Search any question & find its solution
Question:
Answered & Verified by Expert
A child is on a merry-go-round, standing at a distance $2 \mathrm{~m}$ 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? (Talse, $g=10 \mathrm{~m} / \mathrm{s}^2$ )
Options:
Solution:
2365 Upvotes
Verified Answer
The correct answer is:
$2 \mathrm{rad} / \mathrm{s}$
At critical speed, centrifugal force on child must be equal to maximum static friction
$$
m \omega^2 r=\mu N
$$
where radius, $r=2 \mathrm{~m}$,
coefficient of static friction, $\mu=0.8$
and
$$
M=\text { mass of child. }
$$
Reaction force, $N=m g$
So,
$$
\begin{gathered}
m \omega^2 r=\mu m g \quad \Rightarrow \omega=\sqrt{\frac{\mu g}{r}} \\
\omega=\sqrt{\frac{0.8 \times 10}{2}} \Rightarrow \omega=2 \mathrm{rad} / \mathrm{s}
\end{gathered}
$$
$$
m \omega^2 r=\mu N
$$
where radius, $r=2 \mathrm{~m}$,
coefficient of static friction, $\mu=0.8$
and
$$
M=\text { mass of child. }
$$
Reaction force, $N=m g$
So,
$$
\begin{gathered}
m \omega^2 r=\mu m g \quad \Rightarrow \omega=\sqrt{\frac{\mu g}{r}} \\
\omega=\sqrt{\frac{0.8 \times 10}{2}} \Rightarrow \omega=2 \mathrm{rad} / \mathrm{s}
\end{gathered}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.