Question 62: Work, Energy and Power & Circular Motion

Question

A bob of mass '$$m$$' is suspended by a light string of length '$$L$$'. It is imparted a minimum horizontal velocity at the lowest point $$A$$ such that it just completes half circle reaching the top most position B. The ratio of kinetic energies $$\frac{(K . E)_A}{(K . E)_B}$$ is :

Accepted Answer

Apply energy conservation between A & B
$$\begin{aligned}
& \frac{1}{2} \mathrm{mV}_{\mathrm{L}}^2=\frac{1}{2} \mathrm{mV}_{\mathrm{H}}^2+\mathrm{mg}(2 \mathrm{~L}) \
& \because \mathrm{V}_{\mathrm{L}}=\sqrt{5 \mathrm{gL}}
\end{aligned}$$
So, $$\mathrm{V}_{\mathrm{H}}=\sqrt{\mathrm{gL}}$$
$$\frac{(\mathrm{K} . \mathrm{E})_{\mathrm{A}}}{(\mathrm{K} . \mathrm{E})_{\mathrm{B}}}=\frac{\frac{1}{2} \mathrm{~m}(\sqrt{5 \mathrm{gL}})^2}{\frac{1}{2} \mathrm{~m}(\sqrt{\mathrm{gL}})^2}=\frac{5}{1}$$

Answer

5 : 1

Answer

3 : 2

Answer

1 : 5

Answer

2 : 5

Work, Energy and Power & Circular Motion

A bob of mass '$m' is suspended by a light string of length 'L'. It is imparted a minimum horizontal velocity at the lowest point A such that it just completes half circle reaching the top most position B. The ratio of kinetic energies \frac{(K . E)_A}{(K . E)_B}$ is :

Choose the correct answer: