Question

A small bead of mass $m=1\text{kg}$ is carried by a circular hoop having centre at $C$ and radius $r=1\text{m}$ which rotates about a fixed vertical axis (as shown). The coefficient of friction between bead and hoop is $\mu =0.5$. The maximum angular speed of the hoop in $\text{rad/s}$ for which the bead does not have relative motion with respect to hoop: initial position of bead is shown in figure. Answer upto 2 decimal places.

Answer 1

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.
$m{\omega }^2\left(r\sin {45}^{\circ }\right)\cos {45}^{\circ }=mg\sin {45}^{\circ }+\mu N\dots \dots \left(1\right)$
where $N=mg\cos {45}^{\circ }+m{\omega }^2\left(r\sin {45}^{\circ }\right)\sin {45}^{\circ }\dots \dots \left(2\right)$
From (1) and (2) we get
$\omega =\sqrt{30\sqrt{2}}\text{rad/s}=6.51\text{rad/s}$.