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Question Number 164345 by mr W last updated on 16/Jan/22

Commented by mr W last updated on 18/Jan/22

$r o d w i t h m a s s m a n d l e n g t h b r e s t s$ $i n a f r i c t i o n l e s s p a r a b o l i c c u p a s$ $s h o w n . f i n d t h e t i m e p e r i o d o f$ $o s c i l l a t i o n s o f t h e r o d .$
	Answered by ajfour last updated on 16/Jan/22

	$p a r t (I) : f i n d e q u i l i b r i u m p o s i t i o n$ $y = x^{2}, y = x \tan θ + c = s x + c$ $x^{2} = s x + c$ $x = \frac{s}{2} \pm \sqrt{\frac{s^{2}}{4} + c}$ $2 \sqrt{\frac{s^{2}}{4} + c} \sqrt{1 + s^{2}} = b = 2 a$ $\Rightarrow 4 (\frac{s^{2}}{4} + c) (1 + s^{2}) = b^{2}$ $\Rightarrow c = \frac{a^{2}}{1 + s^{2}} - \frac{s^{2}}{4}$ $Y = y_{c m} = \frac{s^{2}}{4} + \frac{a^{2}}{1 + s^{2}}$ $\frac{d Y}{d s} = \frac{s}{2} - \frac{2 a^{2} s}{{(1 + s^{2})}^{2}} = 0$ $\Rightarrow 1 + s^{2} = 2 a$ $θ = \tan^{- 1} \sqrt{2 a - 1}$ $x = \frac{\sqrt{2 a - 1} \pm \sqrt{2 a}}{2}$
	Commented by mr W last updated on 17/Jan/22

	$Y = y_{c m} = y o f c e n t e r o f m a s s ?$
	Commented by ajfour last updated on 18/Jan/22
	yes sir.
	Commented by mr W last updated on 18/Jan/22

	$g r e a t! w e c a n g e t t h e e q u i l i b r i u m$ $p o s i t i o n a l s o t h r o u g h \frac{d Y}{d s}, i . e . t h e$ $c o m i s l o w e s t (o r h i g h e s t) .$
	Answered by mr W last updated on 17/Jan/22

	Commented by mr W last updated on 18/Jan/22

	$s e e a l s o Q 164073$ $p a r a b o l a y = \frac{x^{2}}{a} w i t h a = 1 h e r e$ $l e n g t h o f r o d i s b .$ $C = c e n t e r o f m a s s$ $s a y t h e e n d p o i n t s o f t h e r o d a r e$ $A (- p, \frac{p^{2}}{a}), B (q, \frac{q^{2}}{a})$ $\tan ϕ = - \frac{d y}{d x} ∣_{x = - p} = \frac{2 p}{a}$ $\tan φ = \frac{d y}{d x} ∣_{x = q} = \frac{2 q}{a}$ $\tan γ = \frac{\frac{q^{2}}{a} - \frac{p^{2}}{a}}{q + p} = \frac{q^{2} - p^{2}}{a (q + p)} = \frac{q - p}{a}$ $α = \frac{π}{2} - ϕ - γ$ $β = \frac{π}{2} - φ + γ$ $\frac{S C}{A C} = \frac{\sin α}{\sin ϕ} = \frac{\sin (\frac{π}{2} - ϕ - γ)}{\sin ϕ} = \frac{\cos (ϕ + γ)}{\sin ϕ}$ $\frac{S C}{C B} = \frac{\sin β}{\sin φ} = \frac{\sin (\frac{π}{2} - φ + γ)}{\sin φ} = \frac{\cos (φ - γ)}{\sin φ}$ $s i n c e A C = C B = \frac{b}{2},$ $\frac{\cos (ϕ + γ)}{\sin ϕ} = \frac{\cos (φ - γ)}{\sin φ}$ $\frac{\cos ϕ \cos γ - \sin ϕ \sin γ}{\sin ϕ} = \frac{\cos φ \cos γ + \sin φ \sin γ}{\sin φ}$ $\frac{\cos γ}{\tan ϕ} - \sin γ = \frac{\cos γ}{\tan φ} + \sin γ$ $\frac{1}{\tan ϕ} - \frac{1}{\tan φ} = 2 \tan γ$ $\frac{a}{2 p} - \frac{a}{2 q} = \frac{2 (q - p)}{a}$ $\frac{q - p}{p q} = \frac{4 (q - p)}{a^{2}}$ $\Rightarrow q - p = 0 \Rightarrow p = q \Rightarrow h o r i z o n t a l p o s i t i o n$ $o r$ $\Rightarrow \frac{1}{p q} = \frac{4}{a^{2}}$ $\Rightarrow p q = \frac{a^{2}}{4} \Rightarrow i n c l i n e d p o s i t i o n$ $w e c o n s i d e r t h e i n c l i n e d p o s i t i o n$ $o f r o d, w h i c h i s g e n e r a l l y t h e o n l y$ $s t a b l e e q u i l i b r i u m p o s i t i o n .$ ${(q + p)}^{2} + {(\frac{q^{2}}{a} - \frac{p^{2}}{a})}^{2} = b^{2}$ ${(q + p)}^{2} [1 + \frac{{(q - p)}^{2}}{a^{2}}] = b^{2}$ ${(q + p)}^{2} [1 + \frac{{(q + p)}^{2} - 4 p q}{a^{2}}] = b^{2}$ ${(q + p)}^{2} [1 + \frac{{(q + p)}^{2} - a^{2}}{a^{2}}] = b^{2}$ ${(q + p)}^{4} = a^{2} b^{2}$ $\Rightarrow p + q = \sqrt{a b}$ $p a n d q a r e r o o t s o f$ $z^{2} - \sqrt{a b} z + \frac{a^{2}}{4} = 0$ $\Rightarrow p = \frac{\sqrt{a b} - \sqrt{a (b - a)}}{2}$ $\Rightarrow q = \frac{\sqrt{a b} + \sqrt{a (b - a)}}{2}$ $\tan (φ + ϕ) = \frac{\frac{2 p}{a} + \frac{2 q}{a}}{1 - \frac{2 p}{a} \times \frac{2 q}{a}} = \frac{2 a (p + q)}{a^{2} - 4 p q} = \frac{2 a (p + q)}{a^{2} - a^{2}}$ $\Rightarrow φ + ϕ = 90 ° \Rightarrow S A ⊥ S B$ $\Rightarrow S C = \frac{A B}{2} = \frac{b}{2}$ $h = S C \times θ = \frac{b θ}{2}$ $I_{S} = \frac{{m b}^{2}}{12} + m {(\frac{b}{2})}^{2} = \frac{{m b}^{2}}{3}$ $I_{S} \frac{d^{2} θ}{{d t}^{2}} = - m g h$ $\frac{{m b}^{2}}{3} \frac{d^{2} θ}{{d t}^{2}} = - m g \frac{b θ}{2}$ $\frac{d^{2} θ}{{d t}^{2}} + \frac{3 g}{2 b} θ = 0$ $ω = \sqrt{\frac{3 g}{2 b}}$ $\Rightarrow T = \frac{2 π}{ω} = 2 π \sqrt{\frac{2 b}{3 g}}$
	Commented by mr W last updated on 18/Jan/22

	$T i s o n l y d e p e n d e n t f r o m t h e l e n g t h$ $o f t h e r o d, n o t f r o m t h e s h a p e o f$ $p a r a b o l a (p a r a m e t e r a) . t h i s i s$ $d u e t o S A ⊥ S C a n d t h e r e f o r e S C = \frac{b}{2} .$ $t h e o s c i l l a t i o n s o f t h e r o d i n a$ $p a r a b o l i c c u p o f a n y s h a p e a r e a l w a y s$ $t h e s a m e a s o f f o l l o w i n g c o m p o u n d$ $p e n d u l u m w h o s e p e r i o d i s$ $T = 2 π \sqrt{\frac{I}{m g L}}$ $I = \frac{{m b}^{2}}{3}, L = \frac{b}{2}$
	Commented by mr W last updated on 18/Jan/22

	Commented by Tawa11 last updated on 18/Jan/22

	$Great sirs$
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