Math Question and Answers


Question and Answers Forum

All Questions Topic List
Mechanics Questions
Previous in All Question Next in All Question
Previous in Mechanics Next in Mechanics
Question Number 164073 by mr W last updated on 13/Jan/22

Commented by mr W last updated on 13/Jan/22

$a r e t h e r e o t h e r e q u i l i b r i u m p o s i t i o n s$ $f o r t h e r o d ?$
	Answered by ajfour last updated on 13/Jan/22

	Commented by mr W last updated on 14/Jan/22

	$n o s i r . i m e a n i t s h o u l d b e p o s s i b l e$ $t h a t t h e t h r e e f o r c e s o n t h e r o d a r e$ $i n e q u i l i b r i u m i n a w a y w i t h o u t$ $f r i c t i o n l i k e t h i s :$
	Commented by mr W last updated on 14/Jan/22

	$a c c o r d i n g t o o u r e x p e r i e n c e i n r e a l$ $l i f e, t h e r o d c a n a l s o r e s t i n t h e b o w l$ $i n a i n c l i n e d p o s i t i o n, w h i c h i s e v e n$ $m o r e s t a b l e . i n f a c t t h e h o r i z o n t a l$ $p o s i t i o n i s e v e n i n s t a b l e, b e c a u s e w e$ $c a n h a r d l y b r i n g t h e r o d t o r e s t i n a$ $h o r i z o n t a l p o s i t i o n . b u t w e o n l y n e e d$ $t o r e l e a s e t h e r o d i n t h e b o w l, i t w i l l$ $c o m e t o r e s t b y i t s e l f i n a n i n c l i n e d$ $p o s i t i o n .$
	Commented by ajfour last updated on 13/Jan/22

	$f r i c t i o n n e e d b e t h e r e, t h e n o n l y$ $i t c a n r e s t i n i n c l i n e d w a y i s w h a t$ $i c o n c l u d e, i s t h a t w h a t y o u t o o$ $m e a n, s i r ?$
	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

	$p a r a b o l a y = \frac{x^{2}}{a} w i t h a = 1 h e r e$ $l e n g t h o f r o d i s b .$ $C = c e n t e r o f m a s s$ $t h e r e a r e t h r e e f o r c e s a c t i n g o n r o d :$ $n o r m a l c o n t a c t f o r c e s N_{1} a n d N_{2},$ $a n d g r a v i t y o f r o d m g$ $w h e n t h e r o d i s i n e q u i l i b r i u m, a l l$ $t h e s e t h r e e f o r c e s m u s t i n t e r s e c t a t$ $o n e p o i n t S .$ $s a y t h e e n d p o i n t s o f t h e r o d a r e$ $A (p, \frac{p^{2}}{a}), B (q, \frac{q^{2}}{a})$ $\tan ϕ = - \frac{d y}{d x} ∣_{x = p} = - \frac{2 p}{a}$ $\tan φ = \frac{d y}{d x} ∣_{x = q} = \frac{2 q}{a}$ $\tan θ = \frac{\frac{q^{2}}{a} - \frac{p^{2}}{a}}{q - p} = \frac{q^{2} - p^{2}}{a (q - p)} = \frac{q + p}{a}$ $α = \frac{π}{2} - ϕ - θ$ $β = \frac{π}{2} - φ + θ$ $\frac{S C}{A C} = \frac{\sin α}{\sin ϕ} = \frac{\sin (\frac{π}{2} - ϕ - θ)}{\sin ϕ} = \frac{\cos (ϕ + θ)}{\sin ϕ}$ $\frac{S C}{C B} = \frac{\sin β}{\sin φ} = \frac{\sin (\frac{π}{2} - φ + θ)}{\sin φ} = \frac{\cos (φ - θ)}{\sin φ}$ $s i n c e A C = C B = \frac{b}{2},$ $\frac{\cos (ϕ + θ)}{\sin ϕ} = \frac{\cos (φ - θ)}{\sin φ}$ $\frac{\cos ϕ \cos θ - \sin ϕ \sin θ}{\sin ϕ} = \frac{\cos φ \cos θ + \sin φ \sin θ}{\sin φ}$ $\frac{\cos θ}{\tan ϕ} - \sin θ = \frac{\cos θ}{\tan φ} + \sin θ$ $\frac{1}{\tan ϕ} - \frac{1}{\tan φ} = 2 \tan θ$ $- \frac{a}{2 p} - \frac{a}{2 q} = \frac{2 (p + q)}{a}$ $- \frac{p + q}{p q} = \frac{4 (p + q)}{a^{2}}$ $\Rightarrow p + q = 0 \Rightarrow p = - q \Rightarrow h o r i z o n t a l p o s i t i o n$ $o r$ $- \frac{1}{p q} = \frac{4}{a^{2}}$ $\Rightarrow p q = - \frac{a^{2}}{4} o r (- p) q = \frac{a^{2}}{4}$ ${(q - p)}^{2} + {(\frac{q^{2}}{a} - \frac{p^{2}}{a})}^{2} = b^{2}$ ${(q - p)}^{2} + \frac{{(q - p)}^{2} {(q + p)}^{2}}{a^{2}} = b^{2}$ ${(q - p)}^{2} [1 + \frac{{(q + p)}^{2}}{a^{2}}] = b^{2}$ ${(q - p)}^{2} [1 + \frac{{(q - p)}^{2} - 4 (- p) q}{a^{2}}] = b^{2}$ ${(q - p)}^{2} [1 + \frac{{(q - p)}^{2} - a^{2}}{a^{2}}] = b^{2}$ ${(q - p)}^{4} = a^{2} b^{2}$ $\Rightarrow q - p = \sqrt{a b}$ $q a n d - p a r e r o o t s o f$ $z^{2} - \sqrt{a b} z + \frac{a^{2}}{4} = 0$ $\Rightarrow - p = \frac{\sqrt{a b} - \sqrt{a (b - a)}}{2}$ $\Rightarrow p = - \frac{\sqrt{a b} - \sqrt{a (b - a)}}{2}$ $\Rightarrow q = \frac{\sqrt{a b} + \sqrt{a (b - a)}}{2}$ $t h a t m e a n s g e n e r a l l y t h e r e e x i s t$ $t w o p o s s i b l e e q u i l i b r i u m p o s i t i o n s$ $f o r a r o d i n a b o w l :$ $p o s i t i o n 1 : h o r i z o n t a l$ $q = - p = \frac{b}{2}$ $p o s i t i o n 2 : i n c l i n e d$ $p = - \frac{\sqrt{a b} - \sqrt{a (b - a)}}{2}$ $q = \frac{\sqrt{a b} + \sqrt{a (b - a)}}{2}$ $o n l y f o r a s h o r t r o d w i t h b ⩽ a, t h e$ $h o r i z o n t a l p o s i t i o n i s t h e o n l y$ $p o s s i b l e e q u i l i b r i u m p o s i t i o n .$
	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

	$e x c e l l e n t, t h a n k y o u s i r .$ $I s h a l l a t t e m p t t o o . .$
	Commented by Tawa11 last updated on 15/Jan/22

	$Great sir$
	Commented by mr W last updated on 17/Jan/22

	Commented by mr W last updated on 16/Jan/22

	$w e s e e t h a t t h e i n c l i n e d e q u i l i b r i u m$ $p o s i t i o n o f t h e r o d i s t h e p o s i t i o n$ $w h e r e t h e c e n t e r o f m a s s o f t h e r o d$ $i s l o w e s t . t h e r e f o r e {i t}^{'} s a l s o t h e m o s t$ $s t a b l e p o s i t i o n .$ $s e e a l s o Q 164178 .$
	Commented by ajfour last updated on 16/Jan/22

	$T h i s i s r e a l l y r e a l l y E x c e l l e n t {s o l}^{n} s i r .$
	Commented by mr W last updated on 16/Jan/22

	$t h a n k s f o r r e v i e w i n g s i r!$
	Commented by mr W last updated on 17/Jan/22

	$i n t e r e s t i n g :$ $a t e q u i l i b r i u m p o s i t i o n s, w e a l w a y s$ $h a v e S A ⊥ S B .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum