In the figure shown,ABC is a uniform wire. If | vedclass

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(C) Let the length of wire $AB = x$ and $BC = y$. Let the linear mass density be $\lambda$.The center of mass of $BC$ is at $(y/2, 0)$ and its mass is

Vedclass · Accepted Answer

$1.37$

Vedclass · Answer

(C) Let the length of wire $AB = x$ and $BC = y$. Let the linear mass density be $\lambda$.The center of mass of $BC$ is at $(y/2, 0)$ and its mass is $m_1 = \lambda y$.The center of mass of $AB$ is at $(x/2 \cos 60^{\circ}, x/2 \sin 60^{\circ}) = (x/4, x\sqrt{3}/4)$ and its mass is $m_2 = \lambda x$.The $x$-coordinate of the center of mass of the system is given by:$X_{cm} = \frac{m_1(y/2) + m_2(x/4)}{m_1 + m_2} = \frac{\lambda y(y/2) + \lambda x(x/4)}{\lambda(x + y)} = \frac{y^2/2 + x^2/4}{x + y}$.Since the center of mass lies vertically below $A$,its $x$-coordinate must be equal to the $x$-coordinate of point $A$,which is $x \cos 60^{\circ} = x/2$.Equating the two:$\frac{y^2/2 + x^2/4}{x + y} = \frac{x}{2} \Rightarrow y^2/2 + x^2/4 = x^2/2 + xy/2$.Multiplying by $4$:$2y^2 + x^2 = 2x^2 + 2xy \Rightarrow 2y^2 - 2xy - x^2 = 0$.Dividing by $x^2$ and letting $r = y/x$:$2r^2 - 2r - 1 = 0$.Using the quadratic formula $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$r = \frac{2 \pm \sqrt{4 - 4(2)(-1)}}{4} = \frac{2 \pm \sqrt{12}}{4} = \frac{2 \pm 2\sqrt{3}}{4} = \frac{1 \pm \sqrt{3}}{2}$.Since $r$ must be positive,$r = \frac{1 + \sqrt{3}}{2} \approx \frac{1 + 1.732}{2} = 1.366 \approx 1.37$.

In the figure shown,$ABC$ is a uniform wire. If the center of mass of the wire lies vertically below point $A$,then $\frac{BC}{AB}$ is close to:

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