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Question-164073




Question Number 164073 by mr W last updated on 13/Jan/22
Commented by mr W last updated on 13/Jan/22
are there other equilibrium positions  for the rod?
$${are}\:{there}\:{other}\:{equilibrium}\:{positions} \\ $$$${for}\:{the}\:{rod}? \\ $$
Answered by ajfour last updated on 13/Jan/22
Commented by mr W last updated on 14/Jan/22
no sir. i mean it should be possible  that the three forces on the rod are  in equilibrium in a way without   friction like this:
$${no}\:{sir}.\:{i}\:{mean}\:{it}\:{should}\:{be}\:{possible} \\ $$$${that}\:{the}\:{three}\:{forces}\:{on}\:{the}\:{rod}\:{are} \\ $$$${in}\:{equilibrium}\:{in}\:{a}\:{way}\:{without}\: \\ $$$${friction}\:{like}\:{this}: \\ $$
Commented by mr W last updated on 14/Jan/22
according to our experience in real  life, the rod can also rest in the bowl  in a inclined position, which is even  more stable. in fact the horizontal   position is even instable, because we   can hardly bring the rod to rest in a   horizontal position. but we only need  to release the rod in the bowl, it will   come to rest by itself in an inclined   position.
$${according}\:{to}\:{our}\:{experience}\:{in}\:{real} \\ $$$${life},\:{the}\:{rod}\:{can}\:{also}\:{rest}\:{in}\:{the}\:{bowl} \\ $$$${in}\:{a}\:{inclined}\:{position},\:{which}\:{is}\:{even} \\ $$$${more}\:{stable}.\:{in}\:{fact}\:{the}\:{horizontal}\: \\ $$$${position}\:{is}\:{even}\:{instable},\:{because}\:{we}\: \\ $$$${can}\:{hardly}\:{bring}\:{the}\:{rod}\:{to}\:{rest}\:{in}\:{a}\: \\ $$$${horizontal}\:{position}.\:{but}\:{we}\:{only}\:{need} \\ $$$${to}\:{release}\:{the}\:{rod}\:{in}\:{the}\:{bowl},\:{it}\:{will}\: \\ $$$${come}\:{to}\:{rest}\:{by}\:{itself}\:{in}\:{an}\:{inclined}\: \\ $$$${position}. \\ $$
Commented by ajfour last updated on 13/Jan/22
friction need be there, then only  it can rest in inclined way is what  i conclude, is that what you too   mean, sir?
$${friction}\:{need}\:{be}\:{there},\:{then}\:{only} \\ $$$${it}\:{can}\:{rest}\:{in}\:{inclined}\:{way}\:{is}\:{what} \\ $$$${i}\:{conclude},\:{is}\:{that}\:{what}\:{you}\:{too}\: \\ $$$${mean},\:{sir}? \\ $$
Commented by mr W last updated on 14/Jan/22
Answered by mr W last updated on 14/Jan/22
Commented by mr W last updated on 14/Jan/22
parabola y=(x^2 /a) with a=1 here  length of rod is b.  C=center of mass  there are three forces acting on rod:  normal contact forces N_1  and N_2 ,  and gravity of rod mg  when the rod is in equilibrium, all  these three forces must intersect at  one point S.  say the end points of the rod are  A(p, (p^2 /a)), B(q, (q^2 /a))  tan φ=−(dy/dx)∣_(x=p) =−((2p)/a)  tan ϕ=(dy/dx)∣_(x=q) =((2q)/a)  tan θ=(((q^2 /a)−(p^2 /a))/(q−p))=((q^2 −p^2 )/(a(q−p)))=((q+p)/a)  α=(π/2)−φ−θ  β=(π/2)−ϕ+θ  ((SC)/(AC))=((sin α)/(sin φ))=((sin ((π/2)−φ−θ))/(sin φ))=((cos (φ+θ))/(sin φ))  ((SC)/(CB))=((sin β)/(sin ϕ))=((sin ((π/2)−ϕ+θ))/(sin ϕ))=((cos (ϕ−θ))/(sin ϕ))  since AC=CB=(b/2),  ((cos (φ+θ))/(sin φ))=((cos (ϕ−θ))/(sin ϕ))  ((cos φ cos θ−sin φ sin θ)/(sin φ))=((cos ϕ cos θ+sin ϕ sin θ)/(sin ϕ))  ((cos θ)/(tan φ))−sin θ=((cos θ)/(tan ϕ))+sin θ  (1/(tan φ))−(1/(tan ϕ))=2 tan θ  −(a/(2p))−(a/(2q))=((2(p+q))/a)  −((p+q)/(pq))=((4(p+q))/a^2 )  ⇒p+q=0 ⇒p=−q ⇒horizontal position  or  −(1/(pq))=(4/a^2 )  ⇒pq=−(a^2 /4) or (−p)q=(a^2 /4)  (q−p)^2 +((q^2 /a)−(p^2 /a))^2 =b^2   (q−p)^2 +(((q−p)^2 (q+p)^2 )/a^2 )=b^2   (q−p)^2 [1+(((q+p)^2 )/a^2 )]=b^2   (q−p)^2 [1+(((q−p)^2 −4(−p)q)/a^2 )]=b^2   (q−p)^2 [1+(((q−p)^2 −a^2 )/a^2 )]=b^2   (q−p)^4 =a^2 b^2   ⇒q−p=(√(ab))  q and −p are roots of  z^2 −(√(ab))z+(a^2 /4)=0  ⇒−p=(((√(ab))−(√(a(b−a))))/2)  ⇒p=−(((√(ab))−(√(a(b−a))))/2)  ⇒q=(((√(ab))+(√(a(b−a))))/2)  that means generally there exist   two possible equilibrium positions   for a rod in a bowl:  position 1: horizontal          q=−p=(b/2)  position 2: inclined        p=−(((√(ab))−(√(a(b−a))))/2)        q=(((√(ab))+(√(a(b−a))))/2)  only for a short rod with b≤a, the  horizontal position is the only  possible equilibrium position.
$${parabola}\:{y}=\frac{{x}^{\mathrm{2}} }{{a}}\:{with}\:{a}=\mathrm{1}\:{here} \\ $$$${length}\:{of}\:{rod}\:{is}\:{b}. \\ $$$${C}={center}\:{of}\:{mass} \\ $$$${there}\:{are}\:{three}\:{forces}\:{acting}\:{on}\:{rod}: \\ $$$${normal}\:{contact}\:{forces}\:{N}_{\mathrm{1}} \:{and}\:{N}_{\mathrm{2}} , \\ $$$${and}\:{gravity}\:{of}\:{rod}\:{mg} \\ $$$${when}\:{the}\:{rod}\:{is}\:{in}\:{equilibrium},\:{all} \\ $$$${these}\:{three}\:{forces}\:{must}\:{intersect}\:{at} \\ $$$${one}\:{point}\:{S}. \\ $$$${say}\:{the}\:{end}\:{points}\:{of}\:{the}\:{rod}\:{are} \\ $$$${A}\left({p},\:\frac{{p}^{\mathrm{2}} }{{a}}\right),\:{B}\left({q},\:\frac{{q}^{\mathrm{2}} }{{a}}\right) \\ $$$$\mathrm{tan}\:\phi=−\frac{{dy}}{{dx}}\mid_{{x}={p}} =−\frac{\mathrm{2}{p}}{{a}} \\ $$$$\mathrm{tan}\:\varphi=\frac{{dy}}{{dx}}\mid_{{x}={q}} =\frac{\mathrm{2}{q}}{{a}} \\ $$$$\mathrm{tan}\:\theta=\frac{\frac{{q}^{\mathrm{2}} }{{a}}−\frac{{p}^{\mathrm{2}} }{{a}}}{{q}−{p}}=\frac{{q}^{\mathrm{2}} −{p}^{\mathrm{2}} }{{a}\left({q}−{p}\right)}=\frac{{q}+{p}}{{a}} \\ $$$$\alpha=\frac{\pi}{\mathrm{2}}−\phi−\theta \\ $$$$\beta=\frac{\pi}{\mathrm{2}}−\varphi+\theta \\ $$$$\frac{{SC}}{{AC}}=\frac{\mathrm{sin}\:\alpha}{\mathrm{sin}\:\phi}=\frac{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\phi−\theta\right)}{\mathrm{sin}\:\phi}=\frac{\mathrm{cos}\:\left(\phi+\theta\right)}{\mathrm{sin}\:\phi} \\ $$$$\frac{{SC}}{{CB}}=\frac{\mathrm{sin}\:\beta}{\mathrm{sin}\:\varphi}=\frac{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\varphi+\theta\right)}{\mathrm{sin}\:\varphi}=\frac{\mathrm{cos}\:\left(\varphi−\theta\right)}{\mathrm{sin}\:\varphi} \\ $$$${since}\:{AC}={CB}=\frac{{b}}{\mathrm{2}}, \\ $$$$\frac{\mathrm{cos}\:\left(\phi+\theta\right)}{\mathrm{sin}\:\phi}=\frac{\mathrm{cos}\:\left(\varphi−\theta\right)}{\mathrm{sin}\:\varphi} \\ $$$$\frac{\mathrm{cos}\:\phi\:\mathrm{cos}\:\theta−\mathrm{sin}\:\phi\:\mathrm{sin}\:\theta}{\mathrm{sin}\:\phi}=\frac{\mathrm{cos}\:\varphi\:\mathrm{cos}\:\theta+\mathrm{sin}\:\varphi\:\mathrm{sin}\:\theta}{\mathrm{sin}\:\varphi} \\ $$$$\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\phi}−\mathrm{sin}\:\theta=\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\varphi}+\mathrm{sin}\:\theta \\ $$$$\frac{\mathrm{1}}{\mathrm{tan}\:\phi}−\frac{\mathrm{1}}{\mathrm{tan}\:\varphi}=\mathrm{2}\:\mathrm{tan}\:\theta \\ $$$$−\frac{{a}}{\mathrm{2}{p}}−\frac{{a}}{\mathrm{2}{q}}=\frac{\mathrm{2}\left({p}+{q}\right)}{{a}} \\ $$$$−\frac{{p}+{q}}{{pq}}=\frac{\mathrm{4}\left({p}+{q}\right)}{{a}^{\mathrm{2}} } \\ $$$$\Rightarrow{p}+{q}=\mathrm{0}\:\Rightarrow{p}=−{q}\:\Rightarrow{horizontal}\:{position} \\ $$$${or} \\ $$$$−\frac{\mathrm{1}}{{pq}}=\frac{\mathrm{4}}{{a}^{\mathrm{2}} } \\ $$$$\Rightarrow{pq}=−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}\:{or}\:\left(−{p}\right){q}=\frac{{a}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\left({q}−{p}\right)^{\mathrm{2}} +\left(\frac{{q}^{\mathrm{2}} }{{a}}−\frac{{p}^{\mathrm{2}} }{{a}}\right)^{\mathrm{2}} ={b}^{\mathrm{2}} \\ $$$$\left({q}−{p}\right)^{\mathrm{2}} +\frac{\left({q}−{p}\right)^{\mathrm{2}} \left({q}+{p}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }={b}^{\mathrm{2}} \\ $$$$\left({q}−{p}\right)^{\mathrm{2}} \left[\mathrm{1}+\frac{\left({q}+{p}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right]={b}^{\mathrm{2}} \\ $$$$\left({q}−{p}\right)^{\mathrm{2}} \left[\mathrm{1}+\frac{\left({q}−{p}\right)^{\mathrm{2}} −\mathrm{4}\left(−{p}\right){q}}{{a}^{\mathrm{2}} }\right]={b}^{\mathrm{2}} \\ $$$$\left({q}−{p}\right)^{\mathrm{2}} \left[\mathrm{1}+\frac{\left({q}−{p}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right]={b}^{\mathrm{2}} \\ $$$$\left({q}−{p}\right)^{\mathrm{4}} ={a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$$\Rightarrow{q}−{p}=\sqrt{{ab}} \\ $$$${q}\:{and}\:−{p}\:{are}\:{roots}\:{of} \\ $$$${z}^{\mathrm{2}} −\sqrt{{ab}}{z}+\frac{{a}^{\mathrm{2}} }{\mathrm{4}}=\mathrm{0} \\ $$$$\Rightarrow−{p}=\frac{\sqrt{{ab}}−\sqrt{{a}\left({b}−{a}\right)}}{\mathrm{2}} \\ $$$$\Rightarrow{p}=−\frac{\sqrt{{ab}}−\sqrt{{a}\left({b}−{a}\right)}}{\mathrm{2}} \\ $$$$\Rightarrow{q}=\frac{\sqrt{{ab}}+\sqrt{{a}\left({b}−{a}\right)}}{\mathrm{2}} \\ $$$${that}\:{means}\:{generally}\:{there}\:{exist}\: \\ $$$${two}\:{possible}\:{equilibrium}\:{positions}\: \\ $$$${for}\:{a}\:{rod}\:{in}\:{a}\:{bowl}: \\ $$$${position}\:\mathrm{1}:\:{horizontal}\: \\ $$$$\:\:\:\:\:\:\:{q}=−{p}=\frac{{b}}{\mathrm{2}} \\ $$$${position}\:\mathrm{2}:\:{inclined} \\ $$$$\:\:\:\:\:\:{p}=−\frac{\sqrt{{ab}}−\sqrt{{a}\left({b}−{a}\right)}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:{q}=\frac{\sqrt{{ab}}+\sqrt{{a}\left({b}−{a}\right)}}{\mathrm{2}} \\ $$$${only}\:{for}\:{a}\:{short}\:{rod}\:{with}\:{b}\leqslant{a},\:{the} \\ $$$${horizontal}\:{position}\:{is}\:{the}\:{only} \\ $$$${possible}\:{equilibrium}\:{position}. \\ $$
Commented by mr W last updated on 14/Jan/22
Commented by mr W last updated on 14/Jan/22
Commented by ajfour last updated on 14/Jan/22
excellent, thank you sir.  I shall attempt too..
$${excellent},\:{thank}\:{you}\:{sir}. \\ $$$${I}\:{shall}\:{attempt}\:{too}.. \\ $$
Commented by Tawa11 last updated on 15/Jan/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by mr W last updated on 17/Jan/22
Commented by mr W last updated on 16/Jan/22
we see that the inclined equilibrium  position of the rod is the position  where the center of mass of the rod  is lowest. therefore it′s also  the most  stable position.  see also Q164178.
$${we}\:{see}\:{that}\:{the}\:{inclined}\:{equilibrium} \\ $$$${position}\:{of}\:{the}\:{rod}\:{is}\:{the}\:{position} \\ $$$${where}\:{the}\:{center}\:{of}\:{mass}\:{of}\:{the}\:{rod} \\ $$$${is}\:{lowest}.\:{therefore}\:{it}'{s}\:{also}\:\:{the}\:{most} \\ $$$${stable}\:{position}. \\ $$$${see}\:{also}\:{Q}\mathrm{164178}. \\ $$
Commented by ajfour last updated on 16/Jan/22
This is really really Excellent sol^n  sir.
$${This}\:{is}\:{really}\:{really}\:{Excellent}\:{sol}^{{n}} \:{sir}. \\ $$
Commented by mr W last updated on 16/Jan/22
thanks for reviewing sir!
$${thanks}\:{for}\:{reviewing}\:{sir}! \\ $$
Commented by mr W last updated on 17/Jan/22
interesting:  at equilibrium positions, we always  have SA⊥SB.
$${interesting}: \\ $$$${at}\:{equilibrium}\:{positions},\:{we}\:{always} \\ $$$${have}\:{SA}\bot{SB}. \\ $$

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