Question

A uniform rod $AB$ of mass $m$ and length $l$ is suspended by two massless and inextensible strings $AC$ and $BD$ whose ends $C$ and $D$ are fixed as shown. The tension in the string $BD$ immediately after the string at $A$ is cut is $\frac{15mg}{50+x}$. Find the value of $x$.

Answer 1

Let the tension in the rope BD just after cutting the rope AD be T .
Let us first find the acceleration and angular acceleration of COM .
$\theta$ = 37${}^{\circ }$
Horizontal force equation :
$m{a}_{h,cm}=Tcos(\theta$)
$a_{h,cm}=\frac{Tcos\left(\theta \right)}{m}$
Vertical force equation :
$m{a}_{v,cm}=mg-Tsin(\theta$)
$a_{v,cm}=g-\frac{Tsin\left(\theta \right)}{m}$
Torque equation about center ,
$Tsin\left(\theta \right)\frac{l}{2}=\frac{m{l}^2}{12}\alpha$
$\alpha =\frac{6Tsin\left(\theta \right)}{ml}$
Since B is attached to the rope , it will perform circular motion about point D .
The acceleration of B towards the center $=\frac{v^2}{R}$.
At the instant when the rope is cut, the velocity of B is zero . Therefore acceleration of B along the rope is zero .
Vertical acceleration of B :
$a_{v,b}=a_{v,cm}-\frac{l\alpha }{2}$
$a_{v,b}=g-\frac{Tsin\left(\theta \right)}{m}-\frac{l6Tsin\left(\theta \right)}{2ml}=g-\frac{4Tsin\left(\theta \right)}{m}$
Horizontal acceleration of B :
$a_{h,b}=a_{h,cm}=\frac{Tcos\left(\theta \right)}{m}$
Net acceleration along the rope :
$0=a_{h,b}cos\left(\theta \right)-a_{v,b}sin(\theta$)
$0=\left(\frac{Tcos\left(\theta \right)}{m}\right)cos\left(\theta \right)-(g-\frac{4Tsin\left(\theta \right)}{m})sin(\theta$)
$T\left(4si{n}^2\right(\theta )+co{s}^2(\theta \left)\right)=mgsin(\theta$)
$T=\frac{mgsin\left(\theta \right)}{3si{n}^2\left(\theta \right)+1}$
given $\theta ={37}^{\circ }sin\left(\theta \right)=\frac{3}{5}$
$T=\frac{mg\frac{3}{5}}{3(\frac{3}{5}{)}^2+1}$
$T=\frac{15mg}{52}$
$\therefore x=52$