Question

A particle of mass $0.1\text{ }kg$ is rotated at the end of a string in a vertical circle of radius $1.0\text{ }m$ at a constant speed of $5\text{m se}{{\text{c}}^{-1}}$. The tension in the string at the highest point is
(A) $0.5\text{ N}$
(B) $1.0\text{ N}$
(C) $1.5\text{ N}$
(D) $15\text{ N}$

Answer 1

Hint At the highest point, the tension in the string and weight of the body together provides the necessary centripetal force required for rotation of the body in a circular path. Hence, $\text{T +}\,\text{mg = }\dfrac{\text{m}{{\text{v}}^{\text{2}}}}{\text{r}}$ is used.

Formula Used At highest point,
$\text{T +}\,\text{mg = }\dfrac{\text{m}{{\text{v}}^{\text{2}}}}{\text{r}}$
or, $\text{T = }\dfrac{\text{m}{{\text{v}}^{2}}}{\text{r}}-\text{mg}$

Complete Step by Step Solution
Given:
$\begin{aligned} & \text{m = 0}\text{.1 kg} \\\ & \text{r = 1m} \\\ & \text{v = }{5\text{m}}/{\text{sec}}\; \\\ \end{aligned}$
As the body is rotated at the end of the string in a vertical circle,
At the highest point, the tension in the string and the weight of the body together provides the necessary centripetal force.
Therefore,
$\begin{aligned} & \text{T +}\,\text{mg = }\dfrac{\text{m}{{\text{v}}^{\text{2}}}}{\text{r}} \\\ & \text{T = }\dfrac{\text{m}{{\text{v}}^{2}}}{\text{r}}-\text{mg} \\\ & =\dfrac{0.1\times 5\times 5}{1}-0.1\times 10 \\\ & =2.5-1 \\\ & \text{T = 1}\text{.5 N} \\\ \end{aligned}$

Tension in the string at the highest point is 1.5 N (option C)

Note: If the body is at lowest point, then a part of tension ${{\text{T}}_{1}}$ balances the weight of the body and the remaining part provides the necessary centripetal force. Hence, formula used will be${{\text{T}}_{1}}-\text{mg = }\dfrac{\text{m}{{\text{v}}^{2}}}{\text{r}}$
The conditions when the body is at highest point and at the lowest point must be understood properly so as to solve the problem correctly.