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 a half circle, reaching the topmost position \( B \). The ratio of kinetic energies \( \frac{(\text{K.E.})_A}{(\text{K.E.})_B} \) is:
- 3 : 2
- 5 : 1
- 2 : 5
- 1 : 5
The Correct Option is B
Approach Solution - 1
To solve the problem of finding the ratio of kinetic energies \( \frac{(\text{K.E.})_A}{(\text{K.E.})_B} \), we need to understand the physics involved in a simple pendulum executing circular motion.
The key points to consider are:
- At the lowest point \( A \), the bob has maximum kinetic energy and zero potential energy.
- At the highest point \( B \), the bob has maximum potential energy and minimum kinetic energy, just enough to maintain circular motion.
Let's go through the solution step-by-step:
- The critical condition for the bob to just reach point \( B \) is that its velocity \( v_B \) at point \( B \) should be such that it maintains the tension in the string. The minimum velocity required can be derived using the centripetal force equation:
- The kinetic energy at point \( B \) is given by:
- Using conservation of energy from point \( A \) to point \( B \), the total energy at point \( A \) equals the total energy at point \( B \):
- Substituting the expression for \( v_B^2 \):
- Simplifying the equation to find \( v_A^2 \):
- The kinetic energy at point \( A \) is therefore:
- Now, calculate the ratio of kinetic energies:
The correct answer is therefore \(\frac{5}{1}\).
Approach Solution -2
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
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