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Question Number 21604 by Tinkutara last updated on 29/Sep/17

A particle will leave a vertical circle of radius r, when its velocity at the lowest point of the circle (v_L ) is (a) (√(2gr)) (b) (√(5gr)) (c) (√(3gr)) (d) (√(6gr))

$$\mathrm{A}\:\mathrm{particle}\:\mathrm{will}\:\mathrm{leave}\:\mathrm{a}\:\mathrm{vertical}\:\mathrm{circle}\:\mathrm{of} \\ $$$$\mathrm{radius}\:{r},\:\mathrm{when}\:\mathrm{its}\:\mathrm{velocity}\:\mathrm{at}\:\mathrm{the}\:\mathrm{lowest} \\ $$$$\mathrm{point}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\left({v}_{{L}} \right)\:\mathrm{is} \\ $$$$\left({a}\right)\:\sqrt{\mathrm{2}{gr}} \\ $$$$\left({b}\right)\:\sqrt{\mathrm{5}{gr}} \\ $$$$\left({c}\right)\:\sqrt{\mathrm{3}{gr}} \\ $$$$\left({d}\right)\:\sqrt{\mathrm{6}{gr}} \\ $$

Answered by ajfour last updated on 29/Sep/17

at the top if it leaves the circle, then mg=m(v_H ^2 /r) or mv_H ^2 =mgr ...(i) from conservation of energy: 2mgr=(1/2)m(v_L ^2 −v_H ^2 ) ⇒ 4gr = mv_L ^2 −mgr [see (i) ] v_L = (√(5gr)) . whebever v_L < (√(5gr)) particle leaves the vertical circle. So (a) and (c) .

$${at}\:{the}\:{top}\:{if}\:{it}\:{leaves}\:{the}\: \\ $$$${circle},\:{then}\: \\ $$$${mg}={m}\frac{{v}_{{H}} ^{\mathrm{2}} }{{r}}\:\:\:\:{or}\:\:\:{mv}_{{H}} ^{\mathrm{2}} ={mgr}\:\:\:...\left({i}\right) \\ $$$${from}\:{conservation}\:{of}\:{energy}: \\ $$$$\mathrm{2}{mgr}=\frac{\mathrm{1}}{\mathrm{2}}{m}\left({v}_{{L}} ^{\mathrm{2}} −{v}_{{H}} ^{\mathrm{2}} \right) \\ $$$$\Rightarrow\:\:\:\:\mathrm{4}{gr}\:=\:{mv}_{{L}} ^{\mathrm{2}} −{mgr}\:\:\:\:\:\left[{see}\:\left({i}\right)\:\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:{v}_{{L}} =\:\sqrt{\mathrm{5}{gr}}\:. \\ $$$${whebever}\:{v}_{{L}} <\:\sqrt{\mathrm{5}{gr}}\:\:{particle} \\ $$$${leaves}\:{the}\:{vertical}\:{circle}. \\ $$$${So}\:\left({a}\right)\:{and}\:\left({c}\right)\:. \\ $$

Commented by Tinkutara last updated on 29/Sep/17

But answer is given only (c) not (a).

$$\mathrm{But}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{given}\:\mathrm{only}\:\left({c}\right)\:\mathrm{not}\:\left({a}\right). \\ $$

Commented by ajfour last updated on 29/Sep/17

but with v_L =(√(2gr)) it swings only in the lower half vertixal circle; no chance of leaving the circle. so just (c) and not (a). sorry for the mistake.

$${but}\:{with}\:{v}_{{L}} =\sqrt{\mathrm{2}{gr}}\:{it}\:{swings}\:{only} \\ $$$${in}\:{the}\:{lower}\:{half}\:{vertixal}\:{circle}; \\ $$$${no}\:{chance}\:{of}\:{leaving}\:{the}\:{circle}. \\ $$$${so}\:{just}\:\left({c}\right)\:{and}\:{not}\:\left({a}\right).\:{sorry} \\ $$$${for}\:{the}\:{mistake}. \\ $$

Commented by Tinkutara last updated on 29/Sep/17

Thank you very much Sir!

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

Commented by Tinkutara last updated on 29/Sep/17

Can you please guide Q. 21307?

$$\mathrm{Can}\:\mathrm{you}\:\mathrm{please}\:\mathrm{guide}\:\mathrm{Q}.\:\mathrm{21307}? \\ $$

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