A rigid rod of length ell is sliding. At some | vedclass

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(NONE) Let the position of end $A$ be $(x, 0)$ and end $B$ be $(0, y)$. Since the rod has length $\ell$,we have $x^2 + y^2 = \ell^2$.Given that end $A

Vedclass · Accepted Answer

The velocity of end $B$ along the length of the rod is $\frac{v_0 y}{\sqrt{x^2 + y^2}}$.

Vedclass · Answer

(NONE) Let the position of end $A$ be $(x, 0)$ and end $B$ be $(0, y)$. Since the rod has length $\ell$,we have $x^2 + y^2 = \ell^2$.Given that end $A$ moves with constant velocity $v_0$ starting from $x=0$ at $t=0$,we have $x = v_0 t$.Substituting this into the constraint equation: $(v_0 t)^2 + y^2 = \ell^2$.Rearranging gives $y = \sqrt{\ell^2 - (v_0 t)^2}$,which represents a circular arc in the $(y, t)$ plane for $y, t \ge 0$. Thus,options $A$ and $C$ are incorrect.For option $B$,the velocity of end $B$ is $\vec{v}_B = \dot{y} \hat{j}$. The component of this velocity along the rod is $v_{B, 	ext{rod}} = \dot{y} \cos 	heta$,where $\cos 	heta = y/\ell$.Differentiating $x^2 + y^2 = \ell^2$ with respect to $t$ gives $2x \dot{x} + 2y \dot{y} = 0$,so $\dot{y} = -\frac{x}{y} v_0$.Thus,$v_{B, 	ext{rod}} = -\frac{x}{y} v_0 \cdot \frac{y}{\ell} = -\frac{x v_0}{\ell}$. This does not match option $B$.Since the rod is rigid,the velocity of $B$ along the rod must equal the velocity of $A$ along the rod. The velocity of $A$ is $v_0$ at an angle $	heta$ to the rod (where $\sin 	heta = x/\ell$),so $v_{A, 	ext{rod}} = v_0 \sin 	heta = v_0 \frac{x}{\ell}$.None of the given options are correct.

$A$ rigid rod of length $\ell$ is sliding. At some instant,the position of the rod is as shown in the figure. End $A$ has a constant velocity $v_0$ along the $x$-axis. At $t = 0$,the rod is vertical with end $B$ at $y = \ell$. Which of the following statements is correct?

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