta -\frac{ab{\sin }^2\theta }{l^2})}^2(\frac{m_1{a}^2}{\frac{l^2}{b^2}-{\sin }^2\theta }+\frac{m_2{b}^2}{\frac{l^2}{a^2}-{\sin }^2\theta }+\frac{3{Ma}^2{b}^2}{l^2})}$$$$\omega =\sqrt{\frac{4Mg(h_0-\frac{ab\sin \theta }{l})}{\frac{m_1{m}_2{a}^2{b}^2{\sin }^2\theta }{{Ml}^2}+\frac{1}{3}{(\cos \theta -\frac{ab{\sin }^2\theta }{l^2})}^2(\frac{m_1{a}^2}{\frac{l^2}{b^2}-{\sin }^2\theta }+\frac{m_2{b}^2}{\frac{l^2}{a^2}-{\sin }^2\theta }+\frac{3{Ma}^2{b}^2}{l^2})}}$$$$\Rightarrow \omega =\frac{1}{ab}\sqrt{\frac{4g(h_0-\frac{ab\sin \theta }{l})}{\frac{{\mu }_1{\mu }_2{\sin }^2\theta }{l^2}+\frac{1}{3}{(\cos \theta -\frac{ab{\sin }^2\theta }{l^2})}^2(\frac{{\mu }_1}{l^2-b^2{\sin }^2\theta }+\frac{{\mu }_2}{l^2-a^2{\sin }^2\theta }+\frac{3}{l^2})}}$$$$dt=\frac{d\theta }{\omega }$$$$dt=ab\sqrt{\frac{\frac{{\mu }_1{\mu }_2{\sin }^2\theta }{l^2}+\frac{1}{3}{(\cos \theta -\frac{ab{\sin }^2\theta }{l^2})}^2(\frac{{\mu }_1}{l^2-b^2{\sin }^2\theta }+\frac{{\mu }_2}{l^2-a^2{\sin }^2\theta }+\frac{3}{l^2})}{4g(h_0-\frac{ab\sin \theta }{l})}}d\theta$$$$T=ab{\int }_{{\theta }_0}^{\pi }\sqrt{\frac{\frac{{\mu }_1{\mu }_2{\sin }^2\theta }{l^2}+\frac{1}{3}{(\cos \theta -\frac{ab{\sin }^2\theta }{l^2})}^2(\frac{{\mu }_1}{l^2-b^2{\sin }^2\theta }+\frac{{\mu }_2}{l^2-a^2{\sin }^2\theta }+\frac{3}{l^2})}{4g(h_0-\frac{ab\sin \theta }{l})}}d\theta$$$$u_{Py}=a\cos {\phi }_1{\omega }_1=a(\frac{\cos \theta }{l}-\frac{ab{\sin }^2\theta }{l^3})b\omega$$$$\Rightarrow u_{Py}=(\cos \theta -\frac{ab{\sin }^2\theta }{l^2})\frac{\omega ab}{l}$$$$$$$$att=T:$$$$\theta =\pi$$$$\omega =\frac{(a+b)\sqrt{3{gh}_0}}{ab}$$$$u_{Py}=\sqrt{3{gh}_0}>\sqrt{2{gh}_0}$$$$recall:aballreleasedatpointP$$$$hitsthegroundwithaspeed\sqrt{2{gh}_0}.$$$$++++++++++$$$${\omega }_1=(\cos \theta -\frac{ab{\sin }^2\theta }{l^2})\frac{\omega }{\sqrt{\frac{l^2}{b^2}-{\sin }^2\theta }}$$$${\alpha }_1=\frac{d{\omega }_1}{dt}=\frac{d{\omega }_1}{d\theta }\times \frac{d\theta }{dt}=\omega \frac{d{\omega }_1}{d\theta }$$$$-N_{1}a\cos {\phi }_1+m_{1}g\times \frac{a\cos {\phi }_1}{2}=\frac{m_1{a}^2{\alpha }_1}{3}$$$$\frac{N_1}{m_{1}g}=\frac{1}{2}-\frac{a}{3g\cos {\phi }_1}\times \omega \frac{d{\omega }_1}{d\theta }$$$$\frac{N_1}{m_{1}g}=\frac{1}{2}-\frac{a\omega }{3g\sqrt{1-\frac{b^2{\sin }^2\theta }{l^2}}}\times \frac{d{\omega }_1}{d\theta }$$$$similarly$$$$\frac{N_2}{m_{2}g}=\frac{1}{2}-\frac{b\omega }{3g\sqrt{1-\frac{a^2{\sin }^2\theta }{l^2}}}\times \frac{d{\omega }_2}{d\theta }$$

Commented by mr W last updated on 30/Nov/24

Commented by mr W last updated on 30/Nov/24

$$inallthesecasesthepointPhits$$$$thegroundwiththesamespeed$$$$regardlesswhichrelationthe$$$$lengthesandthemassesoftherods$$$$have.$$

Commented by mr W last updated on 30/Nov/24

Commented by mr W last updated on 30/Nov/24

$$howisitpossiblethatthepointP$$$$hitsthegroundwithalargerspeed$$$$thanaballfromthesameheight?$$

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