A body of mass ' $\mathrm{m}$ ' is moving with speed ' $\mathrm{V}$ ' along a circular path of radius ' $r$…
A body of mass ' $\mathrm{m}$ ' is moving with speed ' $\mathrm{V}$ ' along a circular path of radius ' $r$ '. Now the speed is reduced to $\frac{\mathrm{V}}{2}$ and radius is increased to ' $3 r$ '. For this change, initial centripetal force needs to be
increased by $\frac{7}{12}$ times.
increased by $\frac{10}{12}$ times.
increased by $\frac{11}{12}$ times.
increased by $\frac{1}{12}$ times.
Solution
$\begin{aligned} & \text { Centripetal force, } \mathrm{F}=\frac{\mathrm{mv}^2}{\mathrm{r}} \\ & \mathrm{F}_1=\frac{\mathrm{mv}_1^2}{\mathrm{r}_1} \text { and } \mathrm{F}_2=\frac{\mathrm{mv}_2^2}{\mathrm{r}_2} \\ & \frac{\mathrm{F}_2}{\mathrm{~F}_1}=\frac{\mathrm{v}_2^2}{\mathrm{v}_1^2} \cdot \frac{\mathrm{r}_1}{\mathrm{r}_2} \\ & \mathrm{v}_2=\frac{\mathrm{v}_1}{2} \text { and } \mathrm{r}_2=3 \mathrm{r}_1 \\ & \frac{\mathrm{F}_2}{\mathrm{~F}_1}=\left(\frac{1}{2}\right)^2 \cdot \frac{1}{3}=\frac{1}{12} \\ & \mathrm{v}_2=\frac{\mathrm{v}_1}{2} \text { and } \mathrm{r}_2=3 \mathrm{r}_1 \\ & \frac{\mathrm{F}_2}{\mathrm{~F}_1}=\left(\frac{1}{2}\right)^2 \cdot \frac{1}{3}=\frac{1}{12} \\ & \mathrm{~F}_2=\frac{\mathrm{F}_1}{12}\end{aligned}$
Change in force $=F_2-F_1=\frac{F_1}{12}-F_1=-\frac{11}{12} F_1$ Hence force is decreased by $\frac{11}{12}$ times.