Question

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

• 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, $r=2m$,
coefficient of static friction, $\mu =0.8$
and $M=$ mass of child.
Reaction force, $N=mg$
So, $m{\omega }^{2}r=\mu mg$
$\Rightarrow \omega =\sqrt{\frac{\mu g}{r}}$
$\omega =\sqrt{\frac{0.8\times 10}{2}}$
$\Rightarrow \omega =2rad/s$