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Question Number 24447 by Tinkutara last updated on 18/Nov/17
Figure shows two discs of same mass  m. They are rigidly attached to a spring  of stiffness k. The system is in  equilibrium. From this equilibrium  position, the upper disc is pressed  down slowly by a distance x and  released. Find the minimum value of  x, if the lower disc is just lifted off  the ground.
$${Figure}\:{shows}\:{two}\:{discs}\:{of}\:{same}\:{mass} \\ $$$${m}.\:{They}\:{are}\:{rigidly}\:{attached}\:{to}\:{a}\:{spring} \\ $$$${of}\:{stiffness}\:{k}.\:{The}\:{system}\:{is}\:{in} \\ $$$${equilibrium}.\:{From}\:{this}\:{equilibrium} \\ $$$${position},\:{the}\:{upper}\:{disc}\:{is}\:{pressed} \\ $$$${down}\:{slowly}\:{by}\:{a}\:{distance}\:{x}\:{and} \\ $$$${released}.\:{Find}\:{the}\:{minimum}\:{value}\:{of} \\ $$$${x},\:{if}\:{the}\:{lower}\:{disc}\:{is}\:{just}\:{lifted}\:{off} \\ $$$${the}\:{ground}. \\ $$
Commented by Tinkutara last updated on 18/Nov/17
Commented by ajfour last updated on 18/Nov/17
mg=kx_0          ....(i)  E_i =(1/2)k(x_0 +x)^2         ....(ii)  E_f =(1/2)ky^2 +mg(y+x+x_0 )   ..(iii)  ky=mg       ......(iv)  E_i =E_f      gives  (x+x_0 )^2 = y^2  +((2mg)/k)(y+x+x_0 )  using (iv):  (x+x_0 )^2 =(((mg)/k))^2 +((2mg)/k)(((mg)/k)+x+x_0 )  ⇒ (x+x_0 )^2 =3(((mg)/k))^2 +(((2mg)/k))(x+x_0 )  ⇒  (x+x_0  −((mg)/k))^2 = 4(((mg)/k))^2   ⇒ x+x_0  = ((3mg)/k)  recalling from eqn (i): kx_0 =mg  we have  x+((mg)/k) = ((3mg)/k)  Hence     x_(min) =((2mg)/k) .
$${mg}={kx}_{\mathrm{0}} \:\:\:\:\:\:\:\:\:….\left({i}\right) \\ $$$${E}_{{i}} =\frac{\mathrm{1}}{\mathrm{2}}{k}\left({x}_{\mathrm{0}} +{x}\right)^{\mathrm{2}} \:\:\:\:\:\:\:\:….\left({ii}\right) \\ $$$${E}_{{f}} =\frac{\mathrm{1}}{\mathrm{2}}{ky}^{\mathrm{2}} +{mg}\left({y}+{x}+{x}_{\mathrm{0}} \right)\:\:\:..\left({iii}\right) \\ $$$${ky}={mg}\:\:\:\:\:\:\:……\left({iv}\right) \\ $$$${E}_{{i}} ={E}_{{f}} \:\:\:\:\:{gives} \\ $$$$\left({x}+{x}_{\mathrm{0}} \right)^{\mathrm{2}} =\:{y}^{\mathrm{2}} \:+\frac{\mathrm{2}{mg}}{{k}}\left({y}+{x}+{x}_{\mathrm{0}} \right) \\ $$$${using}\:\left({iv}\right): \\ $$$$\left({x}+{x}_{\mathrm{0}} \right)^{\mathrm{2}} =\left(\frac{{mg}}{{k}}\right)^{\mathrm{2}} +\frac{\mathrm{2}{mg}}{{k}}\left(\frac{{mg}}{{k}}+{x}+{x}_{\mathrm{0}} \right) \\ $$$$\Rightarrow\:\left({x}+{x}_{\mathrm{0}} \right)^{\mathrm{2}} =\mathrm{3}\left(\frac{{mg}}{{k}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2}{mg}}{{k}}\right)\left({x}+{x}_{\mathrm{0}} \right) \\ $$$$\Rightarrow\:\:\left({x}+{x}_{\mathrm{0}} \:−\frac{{mg}}{{k}}\right)^{\mathrm{2}} =\:\mathrm{4}\left(\frac{{mg}}{{k}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:{x}+{x}_{\mathrm{0}} \:=\:\frac{\mathrm{3}{mg}}{{k}} \\ $$$${recalling}\:{from}\:{eqn}\:\left({i}\right):\:{kx}_{\mathrm{0}} ={mg} \\ $$$${we}\:{have}\:\:{x}+\frac{{mg}}{{k}}\:=\:\frac{\mathrm{3}{mg}}{{k}} \\ $$$${Hence}\:\:\:\:\:\boldsymbol{{x}}_{{min}} =\frac{\mathrm{2}\boldsymbol{{mg}}}{\boldsymbol{{k}}}\:. \\ $$
Commented by Tinkutara last updated on 18/Nov/17
Why not E_i =(1/2)kx_0 ^2  and how it become  (1/2)(x+x_0 )^2 ? We pressed it down so it  should be subtracted.
$${Why}\:{not}\:{E}_{{i}} =\frac{\mathrm{1}}{\mathrm{2}}{kx}_{\mathrm{0}} ^{\mathrm{2}} \:{and}\:{how}\:{it}\:{become} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left({x}+{x}_{\mathrm{0}} \right)^{\mathrm{2}} ?\:{We}\:{pressed}\:{it}\:{down}\:{so}\:{it} \\ $$$${should}\:{be}\:{subtracted}. \\ $$
Commented by ajfour last updated on 18/Nov/17
the spring was compressed by  x_0  due to weight of upper disc  when it was in equilibrium.  Pressing it down further by x  increases spring potential energy  to     (1/2)k(x+x_0 )^2   . This instant  is taken as the initial position  in my solution. Gravitational  potential energy is taken zero here.  From here upper disc rises   through a height   = y−(−x_0 −x)     = y+x_0 +x  y is the mximum extension in  spring that comes about in the  spring. For minimum x,   upper disc is at its maximum  height then. Spring pulls both  discs towards each other, and  this pull is just sufficient to  make the normal reaction (of  ground) to the lower disc become  zero, that is to say, the lower   disc is lifted off the ground for  any value of x superior to  ((2mg)/k) .
$${the}\:{spring}\:{was}\:{compressed}\:{by} \\ $$$${x}_{\mathrm{0}} \:{due}\:{to}\:{weight}\:{of}\:{upper}\:{disc} \\ $$$${when}\:{it}\:{was}\:{in}\:{equilibrium}. \\ $$$${Pressing}\:{it}\:{down}\:{further}\:{by}\:{x} \\ $$$${increases}\:{spring}\:{potential}\:{energy} \\ $$$${to}\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}{k}\left({x}+{x}_{\mathrm{0}} \right)^{\mathrm{2}} \:\:.\:{This}\:{instant} \\ $$$${is}\:{taken}\:{as}\:{the}\:{initial}\:{position} \\ $$$${in}\:{my}\:{solution}.\:{Gravitational} \\ $$$${potential}\:{energy}\:{is}\:{taken}\:{zero}\:{here}. \\ $$$${From}\:{here}\:{upper}\:{disc}\:{rises}\: \\ $$$${through}\:{a}\:{height}\:\:\:=\:{y}−\left(−{x}_{\mathrm{0}} −{x}\right) \\ $$$$\:\:\:=\:{y}+{x}_{\mathrm{0}} +{x} \\ $$$${y}\:{is}\:{the}\:{mximum}\:{extension}\:{in} \\ $$$${spring}\:{that}\:{comes}\:{about}\:{in}\:{the} \\ $$$${spring}.\:{For}\:{minimum}\:{x},\: \\ $$$${upper}\:{disc}\:{is}\:{at}\:{its}\:{maximum} \\ $$$${height}\:{then}.\:{Spring}\:{pulls}\:{both} \\ $$$${discs}\:{towards}\:{each}\:{other},\:{and} \\ $$$${this}\:{pull}\:{is}\:{just}\:{sufficient}\:{to} \\ $$$${make}\:{the}\:{normal}\:{reaction}\:\left({of}\right. \\ $$$$\left.{ground}\right)\:{to}\:{the}\:{lower}\:{disc}\:{become} \\ $$$${zero},\:{that}\:{is}\:{to}\:{say},\:{the}\:{lower}\: \\ $$$${disc}\:{is}\:{lifted}\:{off}\:{the}\:{ground}\:{for} \\ $$$${any}\:{value}\:{of}\:{x}\:{superior}\:{to}\:\:\frac{\mathrm{2}{mg}}{{k}}\:. \\ $$
Commented by Tinkutara last updated on 18/Nov/17
But why there y−(−x−x_0 )? It should  be y−(x+x_0 ) since spring has already  streached x+x_0  so we have to subtract  this from maximum compression,  i.e. y−x−x_0 . Why wrong?
$${But}\:{why}\:{there}\:{y}−\left(−{x}−{x}_{\mathrm{0}} \right)?\:{It}\:{should} \\ $$$${be}\:{y}−\left({x}+{x}_{\mathrm{0}} \right)\:{since}\:{spring}\:{has}\:{already} \\ $$$${streached}\:{x}+{x}_{\mathrm{0}} \:{so}\:{we}\:{have}\:{to}\:{subtract} \\ $$$${this}\:{from}\:{maximum}\:{compression}, \\ $$$${i}.{e}.\:{y}−{x}−{x}_{\mathrm{0}} .\:{Why}\:{wrong}? \\ $$
Commented by ajfour last updated on 18/Nov/17
Commented by ajfour last updated on 18/Nov/17
y is extension of spring above  equilibrium position;  x_0  and x  are comressions and are below it.
$${y}\:{is}\:{extension}\:{of}\:{spring}\:{above} \\ $$$${equilibrium}\:{position};\:\:{x}_{\mathrm{0}} \:{and}\:{x} \\ $$$${are}\:{comressions}\:{and}\:{are}\:{below}\:{it}. \\ $$
Commented by Tinkutara last updated on 18/Nov/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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