Question

If the radius of curvature of the path of two particles having same speed are in the ratio of $1:\sqrt{2}$, then in order to have masses in the ratio of $2:1$, their centripetal forces should be in the ratio of

• A. $1:2$
• B. $2:1$
• C. $2\sqrt{2}:1$
• D. $3:2\sqrt{2}$

Answer 1

The correct option is C $2\sqrt{2}:1$

Given ratio of radii of curvature $\frac{r_1}{r_2}=\frac{1}{\sqrt{2}}$
Ratio of masses $\frac{m_1}{m_2}=\frac{2}{1}$

From definition,
$F=mr{\omega }^2=\frac{m{v}^2}{r}$
We get,
$\frac{F_1}{F_2}=\frac{\frac{m_1{v}_1^2}{r_1}}{\frac{m_2{v}_2^2}{r_2}}=\frac{m_1{v}_1^2{r}_2}{m_2{v}_2^2{r}_1}$
$\Rightarrow \frac{F_1}{F_2}=\left(\frac{m_1}{m_2}\right){\left(\frac{v_1}{v_2}\right)}^2\left(\frac{r_2}{r_1}\right)$
Given $v_1=v_2=v$ (say)

$\therefore \frac{F_1}{F_2}=2\times \sqrt{2}=2\sqrt{2}$

Hence ratio of their centripetal force will be,
$F_1:F_2=2\sqrt{2}:1$

Option C is the correct answer.