Q. A particle slides from rest from the topmost point of a vertical circle of radius $r$ along a smooth chord making an angle $\theta$ with the vertical. The time of descent is
Motion in a Straight Line
Solution:
$a=g \sin a=g \sin \left(90^{\circ}-\theta\right)$ (see figure)
$=g \cos \theta$
$l=2 R \cos \theta$
Now using $s=u t+\frac{1}{2} a t^{2}$, we get
$l=0 t+\frac{1}{2} g \cos \theta t^{2}$
$\Rightarrow 2 R \cos \theta=\frac{1}{2} g \cos \theta t^{2}$
$\Rightarrow t=2 \sqrt{\frac{R}{g}}$
This is independent of $\theta .$$a=g \sin a=g \sin \left(90^{\circ}-\theta\right)$ (see figure)
$=g \cos \theta$
$l=2 R \cos \theta$
Now using $s=u t+\frac{1}{2} a t^{2}$, we get
$l=0 t+\frac{1}{2} g \cos \theta t^{2}$
$\Rightarrow 2 R \cos \theta=\frac{1}{2} g \cos \theta t^{2}$
$\Rightarrow t=2 \sqrt{\frac{R}{g}}$
This is independent of $\theta .$
