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Question Number 21628 by Tinkutara last updated on 29/Sep/17

One end of a massless spring of constant  100 N/m and natural length 0.5 m is  fixed and the other end is connected to  a particle of mass 0.5 kg lying on a  frictionless horizontal table. The spring  remains horizontal. If the mass is made  to rotate at an angular velocity of 2  rad/s, find the elongation of the spring.

$$\mathrm{One}\:\mathrm{end}\:\mathrm{of}\:\mathrm{a}\:\mathrm{massless}\:\mathrm{spring}\:\mathrm{of}\:\mathrm{constant} \\ $$$$\mathrm{100}\:\mathrm{N}/\mathrm{m}\:\mathrm{and}\:\mathrm{natural}\:\mathrm{length}\:\mathrm{0}.\mathrm{5}\:\mathrm{m}\:\mathrm{is} \\ $$$$\mathrm{fixed}\:\mathrm{and}\:\mathrm{the}\:\mathrm{other}\:\mathrm{end}\:\mathrm{is}\:\mathrm{connected}\:\mathrm{to} \\ $$$$\mathrm{a}\:\mathrm{particle}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{0}.\mathrm{5}\:\mathrm{kg}\:\mathrm{lying}\:\mathrm{on}\:\mathrm{a} \\ $$$$\mathrm{frictionless}\:\mathrm{horizontal}\:\mathrm{table}.\:\mathrm{The}\:\mathrm{spring} \\ $$$$\mathrm{remains}\:\mathrm{horizontal}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{mass}\:\mathrm{is}\:\mathrm{made} \\ $$$$\mathrm{to}\:\mathrm{rotate}\:\mathrm{at}\:\mathrm{an}\:\mathrm{angular}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{2} \\ $$$$\mathrm{rad}/{s},\:\mathrm{find}\:\mathrm{the}\:\mathrm{elongation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{spring}. \\ $$

Answered by ajfour last updated on 30/Sep/17

mω^2 (l+x)=kx  x=((mω^2 l)/(k−mω^2 )) = ((0.5×4×0.5)/(100−0.5×4))  =(1/(98))m ≈ 1.02cm .

$${m}\omega^{\mathrm{2}} \left({l}+{x}\right)={kx} \\ $$$${x}=\frac{{m}\omega^{\mathrm{2}} {l}}{{k}−{m}\omega^{\mathrm{2}} }\:=\:\frac{\mathrm{0}.\mathrm{5}×\mathrm{4}×\mathrm{0}.\mathrm{5}}{\mathrm{100}−\mathrm{0}.\mathrm{5}×\mathrm{4}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{98}}{m}\:\approx\:\mathrm{1}.\mathrm{02}{cm}\:. \\ $$

Commented by Tinkutara last updated on 30/Sep/17

But in 1^(st)  line shouldn′t there be k(l+x)  because now radius is l+x?

$$\mathrm{But}\:\mathrm{in}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{line}\:\mathrm{shouldn}'\mathrm{t}\:\mathrm{there}\:\mathrm{be}\:{k}\left({l}+{x}\right) \\ $$$$\mathrm{because}\:\mathrm{now}\:\mathrm{radius}\:\mathrm{is}\:{l}+{x}? \\ $$

Commented by ajfour last updated on 30/Sep/17

No think again, spring force  is proportional to change in length.

$${No}\:{think}\:{again},\:{spring}\:{force} \\ $$$${is}\:{proportional}\:{to}\:{change}\:{in}\:{length}. \\ $$

Commented by Tinkutara last updated on 30/Sep/17

OK, Thank you very much Sir!

$$\mathrm{OK},\:\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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