Question

A coin placed on a rotating turntable just slips if it is placed at a distance of $4$ cm from centre. If the angular velocity of the turntable is doubled, it will just slip at a distance of :

• A. $1$ cm
• B. $2$ cm
• C. $3$ cm
• D. $8$ cm

Answer 1

Answer: (A)

$1$ cm

When the coin is just on the verge of slipping, it just overcomes the force of static friction ($F_s$) . which acts as centripetal force for being that coin in the circular motion.
So we have
$F_s=m{\omega }^{2}r$
If we double the $\omega$, then $r$ will also self adjust to a new $r^{{}^{\prime }}$, since the static friction will remain same. We have
$m{\omega }^{2}r=m{\left(2\omega \right)}^2{r}^{{}^{\prime }}\Rightarrow r^{{}^{\prime }}=\frac{r}{4}cm=\frac{4}{4}cm=1cm$