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Question Number 22286 by Tinkutara last updated on 14/Oct/17
A wire, which passes through the hole  in a small bead, is bent in the form of  quarter of a circle. The wire is fixed  vertically on ground. The bead is  released from near the top of the wire  and it slides along the wire without  friction. As the bead moves from A to B,  the force it applies on the wire B is  (a) always radially outwards  (b) always radially inwards  (c) radially outwards initially and  radially inwards later  (d) radially inwards initially and  radially outwards later.
$$\mathrm{A}\:\mathrm{wire},\:\mathrm{which}\:\mathrm{passes}\:\mathrm{through}\:\mathrm{the}\:\mathrm{hole} \\ $$$$\mathrm{in}\:\mathrm{a}\:\mathrm{small}\:\mathrm{bead},\:\mathrm{is}\:\mathrm{bent}\:\mathrm{in}\:\mathrm{the}\:\mathrm{form}\:\mathrm{of} \\ $$$$\mathrm{quarter}\:\mathrm{of}\:\mathrm{a}\:\mathrm{circle}.\:\mathrm{The}\:\mathrm{wire}\:\mathrm{is}\:\mathrm{fixed} \\ $$$$\mathrm{vertically}\:\mathrm{on}\:\mathrm{ground}.\:\mathrm{The}\:\mathrm{bead}\:\mathrm{is} \\ $$$$\mathrm{released}\:\mathrm{from}\:\mathrm{near}\:\mathrm{the}\:\mathrm{top}\:\mathrm{of}\:\mathrm{the}\:\mathrm{wire} \\ $$$$\mathrm{and}\:\mathrm{it}\:\mathrm{slides}\:\mathrm{along}\:\mathrm{the}\:\mathrm{wire}\:\mathrm{without} \\ $$$$\mathrm{friction}.\:\mathrm{As}\:\mathrm{the}\:\mathrm{bead}\:\mathrm{moves}\:\mathrm{from}\:{A}\:\mathrm{to}\:{B}, \\ $$$$\mathrm{the}\:\mathrm{force}\:\mathrm{it}\:\mathrm{applies}\:\mathrm{on}\:\mathrm{the}\:\mathrm{wire}\:{B}\:\mathrm{is} \\ $$$$\left({a}\right)\:\mathrm{always}\:\mathrm{radially}\:\mathrm{outwards} \\ $$$$\left({b}\right)\:\mathrm{always}\:\mathrm{radially}\:\mathrm{inwards} \\ $$$$\left({c}\right)\:\mathrm{radially}\:\mathrm{outwards}\:\mathrm{initially}\:\mathrm{and} \\ $$$$\mathrm{radially}\:\mathrm{inwards}\:\mathrm{later} \\ $$$$\left({d}\right)\:\mathrm{radially}\:\mathrm{inwards}\:\mathrm{initially}\:\mathrm{and} \\ $$$$\mathrm{radially}\:\mathrm{outwards}\:\mathrm{later}. \\ $$
Commented by Tinkutara last updated on 14/Oct/17
Commented by ajfour last updated on 15/Oct/17
as speed increases centripetal  force increases so Normal force  in bead increase radially inwards.  Initially when speed is zero or  small Normal on bead is clearly  radially outwards.  So force of bead on wire is  initially radially inwards and  radially outwards later.
$${as}\:{speed}\:{increases}\:{centripetal} \\ $$$${force}\:{increases}\:{so}\:{Normal}\:{force} \\ $$$${in}\:{bead}\:{increase}\:{radially}\:{inwards}. \\ $$$${Initially}\:{when}\:{speed}\:{is}\:{zero}\:{or} \\ $$$${small}\:{Normal}\:{on}\:{bead}\:{is}\:{clearly} \\ $$$${radially}\:{outwards}. \\ $$$${So}\:{force}\:{of}\:{bead}\:{on}\:{wire}\:{is} \\ $$$${initially}\:{radially}\:{inwards}\:{and} \\ $$$${radially}\:{outwards}\:{later}. \\ $$
Commented by math solver last updated on 16/Oct/17
hey a doubt : what is the direction  of centripetal force ? i mean inwards or  or outwards (radially) ????
$${hey}\:{a}\:{doubt}\::\:{what}\:{is}\:{the}\:{direction} \\ $$$${of}\:{centripetal}\:{force}\:?\:{i}\:{mean}\:{inwards}\:{or} \\ $$$${or}\:{outwards}\:\left({radially}\right)\:???? \\ $$
Commented by Tinkutara last updated on 16/Oct/17
Always inwards.
$$\mathrm{Always}\:\mathrm{inwards}. \\ $$
Commented by math solver last updated on 16/Oct/17
ok . and what about centrifugal  force??
$${ok}\:.\:{and}\:{what}\:{about}\:{centrifugal} \\ $$$${force}?? \\ $$
Answered by ajfour last updated on 14/Oct/17
 When bead is at∠ θ from horizontal,  mgsin θ−N(radially outwards on                               bead)=mv^2 /r  while    (1/2)mv^2 =mgr(1−sin θ)  ⇒   mgsin θ−N=2mg(1−sin θ)  ⇒  N=mg(3sin θ−2) > 0  until sin θ > (2/3) .  hence until the ∠ θ is greater than  θ=sin^(−1) ((2/3)) , Normal on bead is  radially outwards, and radially  inwards later when it acquires  a higher speed.  So by Newton′s 3^(rd ) law force on  wire is radially inwards for   sin^(−1) ((2/3)) < θ ≤ (π/2) and radially  outwards later.
$$\:{When}\:{bead}\:{is}\:{at}\angle\:\theta\:{from}\:{horizontal}, \\ $$$${mg}\mathrm{sin}\:\theta−{N}\left({radially}\:{outwards}\:{on}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{bead}\right)={mv}^{\mathrm{2}} /{r} \\ $$$${while}\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} ={mgr}\left(\mathrm{1}−\mathrm{sin}\:\theta\right) \\ $$$$\Rightarrow\:\:\:{mg}\mathrm{sin}\:\theta−{N}=\mathrm{2}{mg}\left(\mathrm{1}−\mathrm{sin}\:\theta\right) \\ $$$$\Rightarrow\:\:{N}={mg}\left(\mathrm{3sin}\:\theta−\mathrm{2}\right)\:>\:\mathrm{0} \\ $$$${until}\:\mathrm{sin}\:\theta\:>\:\frac{\mathrm{2}}{\mathrm{3}}\:. \\ $$$${hence}\:{until}\:{the}\:\angle\:\theta\:{is}\:{greater}\:{than} \\ $$$$\theta=\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{3}}\right)\:,\:{Normal}\:{on}\:{bead}\:{is} \\ $$$${radially}\:{outwards},\:{and}\:{radially} \\ $$$${inwards}\:{later}\:{when}\:{it}\:{acquires} \\ $$$${a}\:{higher}\:{speed}. \\ $$$${So}\:{by}\:{Newton}'{s}\:\mathrm{3}^{{rd}\:} {law}\:{force}\:{on} \\ $$$${wire}\:{is}\:{radially}\:{inwards}\:{for} \\ $$$$\:\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{3}}\right)\:<\:\theta\:\leqslant\:\frac{\pi}{\mathrm{2}}\:{and}\:{radially} \\ $$$${outwards}\:{later}. \\ $$$$ \\ $$$$ \\ $$
Commented by Tinkutara last updated on 14/Oct/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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