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Question Number 22302 by Tinkutara last updated on 14/Oct/17

A small bead of mass m is given an initial velocity of magnitude v_0 on a horizontal circular wire. If the coefficient of kinetic friction is μ_k , the determine the distance travelled before the collar comes to rest. (Given that radius of circular wire is R).

$$\mathrm{A}\:\mathrm{small}\:\mathrm{bead}\:\mathrm{of}\:\mathrm{mass}\:{m}\:\mathrm{is}\:\mathrm{given}\:\mathrm{an} \\ $$$$\mathrm{initial}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{magnitude}\:{v}_{\mathrm{0}} \:\mathrm{on}\:\mathrm{a} \\ $$$$\mathrm{horizontal}\:\mathrm{circular}\:\mathrm{wire}.\:\mathrm{If}\:\mathrm{the} \\ $$$$\mathrm{coefficient}\:\mathrm{of}\:\mathrm{kinetic}\:\mathrm{friction}\:\mathrm{is}\:\mu_{\mathrm{k}} ,\:\mathrm{the} \\ $$$$\mathrm{determine}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{travelled}\:\mathrm{before} \\ $$$$\mathrm{the}\:\mathrm{collar}\:\mathrm{comes}\:\mathrm{to}\:\mathrm{rest}.\:\left(\mathrm{Given}\:\mathrm{that}\right. \\ $$$$\left.\mathrm{radius}\:\mathrm{of}\:\mathrm{circular}\:\mathrm{wire}\:\mathrm{is}\:{R}\right). \\ $$

Answered by ajfour last updated on 15/Oct/17

f=μ_k N = μ_k (√((((mv^2 )/R))^2 +(mg)^2 )) dW_f =dK −μ_k ((√((((mv^2 )/R))^2 +(mg)^2 )) )Rdθ =mvdv ⇒ −μ_k R∫_0 ^( θ_f ) dθ =∫_v_0 ^( 0) ((vdv)/(√(((v^2 /R))^2 +g^2 ))) μ_k Rθ_f =(R/2)∫_0 ^( v_0 ) ((d((v^2 /R)))/(√(((v^2 /R))^2 +g^2 ))) ⇒distance=Rθ_f = (R/(2μ_k ))ln ∣(v^2 /R)+(√(((v^2 /R))^2 +g^2 )) ∣_0 ^v_0 =(R/(2μ_k ))ln ∣(v_0 ^2 /(Rg))+(√(((v_0 ^2 /(Rg)))^2 +1)) ∣ .

$${f}=\mu_{{k}} {N}\:=\:\mu_{{k}} \sqrt{\left(\frac{{mv}^{\mathrm{2}} }{{R}}\right)^{\mathrm{2}} +\left({mg}\right)^{\mathrm{2}} } \\ $$$${dW}_{{f}} ={dK} \\ $$$$−\mu_{{k}} \left(\sqrt{\left(\frac{{mv}^{\mathrm{2}} }{{R}}\right)^{\mathrm{2}} +\left({mg}\right)^{\mathrm{2}} }\:\right){Rd}\theta\:={mvdv} \\ $$$$\Rightarrow\:−\mu_{{k}} {R}\int_{\mathrm{0}} ^{\:\:\theta_{{f}} } {d}\theta\:=\int_{{v}_{\mathrm{0}} } ^{\:\:\mathrm{0}} \:\frac{{vdv}}{\sqrt{\left(\frac{{v}^{\mathrm{2}} }{{R}}\right)^{\mathrm{2}} +{g}^{\mathrm{2}} }} \\ $$$$\:\:\:\:\mu_{{k}} {R}\theta_{{f}} \:=\frac{{R}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\:{v}_{\mathrm{0}} } \frac{{d}\left(\frac{{v}^{\mathrm{2}} }{{R}}\right)}{\sqrt{\left(\frac{{v}^{\mathrm{2}} }{{R}}\right)^{\mathrm{2}} +{g}^{\mathrm{2}} }} \\ $$$$\Rightarrow{distance}={R}\theta_{{f}} = \\ $$$$\:\:\:\:\:\:\:\:\frac{{R}}{\mathrm{2}\mu_{{k}} }\mathrm{ln}\:\mid\frac{{v}^{\mathrm{2}} }{{R}}+\sqrt{\left(\frac{{v}^{\mathrm{2}} }{{R}}\right)^{\mathrm{2}} +{g}^{\mathrm{2}} }\:\mid_{\mathrm{0}} ^{{v}_{\mathrm{0}} } \\ $$$$\:\:\:\:\:=\frac{{R}}{\mathrm{2}\mu_{{k}} }\mathrm{ln}\:\mid\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} }{{Rg}}+\sqrt{\left(\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} }{{Rg}}\right)^{\mathrm{2}} +\mathrm{1}}\:\mid\:. \\ $$$$ \\ $$

Commented by Tinkutara last updated on 15/Oct/17

Thank you very much Sir!

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

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