Question

A small bead of mass $m=1kg$ is free to move on a circular hoop. The circular hoop has centre at $C$ and radius $r=1m$ and it rotates about a fixed vertical axis. The coefficient of friction between bead and hoop is $\mu =0.5$ . The maximum angular speed of the hoop for which the bead does not have relative motion with respect to the hoop, at the position shown in the figure is: (Take $g=10m{s}^{-2}$ )

• A. ${\left(5\sqrt{2}\right)}^{\frac{1}{2}}$
• B. ${\left(10\sqrt{2}\right)}^{\frac{1}{2}}$
• C. ${\left(15\sqrt{2}\right)}^{\frac{1}{2}}$
• D. ${\left(30\sqrt{2}\right)}^{\frac{1}{2}}$

Answer 1

Answer: (D)

The maximum angular speed of the hoop corresponds to the situation when the bead is just about to slide upwards.
The free-body diagram of the bead is

For the bead not to slide upwards
$m{\omega }^2\left(r\sin 45^{\circ }\right)\cos 45^{\circ }-mg\sin 45^{\circ }<\mu N\dots$ ...(i)
Where $N=mg\cos 45^{\circ }+m{\omega }^2\left(r\sin 45^{\circ }\right)\sin 45^{\circ }\dots$ ...(ii)
From (i) and (ii) we get,
$\omega =\sqrt{30\sqrt{2}}$ $rad{s}^{-1}$