Question

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

• A. Least for $\theta =0$
• B. Maximum for $\theta =0$
• C. Least for $\theta =45^{\circ }$
• D. Independent of $\theta$

Answer 1

Answer: (D)

$a=g\sin a=g\sin \left(90^{\circ }-\theta \right)$ (see figure)
$=g\cos \theta$
$l=2R\cos \theta$
Now using $s=ut+\frac{1}{2}a{t}^2$, we get
$l=0t+\frac{1}{2}g\cos \theta t^2$
$\Rightarrow 2R\cos \theta =\frac{1}{2}g\cos \theta t^2$
$\Rightarrow t=2\sqrt{\frac{R}{g}}$
This is independent of $\theta .$$a=g\sin a=g\sin \left(90^{\circ }-\theta \right)$ (see figure)
$=g\cos \theta$
$l=2R\cos \theta$
Now using $s=ut+\frac{1}{2}a{t}^2$, we get
$l=0t+\frac{1}{2}g\cos \theta t^2$
$\Rightarrow 2R\cos \theta =\frac{1}{2}g\cos \theta t^2$
$\Rightarrow t=2\sqrt{\frac{R}{g}}$
This is independent of $\theta .$