Question

Collars $\text{A}$ and $\text{B}$ slide along the fixed right-angle rods and are connected by a cord of length $L$. Determine the acceleration of collar $\text{B}$ as a function of $y$ if collar $\text{A}$ is given a constant upward velocity $v_A$.

• A. $-\frac{L^2{v}_A^2}{{(L^2-y^2)}^{3/2}}$
• B. $-\frac{L{v}_A^2}{{(L^2-y^2)}^{3/2}}$
• C. $-\frac{L^2{v}_A^2}{{(L-y)}^3}$
• D. $-\frac{L^2{v}_A}{{(L^2-y^2)}^{3/2}}$

Answer 1

The correct option is A $-\frac{L^2{v}_A^2}{{(L^2-y^2)}^{3/2}}$

At the given instant, the velocities and distances of the collars have been shown in figure.


From the right triangle,
$x^2+y^2=L^2$
$\Rightarrow x=(L^2-y^2{)}^{1/2}$
Differentiating both sides w.r.t time:
$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0...\left(i\right)$
$\Rightarrow \frac{dx}{dt}=-\frac{y\frac{dy}{dt}}{x}...\left(ii\right)$

Now using $\dot{x}=\frac{dx}{dt},\dot{y}=\frac{dy}{dt}$, eqn. $\left(ii\right)$ can be written as,
$\Rightarrow \dot{x}=-\frac{y\dot{y}}{x}$
Again differentiating eqn. $\left(i\right)$ w.r.t $t$ gives,
${\left(\frac{dx}{dt}\right)}^2+x\frac{d^{2}x}{d{t}^2}+{\left(\frac{dy}{dt}\right)}^2+y\frac{d^{2}y}{d{t}^2}=0$
$\Rightarrow {\dot{x}}^2+x\ddot{x}+{\dot{y}}^2+y\ddot{y}=0...\left(iii\right)$
where $\dot{y}=\frac{dy}{dt}=v_A=$ constant. It represents the rate of change of distance $y$, for collar $\text{A}$
$\therefore \ddot{y}=\frac{d^{2}y}{d{t}^2}=0$

Hence from Eq $\left(iii\right)$, acceleration of collar $\text{B}$:
$\frac{d^{2}x}{d{t}^2}=-\frac{{\dot{x}}^2+{\dot{y}}^2}{x}=-\frac{{(-\frac{y\dot{y}}{x})}^2+{\dot{y}}^2}{x}$
$\Rightarrow \frac{d^{2}x}{d{t}^2}=-\frac{\left(\frac{y^2\times {\left(\frac{dy}{dt}\right)}^2}{x^2}\right)+{\left(\frac{dy}{dt}\right)}^2}{x}$

Putting the values of $\frac{dy}{dt}=v_A,x^2+y^2=L^2$ gives:
$\frac{d^{2}x}{d{t}^2}=-\frac{{v_A}^2(x^2+y^2)}{x^3}$
$\therefore \frac{d^{2}x}{d{t}^2}=-\frac{L^2{v}_A^2}{{(L^2-y^2)}^{3/2}}$