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Question Number 60697 by ajfour last updated on 24/May/19

Commented by ajfour last updated on 24/May/19

$H o w f a r c a n t h e s l i d e r s l i d e i f i t$ $e j e c t s m a s s a t r a t e \frac{d M}{d t} = R, a t a$ $r e l a t i v e s p e e d u ?$ $(\frac{M_{0}}{2} o f M_{0} {d o e s n}^{'} t b u r n a n d$ $a s s u m e μ s u f f i c i e n t l y l e s s)$
	Answered by mr W last updated on 25/May/19
	$phase 1: t=0 to t_1 , slider ejects material (dM/dt)=−R ⇒M=M_0 −Rt −μMg−u(dM/dt)=Ma −μMg+uR=Ma ⇒a=((uR)/M)−μg a=(dv/dt)=(dv/dM)×(dM/dt)=−R(dv/dM) −R(dv/dM)=((uR)/M)−μg −∫_0 ^( v) Rdv=∫_M_0 ^( M) (((uR)/M)−μg)dM −Rv=[uR ln M−μgM]_M_0 ^M ⇒v=−u ln (M/M_0 )−((μg)/R)(M_0 −M) v=(ds/dt)=(ds/dM)×(dM/dt)=−R(ds/dM) ⇒−R (ds/dM)=−u ln (M/M_0 )−((μg)/R)(M_0 −M) ⇒R∫_0 ^( s) ds=∫_M_0 ^( M) {u ln (M/M_0 )+((μg)/R)(M_0 −M)}dM ⇒Rs=uM_0 [(ln (M/M_0 )−1)(M/M_0 )]_M_0 ^M −((μg)/(2R))[(M_0 −M)^2 ]_M_0 ^M ⇒s=((uM_0 )/R) [(ln (M/M_0 )−1)(M/M_0 )+1]−((μg)/(2R^2 ))(M_0 −M)^2 at t=t_1 =(M_0 /(2R)), M=(M_0 /2): ⇒s_1 =[4(1−ln 2)uR −μgM_0 ](M_0 /(8R^2 )) ⇒v_1 =u ln 2−((μgM_0 )/(2R)) M_1 =(M_0 /2) phase 2: t=t_1 to t_2 , slider doesn′t ejects material after this point it slides s_2 till it stops, and its kinetic energy is completely lost due to friction. (1/2)M_1 v_1 ^2 =μM_1 gs_2 ⇒s_2 =(v_1 ^2 /(2μg))=(1/(2μg))(u ln 2−((μgM_0 )/(2R)))^2 total slide distance l=s_1 +s_2 ⇒l=[4(1−ln 2)uR −μgM_0 ](M_0 /(8R^2 ))+(1/(2μg))(u ln 2−((μgM_0 )/(2R)))^2 ⇒l=(((ln 2)^2 u^2 )/(2μg))−(((2ln 2−1)uM_0 )/(2R))$
	$p h a s e 1 : t = 0 t o t_{1}, s l i d e r e j e c t s m a t e r i a l$ $\frac{d M}{d t} = - R$ $\Rightarrow M = M_{0} - R t$ $- μ M g - u \frac{d M}{d t} = M a$ $- μ M g + u R = M a$ $\Rightarrow a = \frac{u R}{M} - μ g$ $a = \frac{d v}{d t} = \frac{d v}{d M} \times \frac{d M}{d t} = - R \frac{d v}{d M}$ $- R \frac{d v}{d M} = \frac{u R}{M} - μ g$ $- \int_{0}^{v} R d v = \int_{M_{0}}^{M} (\frac{u R}{M} - μ g) d M$ $- R v = {[u R \ln M - μ g M]}_{M_{0}}^{M}$ $\Rightarrow v = - u \ln \frac{M}{M_{0}} - \frac{μ g}{R} (M_{0} - M)$ $v = \frac{d s}{d t} = \frac{d s}{d M} \times \frac{d M}{d t} = - R \frac{d s}{d M}$ $\Rightarrow - R \frac{d s}{d M} = - u \ln \frac{M}{M_{0}} - \frac{μ g}{R} (M_{0} - M)$ $\Rightarrow R \int_{0}^{s} d s = \int_{M_{0}}^{M} {u \ln \frac{M}{M_{0}} + \frac{μ g}{R} (M_{0} - M)} d M$ $\Rightarrow R s = {u M}_{0} {[(\ln \frac{M}{M_{0}} - 1) \frac{M}{M_{0}}]}_{M_{0}}^{M} - \frac{μ g}{2 R} {[{(M_{0} - M)}^{2}]}_{M_{0}}^{M}$ $\Rightarrow s = \frac{{u M}_{0}}{R} [(\ln \frac{M}{M_{0}} - 1) \frac{M}{M_{0}} + 1] - \frac{μ g}{2 R^{2}} {(M_{0} - M)}^{2}$ $a t t = t_{1} = \frac{M_{0}}{2 R}, M = \frac{M_{0}}{2} :$ $\Rightarrow s_{1} = [4 (1 - \ln 2) u R - μ {g M}_{0}] \frac{M_{0}}{8 R^{2}}$ $\Rightarrow v_{1} = u \ln 2 - \frac{μ {g M}_{0}}{2 R}$ $M_{1} = \frac{M_{0}}{2}$ $p h a s e 2 : t = t_{1} t o t_{2}, s l i d e r {d o e s n}^{'} t e j e c t s m a t e r i a l$ $a f t e r t h i s p o i n t i t s l i d e s s_{2} t i l l i t s t o p s,$ $a n d i t s k i n e t i c e n e r g y i s c o m p l e t e l y$ $l o s t d u e t o f r i c t i o n .$ $\frac{1}{2} M_{1} v_{1}^{2} = μ M_{1} {g s}_{2}$ $\Rightarrow s_{2} = \frac{v_{1}^{2}}{2 μ g} = \frac{1}{2 μ g} {(u \ln 2 - \frac{μ {g M}_{0}}{2 R})}^{2}$ $t o t a l s l i d e d i s t a n c e$ $l = s_{1} + s_{2}$ $\Rightarrow l = [4 (1 - \ln 2) u R - μ {g M}_{0}] \frac{M_{0}}{8 R^{2}} + \frac{1}{2 μ g} {(u \ln 2 - \frac{μ {g M}_{0}}{2 R})}^{2}$ $\Rightarrow l = \frac{{(\ln 2)}^{2} u^{2}}{2 μ g} - \frac{(2 \ln 2 - 1) {u M}_{0}}{2 R}$
	Commented by ajfour last updated on 25/May/19

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