Two stones of masses m and 3m are whirled in | vedclass

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(B) Let the mass of the lighter stone be $m_1 = m$ and its radius be $r_1 = r$. Let its tangential speed be $v_1$.Let the mass of the heavier stone be

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) Let the mass of the lighter stone be $m_1 = m$ and its radius be $r_1 = r$. Let its tangential speed be $v_1$.Let the mass of the heavier stone be $m_2 = 3m$ and its radius be $r_2 = \frac{r}{3}$. Let its tangential speed be $v_2$.Given that $v_1 = n v_2$.The centripetal force is given by $F = \frac{mv^2}{r}$.Equating the centripetal forces for both stones: $F_1 = F_2$.$\frac{m_1 v_1^2}{r_1} = \frac{m_2 v_2^2}{r_2}$.Substituting the given values: $\frac{m (n v_2)^2}{r} = \frac{3m v_2^2}{r/3}$.$\frac{m n^2 v_2^2}{r} = \frac{9m v_2^2}{r}$.Canceling common terms $m, v_2^2,$ and $r$ from both sides,we get $n^2 = 9$.Therefore,$n = 3$.

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