Question 66: Work, Energy and Power & Circular Motion

Question

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 :

Accepted Answer

Given $$\mathrm{m}_1=\mathrm{m}_2$$
$$	ext { and } \frac{r_1}{r_2}=\frac{3}{4}$$
As centripetal force $$\mathrm{F}=\frac{\mathrm{mv}^2}{\mathrm{r}}$$
In order to have constant (same in this question) centripetal force
$$\begin{aligned}
& \mathrm{F}_1=\mathrm{F}_2 \
& \frac{\mathrm{m}_1 \mathrm{v}_1^2}{\mathrm{r}_1}=\frac{\mathrm{m}_2 \mathrm{v}_2^2}{\mathrm{r}_2} \
& \Rightarrow \frac{\mathrm{v}_1}{\mathrm{v}_2}=\sqrt{\frac{\mathrm{r}_1}{\mathrm{r}_2}}=\frac{\sqrt{3}}{2}
\end{aligned}$$

Answer

$$1: \sqrt{3}$$

Answer

$$2: \sqrt{3}$$

Answer

$$\sqrt{3}: 2$$

Answer

$$\sqrt{3}: 1$$

Work, Energy and Power & Circular Motion

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 :

Choose the correct answer: