Question

A thin uniform bar of mass m and length 2 L is held at an angle ${30}^0$ with the horizontal by means of two vertical inextensible strings, at each end as shown in the figure. If the string at the right end breaks, leaving the bar to swing, then immediately after string breaks,

• A. the tension in the string at the left end is $T$ = $\frac{mg}{11}$
• B. the tension in the string at the left end is $T$ = $\frac{2mg}{11}$
• C. the angular acceleration of the bar $a$ = $\frac{3\sqrt{3}}{11}$ $\frac{g}{L}$
• D. the angular acceleration of the bar $a$ = $\frac{2\sqrt{3}}{11}$ $\frac{g}{L}$

Answer 1

Answer: (B), (C)

The correct options are
B the tension in the string at the left end is $T$ = $\frac{2mg}{11}$
C the angular acceleration of the bar $a$ = $\frac{3\sqrt{3}}{11}$ $\frac{g}{L}$

Just after cutting the string the tension of right string will disappear and tension in left string will change. let us draw free body diagram of rod immediately after the string breaks.
Let $a_x$ and $b_y$ be the linear accelerations of $COM$ and $\alpha$ the
angular acceleration of the rod about $COM$ as shown.
As no force in horizontal direction. Then $\sum F_x$ = 0
$\therefore$ $a_x$ = 0 ......(i)
$a_y$ = $\frac{\sum F_y}{m}$ = $\frac{mg-T}{m}$ ....(ii)
Applying torque equation about centre of mass
$\tau$ = $I_{cm}\alpha$
$T.\left(L\cos {30}^0\right)$ = $\frac{m(2L{)}^2}{12}.\alpha$
$\Rightarrow$ $\alpha$ =$\frac{\tau }{I}$ = $\frac{TL\cos {30}^0}{\frac{m(2L{)}^2}{12}}$
= $\frac{3\sqrt{3T}}{2mL}$ (II)
Now after the string breaks, acceleration of point A in the vertical direction should be zero.
i.e.,
$a_y$ = $L\alpha cos{30}^0$ (iv)
Solving equations $\left(i\right),\left(ii\right),\left(iii\right)and\left(iv\right)$
$T$ = $\frac{2mg}{11}$ and
$a$ = $\frac{3\sqrt{3}}{11}$ $\frac{g}{L}$