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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-di

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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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