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Question Number 152388 by ajfour last updated on 28/Aug/21

Commented by ajfour last updated on 28/Aug/21

$F i n d t h e m a x i m u m s p e e d u,$ $t h a t c a n b e g i v e n t o t h e s o l i d$ $b a l l, s o t h a t i t g o e s t h r o u g h t h e$ $p a r a b o l o i d d i t c h y = x^{2} - b^{2}$ $a n d e m e r g e s o u t a t s p e e d u,$ $h a v i n g p u r e l y r o l l e d a l l a l o n g .$ $(a s s u m e u p p e r l i m i t o f c o e f f .$ $o f f r i c t i o n t o b e μ e v e r y w h e r e)$
Commented by mr W last updated on 28/Aug/21

$s e e m s t o b e v e r y h a r d$
Commented by ajfour last updated on 28/Aug/21

Commented by ajfour last updated on 29/Aug/21
$say at vertex ball is given a max rolling speed so that it cones over without slipping. so, consider v reversed in diagram above. I=(2/5)mr^2 +mr^2 =I_0 +A (1/2)I(ω_0 ^2 −ω^2 )=mg(b^2 +rcos θ) ⇒ (7/(10))(u^2 −v^2 )=g(b^2 +rcos θ) { v^2 =u^2 −((10g)/7)(b^2 +rcos θ) } ★ differentiating wrt θ [ ((mvdv)/(rdθ))=((5mgsin θ)/7) ] ★ mgcos θ−N=((mv^2 )/r) ⇒ N=m(gcos θ−(v^2 /r)) (normal rxn goes ↓ with θ) mgsin θ−f=m((vdv)/(rdθ)) dynamic μ=(f/N) ≤μ_0 (in Q) mgsin θ−((mdv)/(rdθ))=μm(gcos θ−(v^2 /r)) μ(u,θ)=((grsin θ−((10grsin θ)/7)+u^2 )/((grcos θ−v^2 ))) μ(u,θ)=((grsin θ−((10grsin θ)/7)+u^2 )/(grcos θ−u^2 +((10g)/7)(b^2 +rcos θ))) μ(u,θ)=((7(u^2 /gr)−3sin θ)/(17cos θ+10(b^2 /r)−7(u^2 /gr))) (∂μ/∂θ)=(((−3cos θ){17cos θ+10B−7U}−{7U−3sin θ}(−17sin θ))/([17U+10B−7U]^2 )) (∂μ/∂θ)=0 ⇒ 51cos^2 θ+30Bcos θ−21Ucos θ =119Usin θ−51sin^2 θ ⇒ U(21cos θ+119sin θ)=51+30Bcos θ .......∗∗ Now μ(θ)=((7U−3sin θ)/(17cos θ+10B−7U))=μ_0 hence 7(1+μ_0 )(((51+30Bcos θ)/(21cos θ+119sin θ))) = 3sin θ+μ_0 (17cos θ+10B) from above we get θ=δ U=((51+30Bcos δ)/(21cos δ+119sin δ)) U=(u^2 /(rg)) then for θ=0° { v^2 =u^2 −((10g)/7)(b^2 +rcos θ) } ★ gives (v^2 )_Q =rg(((51+30Bcos δ)/(21cos δ+119sin δ))) −((10g)/7)(b^2 +r) _____________________$
$s a y a t v e r t e x b a l l i s g i v e n a$ $m a x r o l l i n g s p e e d s o t h a t i t$ $c o n e s o v e r w i t h o u t s l i p p i n g .$ $s o, c o n s i d e r v r e v e r s e d i n$ $d i a g r a m a b o v e .$ $I = \frac{2}{5} {m r}^{2} + {m r}^{2} = I_{0} + A$ $\frac{1}{2} I (ω_{0}^{2} - ω^{2}) = m g (b^{2} + r \cos θ)$ $\Rightarrow \frac{7}{10} (u^{2} - v^{2}) = g (b^{2} + r \cos θ)$ ${v^{2} = u^{2} - \frac{10 g}{7} (b^{2} + r \cos θ)} ★$ $d i f f e r e n t i a t i n g w r t θ$ $[\frac{m v d v}{r d θ} = \frac{5 m g \sin θ}{7}] ★$ $m g \cos θ - N = \frac{{m v}^{2}}{r}$ $\Rightarrow N = m (g \cos θ - \frac{v^{2}}{r})$ $(n o r m a l r x n g o e s ↓ w i t h θ)$ $m g \sin θ - f = m \frac{v d v}{r d θ}$ $d y n a m i c μ = \frac{f}{N} ⩽ μ_{0} (i n Q)$ $m g \sin θ - \frac{m d v}{r d θ} = μ m (g \cos θ - \frac{v^{2}}{r})$ $μ (u, θ) = \frac{g r \sin θ - \frac{10 g r \sin θ}{7} + u^{2}}{(g r \cos θ - v^{2})}$ $μ (u, θ) = \frac{g r \sin θ - \frac{10 g r \sin θ}{7} + u^{2}}{g r \cos θ - u^{2} + \frac{10 g}{7} (b^{2} + r \cos θ)}$ $μ (u, θ) = \frac{7 (u^{2} / g r) - 3 \sin θ}{17 \cos θ + 10 (b^{2} / r) - 7 (u^{2} / g r)}$ $\frac{\partial μ}{\partial θ} = \frac{(- 3 \cos θ) {17 \cos θ + 10 B - 7 U} - {7 U - 3 \sin θ} (- 17 \sin θ)}{{[17 U + 10 B - 7 U]}^{2}}$ $\frac{\partial μ}{\partial θ} = 0 \Rightarrow$ $51 \cos^{2} θ + 30 B \cos θ - 21 U \cos θ$ $= 119 U \sin θ - 51 \sin^{2} θ$ $\Rightarrow U (21 \cos θ + 119 \sin θ) = 51 + 30 B \cos θ$ $. . . . . . . * *$ $N o w$ $μ (θ) = \frac{7 U - 3 \sin θ}{17 \cos θ + 10 B - 7 U} = μ_{0}$ $h e n c e$ $7 (1 + μ_{0}) (\frac{51 + 30 B \cos θ}{21 \cos θ + 119 \sin θ})$ $= 3 \sin θ + μ_{0} (17 \cos θ + 10 B)$ $f r o m a b o v e w e g e t θ = δ$ $U = \frac{51 + 30 B \cos δ}{21 \cos δ + 119 \sin δ}$ $U = \frac{u^{2}}{r g} t h e n f o r θ = 0 °$ ${v^{2} = u^{2} - \frac{10 g}{7} (b^{2} + r \cos θ)} ★$ $g i v e s$ ${(v^{2})}_{Q} = r g (\frac{51 + 30 B \cos δ}{21 \cos δ + 119 \sin δ})$ $- \frac{10 g}{7} (b^{2} + r)$ $_____________________$
Commented by ajfour last updated on 29/Aug/21

$h a v e a s s u m e d r ⩽ R$ $\frac{d^{2} y}{{d x}^{2}} = 2 & f r o m$ $x^{2} + {(y - R)}^{2} = R^{2}$ $2 x + 2 (y - R) \frac{d y}{d x} = 0$ $\frac{d y}{d x} = \frac{x}{R - y}$ $\frac{d^{2} y}{{d x}^{2}} = \frac{R - y + \frac{x^{2}}{R - y}}{{(R - y)}^{2}} = 2$ $\Rightarrow n o w x = 0, y = 0$ $\Rightarrow R = \frac{1}{2} s o r ⩽ R = \frac{1}{2} u n i t$ $u n i t c o m e s f r o m u n i t o f y$ $o f p a r a b o l a .$ $s a y i f t r a c k i s 1 m d e e p, t h e n$ $w i d t h i s 2 m; w h i l e r ⩽ \frac{1}{2} m .$
Commented by mr W last updated on 29/Aug/21

$i t h i n k i n m g \cos θ - N = \frac{{m v}^{2}}{r}$ $r s h o u l d b e t h e r a d i u s o f t h e l o c u s$ $o f t h e c e n t e r o f t h e b a l l, n o t t h e$ $r a d i u s o f t h e b a l l .$
Commented by ajfour last updated on 29/Aug/21

$b a l l i s t u r n i n g a b o u t t h e e d g e$ $s o c e n t e r o f b a l l g o e s a b o u t t h e$ $e d g e i n a c i r c l e o f r a d i u s r o n l y .$
Commented by mr W last updated on 29/Aug/21

$y e s, a t t h a t p o i n t .$
	Answered by mr W last updated on 28/Aug/21

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