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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
1
B
2
C
3
D
4
Detailed Solution
$(F_C)_{heavier} = (F_C)_{lighter}$
$\frac{2mV^2}{r/2} = \frac{m(nV)^2}{r} \Rightarrow n^2 = 4 \Rightarrow n = 2$
$\frac{2mV^2}{r/2} = \frac{m(nV)^2}{r} \Rightarrow n^2 = 4 \Rightarrow n = 2$