Question

A particle originally at rest at the highest point of a smooth vertical circle is slightly displaced. It will leave the circle at a vertical distance h below the highest point, such that

• A. h = R
• B. h = R/3
• C. h = R/2
• D. h = 2R

Answer 1

Answer: (B)

$\text{mg cos}\theta -\text{N}=\frac{{\text{mv}}^2}{\text{R}}$

to leave the contact N = 0
$\text{mg cos }\theta =\frac{{\left(\text{mv}\right)}^2}{\text{R}}\Rightarrow \text{g cos }\theta =\frac{{\left(\text{v}\right)}^2}{\text{R}}\dots \left(1\right)$
$\text{cos}\theta =\frac{\left(\text{R}-\text{h}\right)}{\text{R}}\dots \left(2\right)$
from energy conservation $\Rightarrow \frac{1}{2}{\left(\text{mv}\right)}^2=\text{mgh}\dots \left(3\right)$
from 1, 2, 3 $\frac{1}{2}\text{gR}\times \frac{\left(\text{R}-\text{h}\right)}{\text{R}}=\text{gh}\Rightarrow \text{R}-\text{h}=2\text{h}\Rightarrow \text{h}=\text{R}/3$