Question

A hemi spherical bowl of radius ${}^{\prime }R=0.1{\text{m}}^{\prime }$ is rotating about its own axis (which is vertical) with an angular velocity ${}^{\prime }{\omega }^{\prime }$ . A particle of mass ${}^{\prime }{10}^{-2}{\text{kg}}^{\prime }$ is also rotating with same ${}^{\prime }{\omega }^{\prime }$ on the friction-less inner surface of the bowl. The particle is at a height ${}^{\prime }{h}^{\prime }$ from the bottom of the bowl.



The relation between ${}^{\prime }{h}^{\prime }$ and ${}^{\prime }{\omega }^{\prime }$ is

• A. $h=\frac{{\omega }^2}{g}$
• B. $h=\frac{R}{2}$
• C. $h=R-\frac{g}{{\omega }^2}$
• D. None

Answer 1

The correct option is C $h=R-\frac{g}{{\omega }^2}$

F.B.D of particle

Applying the equation of dynamics along the centre of circular path, for which radius is,
$r=R\sin \theta$
$\Rightarrow \text{N sin}\theta =m{\omega }^{2}r=m{\omega }^2\text{R sin}\theta$
On comparing we get,
$N=m{\omega }^{2}R$

Applying equilbrium condition in vertical direction( particle is performing circular motion in a horizontal plane),
$\Rightarrow \text{N cos}\theta =mg$
$\Rightarrow \text{cos}\theta =\frac{mg}{mR{\omega }^2}=\frac{g}{R{\omega }^2}$
From the triangle,
$\cos \theta =\frac{R-h}{R}$
$\frac{R-h}{R}=\frac{g}{R{\omega }^2}$
$\therefore h=R-\frac{g}{{\omega }^2}$