If the radius of curvature of the path of two particles of same mass are in the ratio $3: 4$, then in order to have constant centripetal force, their velocities will be in the ratio of :
Solution
<p>Given $\mathrm{m}_1=\mathrm{m}_2$</p>
<p>$\text { and } \frac{r_1}{r_2}=\frac{3}{4}$</p>
<p>As centripetal force $\mathrm{F}=\frac{\mathrm{mv}^2}{\mathrm{r}}$</p>
<p>In order to have constant (same in this question) centripetal force</p>
<p>$$\begin{aligned}
& \mathrm{F}_1=\mathrm{F}_2 \\
& \frac{\mathrm{m}_1 \mathrm{v}_1^2}{\mathrm{r}_1}=\frac{\mathrm{m}_2 \mathrm{v}_2^2}{\mathrm{r}_2} \\
& \Rightarrow \frac{\mathrm{v}_1}{\mathrm{v}_2}=\sqrt{\frac{\mathrm{r}_1}{\mathrm{r}_2}}=\frac{\sqrt{3}}{2}
\end{aligned}$$</p>
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Torque and Angular Momentum
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