First slide

Force

Question

Two stones of masses m and 2 m 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

Easy

A

4
B

1
C

2
D

3

Solution

Let $ν$ be tangential speed of heavier stone. Then, Centripetal force experienced by lighter stone is
${(F_{c})}_{lighter} = \frac{m (n v)^{2}}{r}$
and that of heavier stone is
${(F_{c})}_{heavier} = \frac{2 m v^{2}}{(r / 2)}$
$But {(F_{c})}_{lighter} = {(F_{c})}_{heavier} (given)$
$∴ \frac{m (n v)^{2}}{r} = \frac{2 m v^{2}}{(r / 2)}$
$\begin{array}{l} n^{2} (\frac{m v^{2}}{r}) = 4 (\frac{m v^{2}}{r}) \\ n^{2} = 4 or n = 2 \end{array}$

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