Question

Two stones of masses $m$ and $2m$ are whirled in horizontal circles, the heavier one in a radius $\frac{r}{2}$ and the lighter one in radius $r$. The tangential speed of lighter stone is $n$ times that of the value of heavier stone when they experience same centripetal forces. The value of $n$ is

• A. 4
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (C)

Let $v$ be tangential speed of heavier stone.
Then, centripetal force experienced by lighter stone is $\left(F_{c}\right)_{\text {lighter }}=\frac{m(n v)^{2}}{r}$
and that of heavier stone is
$\left(F_{c}\right)_{\text {heavier }}=\frac{2 m v^{2}}{(r / 2)}$
But $\left(F_{c}\right)_{\text {lighter }}=\left(F_{c}\right)_{\text {heavier }}$ (given)
$\therefore \frac{m(n v)^{2}}{r}=\frac{2 m v^{2}}{(r / 2)}$
$n^{2}\left(\frac{m v^{2}}{r}\right)=4\left(\frac{m v^{2}}{r}\right)$
$n^{2}=4$ or $n=2$