Question

A child is on a merry-go-round, standing at a distance $2\, m$ from the centre. The coefficient of static friction between the child and the surface of merry-go-round is $0.8 .$ At what maximum angular velocity can the merry-go-round be rotated before the child slips? (Take, $g=10\, m / s ^{2}$ )

• A. 0.5 rad / s
• B. 1 rad / s
• C. 2 rad / s
• D. 4 rad / s

Answer 1

Answer: (C)

At critical speed, centrifugal force on child must be equal to maximum static friction
$m \omega^{2} r=\mu N$
where radius, $r=2\, m$,
coefficient of static friction, $\mu=0.8$
and $\quad M=$ mass of child.
Reaction force, $N=m g$
So, $m \omega^{2} r=\mu m g$
$\Rightarrow \omega=\sqrt{\frac{\mu g}{r}}$
$\omega=\sqrt{\frac{0.8 \times 10}{2}}$
$\Rightarrow \omega=2\, rad / s$