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Question-50101




Question Number 50101 by ajfour last updated on 13/Dec/18
Commented by ajfour last updated on 14/Dec/18
If system of ring and block be  released as shown, and later the  ring hits the block as shown in  dashed lines, determine a.  Find also the velocity of ring then.  Assume friction sufficient such  that ring rolls purely as it moves.
$${If}\:{system}\:{of}\:{ring}\:{and}\:{block}\:{be} \\ $$$${released}\:{as}\:{shown},\:{and}\:{later}\:{the} \\ $$$${ring}\:{hits}\:{the}\:{block}\:{as}\:{shown}\:{in} \\ $$$${dashed}\:{lines},\:{determine}\:\boldsymbol{{a}}. \\ $$$${Find}\:{also}\:{the}\:{velocity}\:{of}\:{ring}\:{then}. \\ $$$${Assume}\:{friction}\:{sufficient}\:{such} \\ $$$${that}\:{ring}\:{rolls}\:{purely}\:{as}\:{it}\:{moves}. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 14/Dec/18
pls mention the source of question and also  attach  the final answer...so that we can  follow the right path to get the answer...  because question it self is not clear...  1) if the thread gets unwinded ...the ring has  a tendency to rotate anticlockwise...  further weight of block Mg cause the ring  to get unwinded...
$${pls}\:{mention}\:{the}\:{source}\:{of}\:{question}\:{and}\:{also} \\ $$$${attach}\:\:{the}\:{final}\:{answer}…{so}\:{that}\:{we}\:{can} \\ $$$${follow}\:{the}\:{right}\:{path}\:{to}\:{get}\:{the}\:{answer}… \\ $$$${because}\:{question}\:{it}\:{self}\:{is}\:{not}\:{clear}… \\ $$$$\left.\mathrm{1}\right)\:{if}\:{the}\:{thread}\:{gets}\:{unwinded}\:…{the}\:{ring}\:{has} \\ $$$${a}\:{tendency}\:{to}\:{rotate}\:{anticlockwise}… \\ $$$${further}\:{weight}\:{of}\:{block}\:{Mg}\:{cause}\:{the}\:{ring} \\ $$$${to}\:{get}\:{unwinded}… \\ $$$$ \\ $$
Commented by ajfour last updated on 14/Dec/18
which way do you think net torque   on ring shall be, Tanmay Sir;  i seldom post questions from   sources.
$${which}\:{way}\:{do}\:{you}\:{think}\:{net}\:{torque}\: \\ $$$${on}\:{ring}\:{shall}\:{be},\:{Tanmay}\:{Sir}; \\ $$$${i}\:{seldom}\:{post}\:{questions}\:{from}\: \\ $$$${sources}. \\ $$
Commented by mr W last updated on 14/Dec/18
to me the question is clear and unique.the  thread keeps in tension.
$${to}\:{me}\:{the}\:{question}\:{is}\:{clear}\:{and}\:{unique}.{the} \\ $$$${thread}\:{keeps}\:{in}\:{tension}. \\ $$
Answered by mr W last updated on 14/Dec/18
let θ=angle between string and ground  length of string at t=0:  a+(R/(tan (θ_0 /2)))=(h/(tan θ_0 ))=((h(1−tan^2  (θ_0 /2)))/(2tan (θ_0 /2)))  δ_0 =tan (θ_0 /2)  a+(R/δ_0 )=((h(1−δ_0 ^2 ))/(2δ_0 ))  a=((h(1−δ_0 ^2 )−2R)/(2δ_0 ))  δ_0 ^2 +2((a/h))δ_0 −(1−((2R)/h))=0  with α=(a/h) and γ=(R/h)  δ_0 ^2 +2αδ_0 −(1−2γ)=0  ⇒δ_0 =(√(1+α^2 −2γ))−α=tan (θ_0 /2)  l_0 =a+(h/(sin θ_0 ))−(R/(tan (θ_0 /2)))  l_0 =a+((h−2R+hδ_0 ^2 )/(2δ_0 ))  length of string at t=t_1 :  R+(R/(tan (θ_1 /2)))=(h/(tan θ_1 ))=((h(1−tan^2  (θ_1 /2)))/(2tan (θ_1 /2)))  R+(R/δ_1 )=((h(1−δ_1 ^2 ))/(2δ_1 ))  R=((h(1−δ_1 ^2 )−2R)/(2δ_1 ))  2Rδ_1 =h−2R−hδ_1 ^2   δ_1 ^2 +2γδ_1 −(1−2γ)=0  ⇒δ_1 =(√(γ^2 +1−2γ))−γ=tan (θ_1 /2)  l_1 =h−R+(h/(sin θ_1 ))−(R/(tan (θ_1 /2)))  l_1 =h−R+((h−2R+hδ_1 ^2 )/(2δ_1 ))  l_1 =l_0 +a−R  h−R+((h−2R+hδ_1 ^2 )/(2δ_1 ))=a+((h−2R+hδ_0 ^2 )/(2δ_0 ))+a−R  h+((h−2R+hδ_1 ^2 )/(2δ_1 ))=2a+((h−2R+hδ_0 ^2 )/(2δ_0 ))  1+((1−2γ+δ_1 ^2 )/(2δ_1 ))=2α+((1−2γ+δ_0 ^2 )/(2δ_0 ))  ((1−2γ)/δ_1 )+δ_1 −((1−2γ)/δ_0 )−δ_0 =2(2α−1)  ((1−2γ)/( (√(γ^2 +1−2γ))−γ))+(√(γ^2 +1−2γ))−γ−((1−2γ)/( (√(1+α^2 −2γ))−α))−(√(1+α^2 −2γ))+α=4α−2  ⇒3α+((1−2γ)/( (√(1+α^2 −2γ))−α))+(√(1+α^2 −2γ))=2−γ+((1−2γ)/( (√(γ^2 +1−2γ))−γ))+(√(γ^2 +1−2γ))  γ is constant, α is unknown  ⇒solve for α=....  ⇒a=αh=.....    at time t:  position of block is y (above ground)  position of ring is x  x+(R/(tan (θ/2)))=(h/(tan θ))=((h(1−tan^2  (θ/2)))/(2tan (θ/2)))  x+(R/δ)=((h(1−δ^2 ))/(2δ))  x=((h−2R−hδ^2 )/(2δ))  λ=(x/h)=((1−2γ−δ^2 )/(2δ))  δ^2 +2λδ−(1−2γ)=0  ⇒δ=(√(1+λ^2 −2γ))−λ  (dδ/dt)=((λ/( (√(1+λ^2 −2γ))))−1)(1/h)×(dx/dt)  l=h−y+(h/(sin θ))−(R/(tan (θ/2)))  l=h−y+((h−2R+hδ^2 )/(2δ))=l_0 +a−x  1−(y/h)+((1−2γ+δ^2 )/(2δ))=2α+((1−2γ+δ_0 ^2 )/(2δ_0 ))−λ  v_y =−(dy/dt)=velocity of block  v=−(dx/dt)=velocity of ring  (v_y /h)+[1−((1−2γ+δ^2 )/(2δ^2 ))](dδ/dt)=(v/h)  (v_y /h)−(((δ^2 −1+2γ)/(2δ^2 )))((λ/( (√(1+λ^2 −2γ))))−1)(v/h)=(v/h)  v_y ={1+(((δ^2 −1+2γ)/(2δ^2 )))((λ/( (√(1+λ^2 −2γ))))−1)}v  at t=t_1 : δ=δ_1 , x=R, λ=(R/h)=γ  v_y ={1+((((√(γ^2 +1−2γ))−γ)^2 −1+2γ)/(2((√(γ^2 +1−2γ))−γ)^2 ))((γ/( (√(1+γ^2 −2γ))))−1)}v  ⇒v_y ={1+((γ(1−2γ))/((1−γ)((√(γ^2 +1−2γ))−γ)))}v    (1/2)Mv^2 +(1/2)(MR^2 )ω^2 +(1/2)Mv_y ^2 =Mg(h−a−R)  2v^2 +v_y ^2 =2gh(1−α−γ)  [2+{1+((γ(1−2γ))/((1−γ)((√(γ^2 +1−2γ))−γ)))}^2 ]v^2 =2gh(1−α−γ)  ⇒v=(√((2gh(1−α−γ))/(2+{1+((γ(1−2γ))/((1−γ)((√(γ^2 +1−2γ))−γ)))}^2 )))
$${let}\:\theta={angle}\:{between}\:{string}\:{and}\:{ground} \\ $$$${length}\:{of}\:{string}\:{at}\:{t}=\mathrm{0}: \\ $$$${a}+\frac{{R}}{\mathrm{tan}\:\frac{\theta_{\mathrm{0}} }{\mathrm{2}}}=\frac{{h}}{\mathrm{tan}\:\theta_{\mathrm{0}} }=\frac{{h}\left(\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\frac{\theta_{\mathrm{0}} }{\mathrm{2}}\right)}{\mathrm{2tan}\:\frac{\theta_{\mathrm{0}} }{\mathrm{2}}} \\ $$$$\delta_{\mathrm{0}} =\mathrm{tan}\:\frac{\theta_{\mathrm{0}} }{\mathrm{2}} \\ $$$${a}+\frac{{R}}{\delta_{\mathrm{0}} }=\frac{{h}\left(\mathrm{1}−\delta_{\mathrm{0}} ^{\mathrm{2}} \right)}{\mathrm{2}\delta_{\mathrm{0}} } \\ $$$${a}=\frac{{h}\left(\mathrm{1}−\delta_{\mathrm{0}} ^{\mathrm{2}} \right)−\mathrm{2}{R}}{\mathrm{2}\delta_{\mathrm{0}} } \\ $$$$\delta_{\mathrm{0}} ^{\mathrm{2}} +\mathrm{2}\left(\frac{{a}}{{h}}\right)\delta_{\mathrm{0}} −\left(\mathrm{1}−\frac{\mathrm{2}{R}}{{h}}\right)=\mathrm{0} \\ $$$${with}\:\alpha=\frac{{a}}{{h}}\:{and}\:\gamma=\frac{{R}}{{h}} \\ $$$$\delta_{\mathrm{0}} ^{\mathrm{2}} +\mathrm{2}\alpha\delta_{\mathrm{0}} −\left(\mathrm{1}−\mathrm{2}\gamma\right)=\mathrm{0} \\ $$$$\Rightarrow\delta_{\mathrm{0}} =\sqrt{\mathrm{1}+\alpha^{\mathrm{2}} −\mathrm{2}\gamma}−\alpha=\mathrm{tan}\:\frac{\theta_{\mathrm{0}} }{\mathrm{2}} \\ $$$${l}_{\mathrm{0}} ={a}+\frac{{h}}{\mathrm{sin}\:\theta_{\mathrm{0}} }−\frac{{R}}{\mathrm{tan}\:\frac{\theta_{\mathrm{0}} }{\mathrm{2}}} \\ $$$${l}_{\mathrm{0}} ={a}+\frac{{h}−\mathrm{2}{R}+{h}\delta_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{2}\delta_{\mathrm{0}} } \\ $$$${length}\:{of}\:{string}\:{at}\:{t}={t}_{\mathrm{1}} : \\ $$$${R}+\frac{{R}}{\mathrm{tan}\:\frac{\theta_{\mathrm{1}} }{\mathrm{2}}}=\frac{{h}}{\mathrm{tan}\:\theta_{\mathrm{1}} }=\frac{{h}\left(\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\frac{\theta_{\mathrm{1}} }{\mathrm{2}}\right)}{\mathrm{2tan}\:\frac{\theta_{\mathrm{1}} }{\mathrm{2}}} \\ $$$${R}+\frac{{R}}{\delta_{\mathrm{1}} }=\frac{{h}\left(\mathrm{1}−\delta_{\mathrm{1}} ^{\mathrm{2}} \right)}{\mathrm{2}\delta_{\mathrm{1}} } \\ $$$${R}=\frac{{h}\left(\mathrm{1}−\delta_{\mathrm{1}} ^{\mathrm{2}} \right)−\mathrm{2}{R}}{\mathrm{2}\delta_{\mathrm{1}} } \\ $$$$\mathrm{2}{R}\delta_{\mathrm{1}} ={h}−\mathrm{2}{R}−{h}\delta_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$\delta_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{2}\gamma\delta_{\mathrm{1}} −\left(\mathrm{1}−\mathrm{2}\gamma\right)=\mathrm{0} \\ $$$$\Rightarrow\delta_{\mathrm{1}} =\sqrt{\gamma^{\mathrm{2}} +\mathrm{1}−\mathrm{2}\gamma}−\gamma=\mathrm{tan}\:\frac{\theta_{\mathrm{1}} }{\mathrm{2}} \\ $$$${l}_{\mathrm{1}} ={h}−{R}+\frac{{h}}{\mathrm{sin}\:\theta_{\mathrm{1}} }−\frac{{R}}{\mathrm{tan}\:\frac{\theta_{\mathrm{1}} }{\mathrm{2}}} \\ $$$${l}_{\mathrm{1}} ={h}−{R}+\frac{{h}−\mathrm{2}{R}+{h}\delta_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{2}\delta_{\mathrm{1}} } \\ $$$${l}_{\mathrm{1}} ={l}_{\mathrm{0}} +{a}−{R} \\ $$$${h}−{R}+\frac{{h}−\mathrm{2}{R}+{h}\delta_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{2}\delta_{\mathrm{1}} }={a}+\frac{{h}−\mathrm{2}{R}+{h}\delta_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{2}\delta_{\mathrm{0}} }+{a}−{R} \\ $$$${h}+\frac{{h}−\mathrm{2}{R}+{h}\delta_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{2}\delta_{\mathrm{1}} }=\mathrm{2}{a}+\frac{{h}−\mathrm{2}{R}+{h}\delta_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{2}\delta_{\mathrm{0}} } \\ $$$$\mathrm{1}+\frac{\mathrm{1}−\mathrm{2}\gamma+\delta_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{2}\delta_{\mathrm{1}} }=\mathrm{2}\alpha+\frac{\mathrm{1}−\mathrm{2}\gamma+\delta_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{2}\delta_{\mathrm{0}} } \\ $$$$\frac{\mathrm{1}−\mathrm{2}\gamma}{\delta_{\mathrm{1}} }+\delta_{\mathrm{1}} −\frac{\mathrm{1}−\mathrm{2}\gamma}{\delta_{\mathrm{0}} }−\delta_{\mathrm{0}} =\mathrm{2}\left(\mathrm{2}\alpha−\mathrm{1}\right) \\ $$$$\frac{\mathrm{1}−\mathrm{2}\gamma}{\:\sqrt{\gamma^{\mathrm{2}} +\mathrm{1}−\mathrm{2}\gamma}−\gamma}+\sqrt{\gamma^{\mathrm{2}} +\mathrm{1}−\mathrm{2}\gamma}−\gamma−\frac{\mathrm{1}−\mathrm{2}\gamma}{\:\sqrt{\mathrm{1}+\alpha^{\mathrm{2}} −\mathrm{2}\gamma}−\alpha}−\sqrt{\mathrm{1}+\alpha^{\mathrm{2}} −\mathrm{2}\gamma}+\alpha=\mathrm{4}\alpha−\mathrm{2} \\ $$$$\Rightarrow\mathrm{3}\alpha+\frac{\mathrm{1}−\mathrm{2}\gamma}{\:\sqrt{\mathrm{1}+\alpha^{\mathrm{2}} −\mathrm{2}\gamma}−\alpha}+\sqrt{\mathrm{1}+\alpha^{\mathrm{2}} −\mathrm{2}\gamma}=\mathrm{2}−\gamma+\frac{\mathrm{1}−\mathrm{2}\gamma}{\:\sqrt{\gamma^{\mathrm{2}} +\mathrm{1}−\mathrm{2}\gamma}−\gamma}+\sqrt{\gamma^{\mathrm{2}} +\mathrm{1}−\mathrm{2}\gamma} \\ $$$$\gamma\:{is}\:{constant},\:\alpha\:{is}\:{unknown} \\ $$$$\Rightarrow{solve}\:{for}\:\alpha=…. \\ $$$$\Rightarrow{a}=\alpha{h}=….. \\ $$$$ \\ $$$${at}\:{time}\:{t}: \\ $$$${position}\:{of}\:{block}\:{is}\:{y}\:\left({above}\:{ground}\right) \\ $$$${position}\:{of}\:{ring}\:{is}\:{x} \\ $$$${x}+\frac{{R}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}=\frac{{h}}{\mathrm{tan}\:\theta}=\frac{{h}\left(\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}\right)}{\mathrm{2tan}\:\frac{\theta}{\mathrm{2}}} \\ $$$${x}+\frac{{R}}{\delta}=\frac{{h}\left(\mathrm{1}−\delta^{\mathrm{2}} \right)}{\mathrm{2}\delta} \\ $$$${x}=\frac{{h}−\mathrm{2}{R}−{h}\delta^{\mathrm{2}} }{\mathrm{2}\delta} \\ $$$$\lambda=\frac{{x}}{{h}}=\frac{\mathrm{1}−\mathrm{2}\gamma−\delta^{\mathrm{2}} }{\mathrm{2}\delta} \\ $$$$\delta^{\mathrm{2}} +\mathrm{2}\lambda\delta−\left(\mathrm{1}−\mathrm{2}\gamma\right)=\mathrm{0} \\ $$$$\Rightarrow\delta=\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} −\mathrm{2}\gamma}−\lambda \\ $$$$\frac{{d}\delta}{{dt}}=\left(\frac{\lambda}{\:\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} −\mathrm{2}\gamma}}−\mathrm{1}\right)\frac{\mathrm{1}}{{h}}×\frac{{dx}}{{dt}} \\ $$$${l}={h}−{y}+\frac{{h}}{\mathrm{sin}\:\theta}−\frac{{R}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}} \\ $$$${l}={h}−{y}+\frac{{h}−\mathrm{2}{R}+{h}\delta^{\mathrm{2}} }{\mathrm{2}\delta}={l}_{\mathrm{0}} +{a}−{x} \\ $$$$\mathrm{1}−\frac{{y}}{{h}}+\frac{\mathrm{1}−\mathrm{2}\gamma+\delta^{\mathrm{2}} }{\mathrm{2}\delta}=\mathrm{2}\alpha+\frac{\mathrm{1}−\mathrm{2}\gamma+\delta_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{2}\delta_{\mathrm{0}} }−\lambda \\ $$$${v}_{{y}} =−\frac{{dy}}{{dt}}={velocity}\:{of}\:{block} \\ $$$${v}=−\frac{{dx}}{{dt}}={velocity}\:{of}\:{ring} \\ $$$$\frac{{v}_{{y}} }{{h}}+\left[\mathrm{1}−\frac{\mathrm{1}−\mathrm{2}\gamma+\delta^{\mathrm{2}} }{\mathrm{2}\delta^{\mathrm{2}} }\right]\frac{{d}\delta}{{dt}}=\frac{{v}}{{h}} \\ $$$$\frac{{v}_{{y}} }{{h}}−\left(\frac{\delta^{\mathrm{2}} −\mathrm{1}+\mathrm{2}\gamma}{\mathrm{2}\delta^{\mathrm{2}} }\right)\left(\frac{\lambda}{\:\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} −\mathrm{2}\gamma}}−\mathrm{1}\right)\frac{{v}}{{h}}=\frac{{v}}{{h}} \\ $$$${v}_{{y}} =\left\{\mathrm{1}+\left(\frac{\delta^{\mathrm{2}} −\mathrm{1}+\mathrm{2}\gamma}{\mathrm{2}\delta^{\mathrm{2}} }\right)\left(\frac{\lambda}{\:\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} −\mathrm{2}\gamma}}−\mathrm{1}\right)\right\}{v} \\ $$$${at}\:{t}={t}_{\mathrm{1}} :\:\delta=\delta_{\mathrm{1}} ,\:{x}={R},\:\lambda=\frac{{R}}{{h}}=\gamma \\ $$$${v}_{{y}} =\left\{\mathrm{1}+\frac{\left(\sqrt{\gamma^{\mathrm{2}} +\mathrm{1}−\mathrm{2}\gamma}−\gamma\right)^{\mathrm{2}} −\mathrm{1}+\mathrm{2}\gamma}{\mathrm{2}\left(\sqrt{\gamma^{\mathrm{2}} +\mathrm{1}−\mathrm{2}\gamma}−\gamma\right)^{\mathrm{2}} }\left(\frac{\gamma}{\:\sqrt{\mathrm{1}+\gamma^{\mathrm{2}} −\mathrm{2}\gamma}}−\mathrm{1}\right)\right\}{v} \\ $$$$\Rightarrow{v}_{{y}} =\left\{\mathrm{1}+\frac{\gamma\left(\mathrm{1}−\mathrm{2}\gamma\right)}{\left(\mathrm{1}−\gamma\right)\left(\sqrt{\gamma^{\mathrm{2}} +\mathrm{1}−\mathrm{2}\gamma}−\gamma\right)}\right\}{v} \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{Mv}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\left({MR}^{\mathrm{2}} \right)\omega^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{Mv}_{{y}} ^{\mathrm{2}} ={Mg}\left({h}−{a}−{R}\right) \\ $$$$\mathrm{2}{v}^{\mathrm{2}} +{v}_{{y}} ^{\mathrm{2}} =\mathrm{2}{gh}\left(\mathrm{1}−\alpha−\gamma\right) \\ $$$$\left[\mathrm{2}+\left\{\mathrm{1}+\frac{\gamma\left(\mathrm{1}−\mathrm{2}\gamma\right)}{\left(\mathrm{1}−\gamma\right)\left(\sqrt{\gamma^{\mathrm{2}} +\mathrm{1}−\mathrm{2}\gamma}−\gamma\right)}\right\}^{\mathrm{2}} \right]{v}^{\mathrm{2}} =\mathrm{2}{gh}\left(\mathrm{1}−\alpha−\gamma\right) \\ $$$$\Rightarrow{v}=\sqrt{\frac{\mathrm{2}{gh}\left(\mathrm{1}−\alpha−\gamma\right)}{\mathrm{2}+\left\{\mathrm{1}+\frac{\gamma\left(\mathrm{1}−\mathrm{2}\gamma\right)}{\left(\mathrm{1}−\gamma\right)\left(\sqrt{\gamma^{\mathrm{2}} +\mathrm{1}−\mathrm{2}\gamma}−\gamma\right)}\right\}^{\mathrm{2}} }} \\ $$
Commented by ajfour last updated on 14/Dec/18
Plentiful of mind and effort Sir;  thanks, let me time to follow it  entirely..
$${Plentiful}\:{of}\:{mind}\:{and}\:{effort}\:{Sir}; \\ $$$${thanks},\:{let}\:{me}\:{time}\:{to}\:{follow}\:{it} \\ $$$${entirely}.. \\ $$
Commented by mr W last updated on 14/Dec/18
Commented by mr W last updated on 14/Dec/18
at t=0: x=a, θ=θ_0   at t=t_1 : x=R, θ=θ_1
$${at}\:{t}=\mathrm{0}:\:{x}={a},\:\theta=\theta_{\mathrm{0}} \\ $$$${at}\:{t}={t}_{\mathrm{1}} :\:{x}={R},\:\theta=\theta_{\mathrm{1}} \\ $$
Commented by ajfour last updated on 16/Dec/18
Very brilliant Sir, AWESOME !
$${Very}\:{brilliant}\:{Sir},\:\mathcal{AWESOME}\:! \\ $$
Answered by ajfour last updated on 14/Dec/18
Commented by ajfour last updated on 14/Dec/18
let initially  θ = α,  x=a,  y=a,  z= l  finally     θ=β,  x=R,  y=h−R, z= s  x+Rcos θ = zsin θ  R+Rsin θ = h−zcos θ  ⇒  R+Rsin θ = h−z(((zsin θ−x)/R))  ⇒   sin θ = ((h−R+((xz)/R))/(R+(z^2 /R)))  ________________________  ⇒   sin α = ((h−R+((al)/R))/(R+(l^2 /R)))          ...(i)  &     sin β = ((h−R+s)/(R+(s^2 /R)))             ...(ii)    l^2  = (h−R−Rsin α)^2 +a^2         ...(iii)    s^2  = (h−R−Rsin β)^2 +R^2       ....(iv)  from length of string ,     (αR+l+a) − (βR+s+R)= a−R                                                                     ...(v)  unknowns are : α, β, l, s, a
$${let}\:{initially}\:\:\theta\:=\:\alpha,\:\:{x}={a},\:\:{y}={a},\:\:{z}=\:{l} \\ $$$${finally}\:\:\:\:\:\theta=\beta,\:\:{x}={R},\:\:{y}={h}−{R},\:{z}=\:{s} \\ $$$${x}+{R}\mathrm{cos}\:\theta\:=\:{z}\mathrm{sin}\:\theta \\ $$$${R}+{R}\mathrm{sin}\:\theta\:=\:{h}−{z}\mathrm{cos}\:\theta \\ $$$$\Rightarrow\:\:{R}+{R}\mathrm{sin}\:\theta\:=\:{h}−{z}\left(\frac{{z}\mathrm{sin}\:\theta−{x}}{{R}}\right) \\ $$$$\Rightarrow\:\:\:\mathrm{sin}\:\theta\:=\:\frac{{h}−{R}+\frac{{xz}}{{R}}}{{R}+\frac{{z}^{\mathrm{2}} }{{R}}} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\Rightarrow\:\:\:\mathrm{sin}\:\alpha\:=\:\frac{{h}−{R}+\frac{{al}}{{R}}}{{R}+\frac{{l}^{\mathrm{2}} }{{R}}}\:\:\:\:\:\:\:\:\:\:…\left({i}\right) \\ $$$$\&\:\:\:\:\:\mathrm{sin}\:\beta\:=\:\frac{{h}−{R}+{s}}{{R}+\frac{{s}^{\mathrm{2}} }{{R}}}\:\:\:\:\:\:\:\:\:\:\:\:\:…\left({ii}\right) \\ $$$$\:\:{l}^{\mathrm{2}} \:=\:\left({h}−{R}−{R}\mathrm{sin}\:\alpha\right)^{\mathrm{2}} +{a}^{\mathrm{2}} \:\:\:\:\:\:\:\:…\left({iii}\right) \\ $$$$\:\:{s}^{\mathrm{2}} \:=\:\left({h}−{R}−{R}\mathrm{sin}\:\beta\right)^{\mathrm{2}} +{R}^{\mathrm{2}} \:\:\:\:\:\:….\left({iv}\right) \\ $$$${from}\:{length}\:{of}\:{string}\:, \\ $$$$\:\:\:\left(\alpha{R}+{l}+{a}\right)\:−\:\left(\beta{R}+{s}+{R}\right)=\:{a}−{R} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\left({v}\right) \\ $$$${unknowns}\:{are}\::\:\alpha,\:\beta,\:{l},\:{s},\:\boldsymbol{{a}} \\ $$

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