Question 277: Work, Energy and Power & Circular Motion

Question

A body of mass 'm' is moving with speed 'V' along a circular path of radius 'r'. Now the speed is reduced to $$\frac{V}{2}$$ and radius is increased to '3r'. For this change, initial centripetal force needs to be

Accepted Answer

Centripetal force $$\mathrm{F}=\frac{\mathrm{mv}^2}{\mathrm{r}}$$
$$\mathrm{F_1=\frac{m v_1^2}{r_1}}$$ and $$\mathrm{F_2=\frac{m v_2^2}{r_2}}$$
$$\begin{aligned}
& 	herefore \frac{F_2}{F_1}=\frac{v_2^2}{v_1^2} \cdot \frac{r_1}{r_2} \
& v_2=\frac{v_1}{2} 	ext { and } r_2=3 r_1 \
& 	herefore \frac{F_2}{F_1}=\left(\frac{1}{2}ight)^2 \cdot \frac{1}{3}=\frac{1}{12} \
& 	herefore F_2=\frac{F_1}{12} \quad 	herefore F_2 < F_1 \
& 	herefore F_1-F_2=F_1-\frac{F_1}{12}=\frac{11}{12} F_1
\end{aligned}$$

Answer

increased by $$\frac{7}{12}$$ times

Answer

increased by $$\frac{10}{12}$$ times

Answer

decreased by $$\frac{11}{12}$$ times

Answer

decreased by $$\frac{1}{12}$$ times

Work, Energy and Power & Circular Motion

A body of mass 'm' is moving with speed 'V' along a circular path of radius 'r'. Now the speed is reduced to $\frac{V}{2}$ and radius is increased to '3r'. For this change, initial centripetal force needs to be

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