Question

A heavy particle is tied to the end A of a string of length $1.6m$. Its other end O is fixed. It revolves as a conical pendulum with the string making ${60}^0$ with the vertical. Then,

• A. its period of revolution is $\frac{4\pi }{7}$ sec
• B. the tension in the string is double the weight of the particle
• C. the velocity of the particle $=2.8\sqrt{3}m/s$
• D. the centripetal acceleration of the particle is $9.8\sqrt{3}m/s^2.$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A its period of revolution is $\frac{4\pi }{7}$ sec
B the tension in the string is double the weight of the particle
C the velocity of the particle $=2.8\sqrt{3}m/s$
D the centripetal acceleration of the particle is $9.8\sqrt{3}m/s^2.$

By diagram we can see that particle rotates in a circle of radius $\frac{\sqrt{3}l}{2}$
So taking horizontal and vertical components of tension $T$, we have,
$T\sin {60}^{\circ }=\frac{T\sqrt{3}}{2}=\frac{m{v}^2}{\left(\frac{\sqrt{3}}{2}l\right)}$ .......$\left(I\right)$
$T\cos {60}^{\circ }=\frac{T}{2}=mg$ .......$\left(II\right)$
$\Rightarrow T=2mg$ $\Rightarrow B$ is correct.
From $\left(I\right)$ and $\left(II\right)$, we have
$v^2=\frac{3gl}{2}$
$\Rightarrow v=\sqrt{\frac{3gl}{2}}=\sqrt{\frac{3\times 9.8\times 1.6}{2}}=2.8\sqrt{3}m/s$ $\Rightarrow C$ is correct.
$a_c=v^2/r=\frac{\left(3gl/2\right)}{l\sqrt{3/2}}=\sqrt{3}g$ $\Rightarrow D$ is correct.
$T=\frac{2\pi r}{v}=\frac{2\pi \left(\frac{\sqrt{3}}{2}l\right)}{2.8\sqrt{3}}$
$\Rightarrow T=\frac{4\pi }{7}s$ $\Rightarrow A$ is correct.