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

$$\mathrm{A}\mathrm{small}\mathrm{bead}\mathrm{of}\mathrm{mass}m\mathrm{is}\mathrm{given}\mathrm{an}$$$$\mathrm{initial}\mathrm{velocity}\mathrm{of}\mathrm{magnitude}{v}_0\mathrm{on}\mathrm{a}$$$$\mathrm{horizontal}\mathrm{circular}\mathrm{wire}.\mathrm{If}\mathrm{the}$$$$\mathrm{coefficient}\mathrm{of}\mathrm{kinetic}\mathrm{friction}\mathrm{is}{\mu }_{\mathrm{k}},\mathrm{the}$$$$\mathrm{determine}\mathrm{the}\mathrm{distance}\mathrm{travelled}\mathrm{before}$$$$\mathrm{the}\mathrm{collar}\mathrm{comes}\mathrm{to}\mathrm{rest}.(\mathrm{Given}\mathrm{that}$$$$\mathrm{radius}\mathrm{of}\mathrm{circular}\mathrm{wire}\mathrm{is}R).$$

Answered by ajfour last updated on 15/Oct/17

$$f={\mu }_{k}N={\mu }_k\sqrt{{\left(\frac{{mv}^2}{R}\right)}^2+{\left(mg\right)}^2}$$$${dW}_f=dK$$$$-{\mu }_k\left(\sqrt{{\left(\frac{{mv}^2}{R}\right)}^2+{\left(mg\right)}^2}\right)Rd\theta =mvdv$$$$\Rightarrow -{\mu }_{k}R{\int }_0^{{\theta }_f}d\theta ={\int }_{v_0}^0\frac{vdv}{\sqrt{{\left(\frac{v^2}{R}\right)}^2+g^2}}$$$${\mu }_{k}R{\theta }_f=\frac{R}{2}{\int }_0^{v_0}\frac{d\left(\frac{v^2}{R}\right)}{\sqrt{{\left(\frac{v^2}{R}\right)}^2+g^2}}$$$$\Rightarrow distance=R{\theta }_f=$$$$\frac{R}{2{\mu }_k}\ln \mid \frac{v^2}{R}+\sqrt{{\left(\frac{v^2}{R}\right)}^2+g^2}{\mid }_0^{v_0}$$$$=\frac{R}{2{\mu }_k}\ln \mid \frac{v_0^2}{Rg}+\sqrt{{\left(\frac{v_0^2}{Rg}\right)}^2+1}\mid .$$$$$$

Commented by Tinkutara last updated on 15/Oct/17

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

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