Question

A hemi-spherical bowl of radius $R=0.1\text{m}$ is rotating about its own axis (vertical) with uniform angular speed ${}^{\prime }{\omega }^{\prime }$. A particle of mass ${10}^{-2}\text{kg}$ on the smooth inner surface of the bowl is also rotating with the same $\omega$ at a height $h$ from the bottom of the bowl. What is the minimum value of ${}^{\prime }{\omega }^{\prime }$ to have a non-zero value of ${}^{\prime }{h}^{\prime }$, for which the particle remains at rest with respect to the bowl?
(Take $g=10{\text{m/s}}^2$)

• A. $20\text{rad/s}$
• B. $10\text{rad/s}$
• C. $5\text{rad/s}$
• D. $1\text{rad/s}$

Answer 1

The correct option is B $10\text{rad/s}$


Equation of dynamics along centre of circular path:
$F_{net}=m\times a=m\times r{\omega }^2$.......(1)
Since $F_{net}=N\sin \theta$
$\Rightarrow N\sin \theta =mr{\omega }^2$..........(2)
For equilibrium in vertical direction,
$N\cos \theta =mg$..........(3)
From geometry we get, $r=R\sin \theta$...........(4)
Combining equation (2) and (4) gives:
$N=m{\omega }^{2}R$
From equation (3) we get:
$\cos \theta =\frac{mg}{mR{\omega }^2}=\frac{g}{R{\omega }^2}$
From the geometry, $\cos \theta =\frac{R-h}{R}$
$\frac{R-h}{R}=\frac{g}{R{\omega }^2}$
$\Rightarrow h=R-\frac{g}{{\omega }^2}$
For non-zero height,
$\therefore$ $h\ge 0\Rightarrow R\ge \frac{g}{{\omega }^2}\Rightarrow \omega \ge \sqrt{\frac{g}{R}}$
$\Rightarrow {\omega }_{\text{min}}=\sqrt{\frac{g}{R}}=\sqrt{\frac{10}{0.1}}=10\text{rad/s}$