Question

In the figure shown below, $u=\sqrt{7gl}$ and mass of the bob is $3\text{kg}$. Find the value of tension in the string when angle $\theta ={180}^{\circ }$ [Take $g=10{\text{m/s}}^2$]

• A. $40\text{N}$
• B. $20\text{N}$
• C. $0\text{N}$
• D. $60\text{N}$

Answer 1

The correct option is D $60\text{N}$


Taking $A$ to be the reference point
At an angle $\theta ={180}^{\circ }$, the height $\left(h\right)=2l$.
Let $v$ be the velocity of bob at the highest point of vertical circular path.
Applying energy conservation at point $A$ and $B$
$(KE{)}_A=(KE{)}_B+(PE{)}_B$
$\Rightarrow \frac{1}{2}m{u}^2=\frac{1}{2}m{v}^2+mg\left(2l\right)$
$\Rightarrow v=\sqrt{u^2-4gl}=\sqrt{7gl-4gl}=\sqrt{3gl}$
Radial acceleration at $B$ $\left(a_n\right)=\frac{v^2}{r}=\frac{(\sqrt{3gl}{)}^2}{l}=3g{\text{m/s}}^2$
Tangential acceleration $\left(a_t\right)=0$
So, total acceleration = radial acceleration.

At highest point ($B$) :
$T+mg=m{a}_r$
$\Rightarrow T=m\left(3g\right)-mg=2mg$
$=2\times 3\times 10$
$\therefore T=60\text{N}$