Menu Close

Question-228146

Tinku Tara 0 Comments
Pin




Question Number 228146 by ajfour last updated on 31/Mar/26
Commented by ajfour last updated on 31/Mar/26
Find ω when rod begins to slip along its length on the edge.
$${Find}\:\omega\:{when}\:{rod}\:{begins}\:{to}\:{slip} \\ $$$${along}\:{its}\:{length}\:{on}\:{the}\:{edge}. \\ $$
Answered by mr W last updated on 31/Mar/26
Commented by mr W last updated on 31/Mar/26
let c=((b−a)/2) I=((m(a+b)^2 )/(12))−m(((b−a)/2))^2 =((m(a^2 +b^2 −ab))/3) Iα=mgc cos θ ((m(a^2 +b^2 −ab))/3)×ω(dω/dθ)=mgc cos θ (((a^2 +b^2 −ab))/(3gc))×ωdω=cos θ dθ (((a^2 +b^2 −ab))/(3gc))∫_0 ^ω ωdω=∫_0 ^θ cos θ dθ (((a^2 +b^2 −ab)ω^2 )/(6gc))=sin θ ⇒ω=(√((6gc sin θ)/(a^2 +b^2 −ab)))=(√((6gλ sin θ)/c)) with λ=(c^2 /(a^2 +b^2 −ab))=(((b−a)^2 )/(4(a^2 +b^2 −ab))) mg cos θ−N=mcα ⇒N=mg cos θ−mc×((3gc cos θ)/((a^2 +b^2 −ab))) ⇒N=mg cos θ(1−((3c^2 )/(a^2 +b^2 −ab)))>0 f−mg sin θ=mcω^2 ⇒f=mg sin θ+mc×((6gc sin θ)/((a^2 +b^2 −ab))) ⇒f=mg sin θ(1+((6c^2 )/(a^2 +b^2 −ab))) such that no slipping, (f/N)≤μ ((mg sin θ(1+((6c^2 )/(a^2 +b^2 −ab))))/(mg cos θ(1−((3c^2 )/(a^2 +b^2 −ab)))))≤μ ⇒tan θ≤((μ(1−((3c^2 )/(a^2 +b^2 −ab))))/(1+((6c^2 )/(a^2 +b^2 −ab))))=((μ(1−3λ))/(1+6λ)) ⇒sin θ_(max) =((μ(1−3λ))/( (√((1+6λ)^2 +μ^2 (1−3λ)^2 )))) ⇒ω_(max) =(√((12gλμ(1−3λ))/( (b−a)(√((1+6λ)^2 +μ^2 (1−3λ)^2 ))))) example: a=3m, b=5m, μ=0.8 ⇒c=((5−3)/2)=1 ⇒λ=(1^2 /(3^2 +5^2 −3×5))=(1/(19)) tan θ_(max) =((0.8(1−(3/(19))))/(1+(6/(19))))=((64)/(125)) ⇒θ_(max) ≈27.1° ω_(max) =(√((6×9.81×1 sin θ_(max) )/(3^2 +5^2 −3×5))) ≈1.118 rad/s
$${let}\:{c}=\frac{{b}−{a}}{\mathrm{2}} \\ $$$${I}=\frac{{m}\left({a}+{b}\right)^{\mathrm{2}} }{\mathrm{12}}−{m}\left(\frac{{b}−{a}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{{m}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{ab}\right)}{\mathrm{3}} \\ $$$${I}\alpha={mgc}\:\mathrm{cos}\:\theta \\ $$$$\frac{{m}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{ab}\right)}{\mathrm{3}}×\omega\frac{{d}\omega}{{d}\theta}={mgc}\:\mathrm{cos}\:\theta \\ $$$$\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{ab}\right)}{\mathrm{3}{gc}}×\omega{d}\omega=\mathrm{cos}\:\theta\:{d}\theta \\ $$$$\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{ab}\right)}{\mathrm{3}{gc}}\int_{\mathrm{0}} ^{\omega} \omega{d}\omega=\int_{\mathrm{0}} ^{\theta} \mathrm{cos}\:\theta\:{d}\theta \\ $$$$\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{ab}\right)\omega^{\mathrm{2}} }{\mathrm{6}{gc}}=\mathrm{sin}\:\theta \\ $$$$\Rightarrow\omega=\sqrt{\frac{\mathrm{6}{gc}\:\mathrm{sin}\:\theta}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{ab}}}=\sqrt{\frac{\mathrm{6}{g}\lambda\:\mathrm{sin}\:\theta}{{c}}} \\ $$$${with}\:\lambda=\frac{{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{ab}}=\frac{\left({b}−{a}\right)^{\mathrm{2}} }{\mathrm{4}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{ab}\right)} \\ $$$$ \\ $$$${mg}\:\mathrm{cos}\:\theta−{N}={mc}\alpha \\ $$$$\Rightarrow{N}={mg}\:\mathrm{cos}\:\theta−{mc}×\frac{\mathrm{3}{gc}\:\mathrm{cos}\:\theta}{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{ab}\right)} \\ $$$$\Rightarrow{N}={mg}\:\mathrm{cos}\:\theta\left(\mathrm{1}−\frac{\mathrm{3}{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{ab}}\right)>\mathrm{0} \\ $$$${f}−{mg}\:\mathrm{sin}\:\theta={mc}\omega^{\mathrm{2}} \\ $$$$\Rightarrow{f}={mg}\:\mathrm{sin}\:\theta+{mc}×\frac{\mathrm{6}{gc}\:\mathrm{sin}\:\theta}{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{ab}\right)} \\ $$$$\Rightarrow{f}={mg}\:\mathrm{sin}\:\theta\left(\mathrm{1}+\frac{\mathrm{6}{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{ab}}\right) \\ $$$${such}\:{that}\:{no}\:{slipping}, \\ $$$$\frac{{f}}{{N}}\leqslant\mu \\ $$$$\frac{{mg}\:\mathrm{sin}\:\theta\left(\mathrm{1}+\frac{\mathrm{6}{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{ab}}\right)}{{mg}\:\mathrm{cos}\:\theta\left(\mathrm{1}−\frac{\mathrm{3}{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{ab}}\right)}\leqslant\mu \\ $$$$\Rightarrow\mathrm{tan}\:\theta\leqslant\frac{\mu\left(\mathrm{1}−\frac{\mathrm{3}{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{ab}}\right)}{\mathrm{1}+\frac{\mathrm{6}{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{ab}}}=\frac{\mu\left(\mathrm{1}−\mathrm{3}\lambda\right)}{\mathrm{1}+\mathrm{6}\lambda} \\ $$$$\Rightarrow\mathrm{sin}\:\theta_{{max}} =\frac{\mu\left(\mathrm{1}−\mathrm{3}\lambda\right)}{\:\sqrt{\left(\mathrm{1}+\mathrm{6}\lambda\right)^{\mathrm{2}} +\mu^{\mathrm{2}} \left(\mathrm{1}−\mathrm{3}\lambda\right)^{\mathrm{2}} }} \\ $$$$\Rightarrow\omega_{{max}} =\sqrt{\frac{\mathrm{12}{g}\lambda\mu\left(\mathrm{1}−\mathrm{3}\lambda\right)}{\:\left({b}−{a}\right)\sqrt{\left(\mathrm{1}+\mathrm{6}\lambda\right)^{\mathrm{2}} +\mu^{\mathrm{2}} \left(\mathrm{1}−\mathrm{3}\lambda\right)^{\mathrm{2}} }}} \\ $$$${example}: \\ $$$${a}=\mathrm{3}{m},\:{b}=\mathrm{5}{m},\:\mu=\mathrm{0}.\mathrm{8} \\ $$$$\Rightarrow{c}=\frac{\mathrm{5}−\mathrm{3}}{\mathrm{2}}=\mathrm{1} \\ $$$$\Rightarrow\lambda=\frac{\mathrm{1}^{\mathrm{2}} }{\mathrm{3}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} −\mathrm{3}×\mathrm{5}}=\frac{\mathrm{1}}{\mathrm{19}} \\ $$$$\mathrm{tan}\:\theta_{{max}} =\frac{\mathrm{0}.\mathrm{8}\left(\mathrm{1}−\frac{\mathrm{3}}{\mathrm{19}}\right)}{\mathrm{1}+\frac{\mathrm{6}}{\mathrm{19}}}=\frac{\mathrm{64}}{\mathrm{125}}\:\Rightarrow\theta_{{max}} \approx\mathrm{27}.\mathrm{1}° \\ $$$$\omega_{{max}} =\sqrt{\frac{\mathrm{6}×\mathrm{9}.\mathrm{81}×\mathrm{1}\:\mathrm{sin}\:\theta_{{max}} }{\mathrm{3}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} −\mathrm{3}×\mathrm{5}}}\:\approx\mathrm{1}.\mathrm{118}\:{rad}/{s} \\ $$
Commented by ajfour last updated on 02/Apr/26
c=((b−a)/2) cmgcosθ=m((L^2 /(12))+c^2 )((ωdω)/dθ) (ω^2 /(2c))((L^2 /(12))+c^2 )=gsin θ μN−mgsin θ=mω^2 c mgcos θ−N=mcα=mc(((ωdω)/dθ)) so μmgcos θ−mgsin θ−m(((2c^2 gsin θ)/((L^2 /(12))+c^2 ))) =μcm(((cgcos θ)/((L^2 /(12))+c^2 ))) dividing by mgcos θ and with (c^2 /((L^2 /(12))+c^2 ))=q μ−tan θ=μq+2qtan θ tan θ=((μ(1−q))/(2q+1)) , 𝛚^2 =((2qgsin 𝛉)/c) with q=(c^2 /((L^2 /(12))+c^2 ))
$${c}=\frac{{b}−{a}}{\mathrm{2}} \\ $$$${cmgcos}\theta={m}\left(\frac{{L}^{\mathrm{2}} }{\mathrm{12}}+{c}^{\mathrm{2}} \right)\frac{\omega{d}\omega}{{d}\theta} \\ $$$$\frac{\omega^{\mathrm{2}} }{\mathrm{2}{c}}\left(\frac{{L}^{\mathrm{2}} }{\mathrm{12}}+{c}^{\mathrm{2}} \right)={g}\mathrm{sin}\:\theta \\ $$$$\mu{N}−{mg}\mathrm{sin}\:\theta={m}\omega^{\mathrm{2}} {c} \\ $$$${mg}\mathrm{cos}\:\theta−{N}={mc}\alpha={mc}\left(\frac{\omega{d}\omega}{{d}\theta}\right) \\ $$$${so} \\ $$$$\mu{mg}\mathrm{cos}\:\theta−{mg}\mathrm{sin}\:\theta−{m}\left(\frac{\mathrm{2}{c}^{\mathrm{2}} {g}\mathrm{sin}\:\theta}{\frac{{L}^{\mathrm{2}} }{\mathrm{12}}+{c}^{\mathrm{2}} }\right) \\ $$$$\:\:\:\:\:=\mu{cm}\left(\frac{{cg}\mathrm{cos}\:\theta}{\frac{{L}^{\mathrm{2}} }{\mathrm{12}}+{c}^{\mathrm{2}} }\right) \\ $$$${dividing}\:{by}\:{mg}\mathrm{cos}\:\theta \\ $$$${and}\:{with}\:\:\:\:\:\frac{{c}^{\mathrm{2}} }{\frac{{L}^{\mathrm{2}} }{\mathrm{12}}+{c}^{\mathrm{2}} }={q} \\ $$$$\mu−\mathrm{tan}\:\theta=\mu{q}+\mathrm{2}{q}\mathrm{tan}\:\theta \\ $$$$\mathrm{tan}\:\theta=\frac{\mu\left(\mathrm{1}−{q}\right)}{\mathrm{2}{q}+\mathrm{1}}\:\:,\:\:\:\boldsymbol{\omega}^{\mathrm{2}} =\frac{\mathrm{2}\boldsymbol{{qg}}\mathrm{sin}\:\boldsymbol{\theta}}{\boldsymbol{{c}}} \\ $$$$\:\:\:\:{with}\:\:\:\:\:{q}=\frac{{c}^{\mathrm{2}} }{\frac{{L}^{\mathrm{2}} }{\mathrm{12}}+{c}^{\mathrm{2}} } \\ $$$$ \\ $$
Commented by mr W last updated on 02/Apr/26
great! thanks!
$${great}!\:{thanks}! \\ $$
Commented by ajfour last updated on 02/Apr/26
Thank you too.
$${Thank}\:{you}\:{too}. \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *