Question

An object is tied to a string and rotated in a vertical circle of radius $r$. Constant speed $\left(v\right)$ is maintained along the trajectory. If $\frac{T_{max}}{T_{min}}=2$, then $\frac{v^2}{rg}=$

• A. 3
• B. 3.00
• C. 3.0

Answer 1

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

At the lowest point, $\frac{m{v}^2}{r}=T_L-mg.......\left(\text{i}\right)$

At the highest point, $\frac{m{v}^2}{r}=T_H+mg.......\left(\text{ii}\right)$

As $\frac{T_{max}}{T_{min}}=\frac{T_L}{T_H}=2$
$\therefore T_L=2T_H$

From Eqs. $\left(\text{i}\right)$ and $\left(\text{ii}\right)$, we have
$2T_H-mg=T_H+mg$
or $T_H=2mg$
From Eq. $\left(ii\right)$, $\frac{m{v}^2}{r}=3mg$
$\Rightarrow \frac{v^2}{rg}=3$