Question

A particle slides from rest from the highest point of a vertical circle of radius r, along a smooth chord. Time of descent of the particle along the chord is

• A. $\sqrt{\frac{r}{4g}}$
• B. $\sqrt{2rg}$
• C. $\sqrt{\frac{2r}{g}}$
• D. $\sqrt{\frac{4r}{g}}$

Answer 1

Answer: (D)

$\sqrt{\frac{4r}{g}}$

Answer 2

Note HH' = 2HO' = 2r cos α

Using s = 1/2 at², we get

2r cos α = 1/2 (g cos α)t²

or t² = $\frac{4r}{g}$ or t = $\sqrt{\frac{4r}{g}}$