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