Question

The length of a conical pendulum is $l=2\text{m}$ and its bob has a mass of $m=0.2\text{kg}$. The string will break if the tension $T$ exceeds $4\text{N}$ and $g=10{\text{m/s}}^2$. Then the smallest angle the string can make with the horizontal is

• A. ${60}^{\circ }$
• B. ${30}^{\circ }$
• C. ${15}^{\circ }$
• D. ${45}^{\circ }$

Answer 1

The correct option is B ${30}^{\circ }$

The diagram of the system is given below.


From FBD of the bob:

$T\cos \theta =mg$ ... (1)
and $T\sin \theta =\frac{m{v}^2}{r}$ ... (2)
($T\sin \theta$ provides necessary centripetal force)

From (1) & (2), $\tan \theta =\frac{v^2}{rg}$ ... (3)

Given that $T_{\text{max}}=4\text{N}$; mass of bob $=0.2\text{kg}$
$\therefore$ From equation (1) and (2) we get
$(T\cos \theta {)}^2+(T\sin \theta {)}^2={\left(\frac{m{v}^2}{r}\right)}^2+(mg{)}^2$
$\Rightarrow T^2={\left(\frac{m{v}^2}{r}\right)}^2+(mg{)}^2$
To find $v$ for $T_{\text{max}}$:
$\frac{m{v}^2}{r}=\sqrt{T_{max}^2-(mg{)}^2}$
$\Rightarrow \frac{v^2}{r}=\frac{1}{m}\sqrt{T_{max}^2-(mg{)}^2}$
$\Rightarrow \frac{v^2}{r}=\frac{1}{0.2}\sqrt{4^2-(0.2\times 10{)}^2}$
$=5\sqrt{4^2-2^2}$
$=10\sqrt{3}$
$\therefore$ From equation (3)
$\tan \theta =\frac{10\sqrt{3}}{10}$
$\Rightarrow \tan \theta =\sqrt{3}$
$\Rightarrow \theta ={60}^{\circ }$
So, the angle from the horizontal will be ${30}^0$