Question

The bob of a $0.2\text{m}$ pendulum describes an arc of circle in a vertical plane. If the tension in the cord is $\sqrt{3}$ times the weight of the bob when the cord makes an angle ${30}^{\circ }$ with the vertical, the magnitude of acceleration of bob in that position will be:

• A. $g$
• B. $\frac{g}{2}$
• C. $\frac{\sqrt{3}g}{2}$
• D. $\frac{g}{4}$

Answer 1

The correct option is A $g$

F.B.D of pendulum bob:



Length of pendulum $=0.2\text{m}$ $=$ Radius of vertical circle.

Net force on bob in tangential direction is,

$F_t$ = $mgsin\theta$

$\Rightarrow$ $m{a}_t$ = $mgsin\theta$

Given, $\theta ={30}^{\circ }$

$a_t=gsin{30}^{\circ }$ $=$ $\frac{g}{2}$

From equation of circular dynamics,

$T-mgcos\theta =\frac{m{v}^2}{r}$

$T-mgcos\theta =m{a}_c$

$a_c=\frac{T-mgcos\theta }{m}$

($\because T=\sqrt{3}mg$)

$\Rightarrow a_c=\frac{\sqrt{3}mg-\frac{mg\sqrt{3}}{2}}{m}=\frac{\sqrt{3}g}{2}$

we know that, total acceleration is given by,

${\overrightarrow{a}}_{net}$ $=$ ${\overrightarrow{a}}_c$ $+$ ${\overrightarrow{a}}_t$

$\mid {\overrightarrow{a}}_{net}\mid =\sqrt{|\overrightarrow{a_c}{|}^2+|\overrightarrow{a_t}{|}^2}$

Substituting the values of $a_c$ and $a_t$ we have,

$\mid {\overrightarrow{a}}_{net}\mid$ $=$ $g$

Hence, option ($a$) is the correct answer.

Why this question ? Key point : Always remember that the net accelaration of a particle performing circular motion is equal to the vector summation of its radial and tangential components.