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Question Number 51539 by ajfour last updated on 28/Dec/18

Commented by ajfour last updated on 28/Dec/18

$R o p e h a s a t o t a l l e n g t h L a n d l i n e a r$ $m a s s d e n s i t y ρ . S y s t e m i s r e l e a s e d$ $a s s h o w n . T h e p u l l e y i s a d i s c . F i n d$ $t i m e i t t a k e s f o r t h e r o p e t o u n w i n d$ $c o m p l e t e l y .$
	Answered by mr W last updated on 28/Dec/18
	$let x=unwinded length of rope with x_0 =a T=tension in rope velocity v=(dx/dt), acceletation b=(dv/dt) m_1 =M+xρ (dm_1 /dt)=ρ(dx/dt)=ρv v_(rel) =−v m_1 g−T+v_(rel) (dm_1 /dt)=m_1 b (M+ρx)g−T−ρv^2 =(M+ρx)(dv/dt) ω=(v/R) α=(b/R)=(1/R)(dv/dt) m_2 =m+ρ(l−x)ρ I=(1/2)mR^2 +ρ(l−x)R^2 =[(m/2)+ρ(l−x)]R^2 (dI/dt)=−ρR^2 (dx/dt)=−ρR^2 v ((d(Iω))/dt)=TR I(dω/dt)+ω(dI/dt)=TR Iα−(v/R)×ρR^2 v=TR Iα−ρRv^2 =TR [(m/2)+ρ(l−x)]R^2 ×(1/R)(dv/dt)−ρRv^2 =TR T=[(m/2)+ρ(l−x)](dv/dt)−ρv^2 ⇒(M+ρx)g−[(m/2)+ρ(l−x)](dv/dt)+ρv^2 −ρv^2 =(M+ρx)(dv/dt) ⇒Mg+ρgx=(M+(m/2)+ρl)(dv/dt) (dv/dt)=(dv/dx)×(dx/dt)=v(dv/dx) ⇒v((M/ρ)+(m/(2ρ))+l)(dv/dx)=gx+(M/ρ)g let c=(M/ρ)+(m/(2ρ))+l, d=(M/ρ) ⇒cv(dv/dx)=g(x+d) c∫_0 ^v vdv=g∫_a ^x (x+d)dx ((cv^2 )/2)=g[(((x^2 −a^2 ))/2)+d(x−a)] ⇒v^2 =(g/c)(x+a+2d)(x−a) ⇒v=(dx/dt)=(√((g/c)(x+a+2d)(x−a))) (dx/(√((x+a+2d)(x−a))))=(√(g/c)) dt ∫_a ^l (dx/(√((x+a+2d)(x−a))))=t(√(g/c)) ∫_0 ^(l−a) (du/(√(u{u+2(d+a)})))=t(√(g/c)) 2[ln ((√(u+2(d+a)))+(√u))]_0 ^(l−a) =t(√(g/c)) 2[ln ((√(l−a+2(d+a)))+(√(l−a)))−ln ((√(2(d+a))))] =t(√(g/c)) t=(√(c/g))ln ((((√(l−a+2(d+a)))+(√(l−a)))^2 )/(2(d+a))) ⇒t=(√(c/g))ln ((l+d+(√((l−a)(l+a+2d))))/(a+d)) with c=(M/ρ)+(m/(2ρ))+l, d=(M/ρ)$
	$l e t x = u n w i n d e d l e n g t h o f r o p e w i t h$ $x_{0} = a$ $T = t e n s i o n i n r o p e$ $v e l o c i t y v = \frac{d x}{d t}, a c c e l e t a t i o n b = \frac{d v}{d t}$ $m_{1} = M + x ρ$ $\frac{{d m}_{1}}{d t} = ρ \frac{d x}{d t} = ρ v$ $v_{r e l} = - v$ $m_{1} g - T + v_{r e l} \frac{{d m}_{1}}{d t} = m_{1} b$ $(M + ρ x) g - T - ρ v^{2} = (M + ρ x) \frac{d v}{d t}$ $ω = \frac{v}{R}$ $α = \frac{b}{R} = \frac{1}{R} \frac{d v}{d t}$ $m_{2} = m + ρ (l - x) ρ$ $I = \frac{1}{2} {m R}^{2} + ρ (l - x) R^{2} = [\frac{m}{2} + ρ (l - x)] R^{2}$ $\frac{d I}{d t} = - ρ R^{2} \frac{d x}{d t} = - ρ R^{2} v$ $\frac{d (I ω)}{d t} = T R$ $I \frac{d ω}{d t} + ω \frac{d I}{d t} = T R$ $I α - \frac{v}{R} \times ρ R^{2} v = T R$ $I α - ρ {R v}^{2} = T R$ $[\frac{m}{2} + ρ (l - x)] R^{2} \times \frac{1}{R} \frac{d v}{d t} - ρ {R v}^{2} = T R$ $T = [\frac{m}{2} + ρ (l - x)] \frac{d v}{d t} - ρ v^{2}$ $\Rightarrow (M + ρ x) g - [\frac{m}{2} + ρ (l - x)] \frac{d v}{d t} + ρ v^{2} - ρ v^{2} = (M + ρ x) \frac{d v}{d t}$ $\Rightarrow M g + ρ g x = (M + \frac{m}{2} + ρ l) \frac{d v}{d t}$ $\frac{d v}{d t} = \frac{d v}{d x} \times \frac{d x}{d t} = v \frac{d v}{d x}$ $\Rightarrow v (\frac{M}{ρ} + \frac{m}{2 ρ} + l) \frac{d v}{d x} = g x + \frac{M}{ρ} g$ $l e t c = \frac{M}{ρ} + \frac{m}{2 ρ} + l, d = \frac{M}{ρ}$ $\Rightarrow c v \frac{d v}{d x} = g (x + d)$ $c \int_{0}^{v} v d v = g \int_{a}^{x} (x + d) d x$ $\frac{{c v}^{2}}{2} = g [\frac{(x^{2} - a^{2})}{2} + d (x - a)]$ $\Rightarrow v^{2} = \frac{g}{c} (x + a + 2 d) (x - a)$ $\Rightarrow v = \frac{d x}{d t} = \sqrt{\frac{g}{c} (x + a + 2 d) (x - a)}$ $\frac{d x}{\sqrt{(x + a + 2 d) (x - a)}} = \sqrt{\frac{g}{c}} d t$ $\int_{a}^{l} \frac{d x}{\sqrt{(x + a + 2 d) (x - a)}} = t \sqrt{\frac{g}{c}}$ $\int_{0}^{l - a} \frac{d u}{\sqrt{u {u + 2 (d + a)}}} = t \sqrt{\frac{g}{c}}$ $2 {[\ln (\sqrt{u + 2 (d + a)} + \sqrt{u})]}_{0}^{l - a} = t \sqrt{\frac{g}{c}}$ $2 [\ln (\sqrt{l - a + 2 (d + a)} + \sqrt{l - a}) - \ln (\sqrt{2 (d + a)})] = t \sqrt{\frac{g}{c}}$ $t = \sqrt{\frac{c}{g}} \ln \frac{{(\sqrt{l - a + 2 (d + a)} + \sqrt{l - a})}^{2}}{2 (d + a)}$ $\Rightarrow t = \sqrt{\frac{c}{g}} \ln \frac{l + d + \sqrt{(l - a) (l + a + 2 d)}}{a + d}$ $w i t h c = \frac{M}{ρ} + \frac{m}{2 ρ} + l, d = \frac{M}{ρ}$
	Commented by ajfour last updated on 29/Dec/18

	$T h a n k s f o r t h e d y n a m i c w a y S i r!$
	Answered by ajfour last updated on 28/Dec/18
	$Energy conservation: Mg(x−a)+ρg(((x^2 −a^2 )/2))=(1/2)Mv^2 + +((ρx)/2)v^2 +(v^2 /2){(m/2)+ρ(l−x)} ⇒ v^2 = ((2Mg(x−a)+ρg(x^2 −a^2 ))/(M+ρl+m/2)) ⇒ ∫_a ^( l) (dx/(√(2Mg(x−a)+ρg(x^2 −a^2 )))) = ∫_0 ^( t) (dt/(√(M+ρl+m/2))) ⇒ (1/(√(ρg)))∫_0 ^( a) (dx/(√(x^2 +((2M)/ρ)x−a^2 −((2Ma)/ρ)))) = (t/(√(M+ρl+m/2))) ⇒∫_a ^( l) (dx/(√((x+(M/ρ))^2 −((M/ρ)+a)^2 ))) = t(√((ρg)/(M+ρl+m/2))) let (M/ρ) = d , ((M+ρl+m/2)/ρ) = c ⇒ ∫_a ^( l) (dx/(√((x+d)^2 −(a+d)^2 ))) = t(√(g/c)) t = (√(c/g))ln ((l+d+(√((l−a)(l+a+2d))))/(a+d)) .$
	$E n e r g y c o n s e r v a t i o n :$ $M g (x - a) + ρ g (\frac{x^{2} - a^{2}}{2}) = \frac{1}{2} {M v}^{2} +$ $+ \frac{ρ x}{2} v^{2} + \frac{v^{2}}{2} {\frac{m}{2} + ρ (l - x)}$ $\Rightarrow v^{2} = \frac{2 M g (x - a) + ρ g (x^{2} - a^{2})}{M + ρ l + m / 2}$ $\Rightarrow \int_{a}^{l} \frac{d x}{\sqrt{2 M g (x - a) + ρ g (x^{2} - a^{2})}} = \int_{0}^{t} \frac{d t}{\sqrt{M + ρ l + m / 2}}$ $\Rightarrow \frac{1}{\sqrt{ρ g}} \int_{0}^{a} \frac{d x}{\sqrt{x^{2} + \frac{2 M}{ρ} x - a^{2} - \frac{2 M a}{ρ}}} = \frac{t}{\sqrt{M + ρ l + m / 2}}$ $\Rightarrow \int_{a}^{l} \frac{d x}{\sqrt{{(x + \frac{M}{ρ})}^{2} - {(\frac{M}{ρ} + a)}^{2}}}$ $= t \sqrt{\frac{ρ g}{M + ρ l + m / 2}}$ $l e t \frac{M}{ρ} = d, \frac{M + ρ l + m / 2}{ρ} = c$ $\Rightarrow \int_{a}^{l} \frac{d x}{\sqrt{{(x + d)}^{2} - {(a + d)}^{2}}} = t \sqrt{\frac{g}{c}}$ $t = \sqrt{\frac{c}{g}} l n \frac{l + d + \sqrt{(l - a) (l + a + 2 d)}}{a + d} .$
	Commented by ajfour last updated on 28/Dec/18

	$t h a n k s f o r c h e c k i n g S i r .$
	Commented by mr W last updated on 28/Dec/18

	$w e h a v e t h e s a m e r e s u l t .$
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