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 72519 by ajfour last updated on 29/Oct/19

Commented by ajfour last updated on 29/Oct/19

I mean with the stone in the hand,  giving it the minimum velocity.

$${I}\:{mean}\:{with}\:{the}\:{stone}\:{in}\:{the}\:{hand}, \\ $$$${giving}\:{it}\:{the}\:{minimum}\:{velocity}. \\ $$

Answered by mr W last updated on 29/Oct/19

Commented by ajfour last updated on 30/Oct/19

b=atan θ−((a^2 g)/(2u^2 ))(1+tan^2 θ)  let  cot α=m, tan θ=t  h−x=mtx−((m^2 x^2 g)/(2u^2 ))(1+t^2 )  u^2 =((m^2 x^2 g(1+t^2 ))/(2(mtx+x−h)))  (2/(m^2 g))((∂u^2 /∂x))=((2x(1+t^2 )(mtx+x−h)−(mt+1)x^2 (1+t^2 ))/((mtx+x−h)^2 ))  =0 ⇒  2(mtx+x−h)=(mt+1)x  ⇒ mtx+x=2h     .....(i)  (2/(m^2 g))((∂u^2 /∂t))=((2tx^2 (mtx+x−h)−x^2 (1+t^2 )mx)/((mtx+x−h)^2 ))  =0 ⇒  2t(mtx+x−h)=mx(1+t^2 )  ⇒ mt^2 x+2tx−mx=2th     .....(ii)  from (i)&(ii)   x=((2h)/(1+mt))=((2th)/(mt^2 +2t−m))  ⇒  mt^2 +2t−m=t+mt^2   ⇒  t=m  ⇒ tan θ=cot α  ⇒  θ=(π/2)−α     x=((2h)/(1+m^2 ))=2hsin^2 α   (as m=cot α)  And from u^2 =((m^2 x^2 g(1+t^2 ))/(2(mtx+x−h)))     u^2 =((m^2 (((4h^2 )/((1+m^2 )^2 )))g(1+m^2 ))/(2h))  ⇒  u^2 =((2m^2 gh)/(1+m^2 )) = 2ghcos^2 α        u=(√(2gh))cos α , x=2hsin^2 α,        𝛉=(𝛑/2)−𝛂   if α=45°  ,  x=h, u=(√(gh)), θ=45°  for α=30° , x=(h/2), u=(√((3gh)/2)) , θ=60°.

$${b}={a}\mathrm{tan}\:\theta−\frac{{a}^{\mathrm{2}} {g}}{\mathrm{2}{u}^{\mathrm{2}} }\left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \theta\right) \\ $$$${let}\:\:\mathrm{cot}\:\alpha={m},\:\mathrm{tan}\:\theta={t} \\ $$$${h}−{x}={mtx}−\frac{{m}^{\mathrm{2}} {x}^{\mathrm{2}} {g}}{\mathrm{2}{u}^{\mathrm{2}} }\left(\mathrm{1}+{t}^{\mathrm{2}} \right) \\ $$$${u}^{\mathrm{2}} =\frac{{m}^{\mathrm{2}} {x}^{\mathrm{2}} {g}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{\mathrm{2}\left({mtx}+{x}−{h}\right)} \\ $$$$\frac{\mathrm{2}}{{m}^{\mathrm{2}} {g}}\left(\frac{\partial{u}^{\mathrm{2}} }{\partial{x}}\right)=\frac{\mathrm{2}{x}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left({mtx}+{x}−{h}\right)−\left({mt}+\mathrm{1}\right){x}^{\mathrm{2}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{\left({mtx}+{x}−{h}\right)^{\mathrm{2}} } \\ $$$$=\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{2}\left({mtx}+{x}−{h}\right)=\left({mt}+\mathrm{1}\right){x} \\ $$$$\Rightarrow\:{mtx}+{x}=\mathrm{2}{h}\:\:\:\:\:.....\left({i}\right) \\ $$$$\frac{\mathrm{2}}{{m}^{\mathrm{2}} {g}}\left(\frac{\partial{u}^{\mathrm{2}} }{\partial{t}}\right)=\frac{\mathrm{2}{tx}^{\mathrm{2}} \left({mtx}+{x}−{h}\right)−{x}^{\mathrm{2}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right){mx}}{\left({mtx}+{x}−{h}\right)^{\mathrm{2}} } \\ $$$$=\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{2}{t}\left({mtx}+{x}−{h}\right)={mx}\left(\mathrm{1}+{t}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\:{mt}^{\mathrm{2}} {x}+\mathrm{2}{tx}−{mx}=\mathrm{2}{th}\:\:\:\:\:.....\left({ii}\right) \\ $$$${from}\:\left({i}\right)\&\left({ii}\right) \\ $$$$\:{x}=\frac{\mathrm{2}{h}}{\mathrm{1}+{mt}}=\frac{\mathrm{2}{th}}{{mt}^{\mathrm{2}} +\mathrm{2}{t}−{m}} \\ $$$$\Rightarrow\:\:{mt}^{\mathrm{2}} +\mathrm{2}{t}−{m}={t}+{mt}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:{t}={m}\:\:\Rightarrow\:\mathrm{tan}\:\theta=\mathrm{cot}\:\alpha \\ $$$$\Rightarrow\:\:\theta=\frac{\pi}{\mathrm{2}}−\alpha \\ $$$$\:\:\:{x}=\frac{\mathrm{2}{h}}{\mathrm{1}+{m}^{\mathrm{2}} }=\mathrm{2}{h}\mathrm{sin}\:^{\mathrm{2}} \alpha\:\:\:\left({as}\:{m}=\mathrm{cot}\:\alpha\right) \\ $$$${And}\:{from}\:{u}^{\mathrm{2}} =\frac{{m}^{\mathrm{2}} {x}^{\mathrm{2}} {g}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{\mathrm{2}\left({mtx}+{x}−{h}\right)} \\ $$$$\:\:\:{u}^{\mathrm{2}} =\frac{{m}^{\mathrm{2}} \left(\frac{\mathrm{4}{h}^{\mathrm{2}} }{\left(\mathrm{1}+{m}^{\mathrm{2}} \right)^{\mathrm{2}} }\right){g}\left(\mathrm{1}+{m}^{\mathrm{2}} \right)}{\mathrm{2}{h}} \\ $$$$\Rightarrow\:\:{u}^{\mathrm{2}} =\frac{\mathrm{2}{m}^{\mathrm{2}} {gh}}{\mathrm{1}+{m}^{\mathrm{2}} }\:=\:\mathrm{2}{gh}\mathrm{cos}\:^{\mathrm{2}} \alpha \\ $$$$\:\:\:\:\:\:\boldsymbol{{u}}=\sqrt{\mathrm{2}\boldsymbol{{gh}}}\mathrm{cos}\:\alpha\:,\:\boldsymbol{{x}}=\mathrm{2}\boldsymbol{{h}}\mathrm{sin}\:^{\mathrm{2}} \alpha, \\ $$$$\:\:\:\:\:\:\boldsymbol{\theta}=\frac{\boldsymbol{\pi}}{\mathrm{2}}−\boldsymbol{\alpha}\: \\ $$$${if}\:\alpha=\mathrm{45}°\:\:,\:\:{x}={h},\:{u}=\sqrt{{gh}},\:\theta=\mathrm{45}° \\ $$$${for}\:\alpha=\mathrm{30}°\:,\:{x}=\frac{{h}}{\mathrm{2}},\:{u}=\sqrt{\frac{\mathrm{3}{gh}}{\mathrm{2}}}\:,\:\theta=\mathrm{60}°. \\ $$

Commented by mr W last updated on 29/Oct/19

a=u cos θ t  b=u sin θ t−(1/2)gt^2 =a tan θ−((ga^2 )/(2u^2  cos^2  θ))  (b/a)=tan θ−((ga(1+tan^2  θ))/(2u^2 ))  u^2 =((ga(1+tan^2  θ))/(2(tan θ−(b/a))))  let U=((2u^2 )/(ga)), t=tan θ, λ=(b/a)  ⇒U=((1+t^2 )/(t−λ))  (dU/dt)=((2t)/(t−λ))−((1+t^2 )/((t−λ)^2 ))=0  ⇒t^2 −2λt−1=0  ⇒t=λ+(√(λ^2 +1))  ⇒t^2 =2λ(λ+(√(λ^2 +1)))+1  Φ=u_(min) ^2 =((ga(λ^2 +1+λ(√(λ^2 +1))))/(√(λ^2 +1)))=ga(λ+(√(λ^2 +1)))  a=(x/(tan α))  b=h−x  Φ=((gx)/(tan α))(((h−x)/x) tan α+(√((((h−x)/x))^2 tan^2  α+1)))  Φ=g[h−x+(√((h−x)^2 +(x^2 /(tan^2  α))))]  (dΦ/dx)=0  ((−h+(x/(sin^2  α)))/(√((h−x)^2 +(x^2 /(tan^2  α)))))=1  ⇒x=2h sin^2  α  ⇒Φ_(min) =2gh cos^2  α  E_(min) =(1/2)mu^2 =(1/2)mΦ_(min) =mgh cos^2  α

$${a}={u}\:\mathrm{cos}\:\theta\:{t} \\ $$$${b}={u}\:\mathrm{sin}\:\theta\:{t}−\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} ={a}\:\mathrm{tan}\:\theta−\frac{{ga}^{\mathrm{2}} }{\mathrm{2}{u}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta} \\ $$$$\frac{{b}}{{a}}=\mathrm{tan}\:\theta−\frac{{ga}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta\right)}{\mathrm{2}{u}^{\mathrm{2}} } \\ $$$${u}^{\mathrm{2}} =\frac{{ga}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta\right)}{\mathrm{2}\left(\mathrm{tan}\:\theta−\frac{{b}}{{a}}\right)} \\ $$$${let}\:{U}=\frac{\mathrm{2}{u}^{\mathrm{2}} }{{ga}},\:{t}=\mathrm{tan}\:\theta,\:\lambda=\frac{{b}}{{a}} \\ $$$$\Rightarrow{U}=\frac{\mathrm{1}+{t}^{\mathrm{2}} }{{t}−\lambda} \\ $$$$\frac{{dU}}{{dt}}=\frac{\mathrm{2}{t}}{{t}−\lambda}−\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\left({t}−\lambda\right)^{\mathrm{2}} }=\mathrm{0} \\ $$$$\Rightarrow{t}^{\mathrm{2}} −\mathrm{2}\lambda{t}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{t}=\lambda+\sqrt{\lambda^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Rightarrow{t}^{\mathrm{2}} =\mathrm{2}\lambda\left(\lambda+\sqrt{\lambda^{\mathrm{2}} +\mathrm{1}}\right)+\mathrm{1} \\ $$$$\Phi={u}_{{min}} ^{\mathrm{2}} =\frac{{ga}\left(\lambda^{\mathrm{2}} +\mathrm{1}+\lambda\sqrt{\lambda^{\mathrm{2}} +\mathrm{1}}\right)}{\sqrt{\lambda^{\mathrm{2}} +\mathrm{1}}}={ga}\left(\lambda+\sqrt{\lambda^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$${a}=\frac{{x}}{\mathrm{tan}\:\alpha} \\ $$$${b}={h}−{x} \\ $$$$\Phi=\frac{{gx}}{\mathrm{tan}\:\alpha}\left(\frac{{h}−{x}}{{x}}\:\mathrm{tan}\:\alpha+\sqrt{\left(\frac{{h}−{x}}{{x}}\right)^{\mathrm{2}} \mathrm{tan}^{\mathrm{2}} \:\alpha+\mathrm{1}}\right) \\ $$$$\Phi={g}\left[{h}−{x}+\sqrt{\left({h}−{x}\right)^{\mathrm{2}} +\frac{{x}^{\mathrm{2}} }{\mathrm{tan}^{\mathrm{2}} \:\alpha}}\right] \\ $$$$\frac{{d}\Phi}{{dx}}=\mathrm{0} \\ $$$$\frac{−{h}+\frac{{x}}{\mathrm{sin}^{\mathrm{2}} \:\alpha}}{\sqrt{\left({h}−{x}\right)^{\mathrm{2}} +\frac{{x}^{\mathrm{2}} }{\mathrm{tan}^{\mathrm{2}} \:\alpha}}}=\mathrm{1} \\ $$$$\Rightarrow{x}=\mathrm{2}{h}\:\mathrm{sin}^{\mathrm{2}} \:\alpha \\ $$$$\Rightarrow\Phi_{{min}} =\mathrm{2}{gh}\:\mathrm{cos}^{\mathrm{2}} \:\alpha \\ $$$${E}_{{min}} =\frac{\mathrm{1}}{\mathrm{2}}{mu}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{m}\Phi_{{min}} ={mgh}\:\mathrm{cos}^{\mathrm{2}} \:\alpha \\ $$

Commented by ajfour last updated on 30/Oct/19

Thanks Sir. i corrected.

$$\mathcal{T}{hanks}\:{Sir}.\:{i}\:{corrected}. \\ $$

Commented by mr W last updated on 30/Oct/19

Commented by mr W last updated on 30/Oct/19

position to hit the lamp with minimum  energy.

$${position}\:{to}\:{hit}\:{the}\:{lamp}\:{with}\:{minimum} \\ $$$${energy}. \\ $$

Commented by ajfour last updated on 30/Oct/19

Great interpretation Sir, thanks  a lot. (should have been a  classical question:)

$${Great}\:{interpretation}\:{Sir},\:{thanks} \\ $$$${a}\:{lot}.\:\left({should}\:{have}\:{been}\:{a}\right. \\ $$$$\left.{classical}\:{question}:\right) \\ $$

Commented by mr W last updated on 30/Oct/19

nice question!

$${nice}\:{question}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com