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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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