Two stones of masses $m$ and $2m$ are whirled in horizontal circles, the heavier one in a radius $r/2$ and the

Question

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

Accepted Answer

1. **Formula:** The centripetal force $F_c$ acting on a body of mass $m$ moving in a circle of radius $R$ with speed $v$ is given by $F_c = \frac{mv^2}{R}$ [NCERT Class 11, Physics Part I, Sec 5.10, Eq 4.16].
2. **Given Data:**
   - Lighter Stone: Mass $m_1 = m$, Radius $R_1 = r$, Speed $v_1$.
   - Heavier Stone: Mass $m_2 = 2m$, Radius $R_2 = r/2$, Speed $v_2$.
   - Condition: Centripetal forces are equal ($F_{c1} = F_{c2}$).
   - Relationship: $v_1 = n v_2$.
3. **Calculation:**
   $$ \frac{m_1 v_1^2}{R_1} = \frac{m_2 v_2^2}{R_2} $$
   $$ \frac{m v_1^2}{r} = \frac{(2m) v_2^2}{(r/2)} $$
   $$ \frac{v_1^2}{r} = \frac{4 v_2^2}{r} $$
   $$ v_1^2 = 4 v_2^2 $$
   $$ v_1 = 2 v_2 $$
4. **Conclusion:** Comparing this with $v_1 = n v_2$, we get $n = 2$.

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