To solve this problem, we apply energy conservation between points \( A \) and \( B \).
Step 1: Energy conservation between \( A \) and \( B \):
\(\frac{1}{2} m V_L^2 = \frac{1}{2} m V_B^2 + mg(2L)\)
where:
- \( V_L \) is the velocity at point \( A \) (the lowest point),
- \( V_B \) is the velocity at point \( B \) (the highest point).
Rearranging the equation:
\(V_L^2 = V_B^2 + 4gL\)
Step 2: Calculate \( V_L \) (Minimum Velocity at \( A \)):
Since the bob must just complete the circular path, the minimum velocity at \( A \) must be such that:
\(V_L = \sqrt{5gL}\)
Step 3: Calculate \( V_B \):
Applying energy conservation:
\(\frac{1}{2} m V_L^2 = \frac{1}{2} m V_B^2 + mg(2L)\)
Substitute \(V_L = \sqrt{5gL}\):
\(\frac{1}{2} m (5gL) = \frac{1}{2} m V_B^2 + 2mgL\)
Simplifying:
\(\frac{5}{2} mgL = \frac{1}{2} m V_B^2 + 2mgL\)
\(\frac{1}{2} m V_B^2 = \frac{5}{2} mgL - 2mgL\)
\(V_B^2 = gL\)
\(V_B = \sqrt{gL}\)
Step 4: Calculate the ratio of kinetic energies:
\(\left(\frac{\text{(K.E.)}_A}{\text{(K.E.)}_B}\right) = \frac{\frac{1}{2} m V_L^2}{\frac{1}{2} m V_B^2} = \frac{V_L^2}{V_B^2} = \frac{5gL}{gL} = 5\)
Thus, the ratio of kinetic energies \( \left(\frac{\text{(K.E.)}_A}{\text{(K.E.)}_B}\right) \) is 5 : 1.
The Correct Answer is: 5 : 1
A body of mass 100 g is moving in a circular path of radius 2 m on a vertical plane as shown in the figure. The velocity of the body at point A is 10 m/s. The ratio of its kinetic energies at point B and C is: (Take acceleration due to gravity as 10 m/s^2)