Question

One end of a string of length $2\text{m}$ is tied to a body of mass $1\text{kg}$ and is whirled about the other end in a vertical circle at an angular speed of $4\text{rad/s}$. The tension in the string when the body is at the bottommost point of its motion will be equal to
Take $(g=10{\text{m/s}}^2)$

• A. $22\text{N}$
• B. $42\text{N}$
• C. $36\text{N}$
• D. $58\text{N}$

Answer 1

The correct option is B $42\text{N}$

Given
$l=2\text{m}$, $m=1\text{kg}$, $\omega =4\text{rad/s}$

By the FBD of the mass at the lowest point

Writing the force equation at lowest point $\text{A}$ we get,

$T-mg=F_c$ where $F_c=\frac{m{v}^2}{l}$, is the centripetal force
here the radius of the circle will be $\left(l\right)$ length of the string.

$\Rightarrow T-mg=\frac{m{v}^2}{l}$

$\Rightarrow T=mg+\frac{m{v}^2}{l}=mg+ml{w}^2$

$=[1\times 10]+[1\times (2)\times (4{)}^2]$
$=10+2\times 16$
$T=42\text{N}$

Hence option B is the correct answer