Question:
A wheel is rotating freely at some angular speed. A second wheel initially at rest and with thrice the rotational inertia of the first, is suddenly coupled to the first wheel. The fraction of the original rotational kinetic energy lost is
A wheel is rotating freely at some angular speed. A second wheel initially at rest and with thrice the rotational inertia of the first, is suddenly coupled to the first wheel. The fraction of the original rotational kinetic energy lost is
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When rotating bodies are coupled together, angular momentum is conserved but rotational kinetic energy is generally not conserved. The loss occurs due to internal friction during coupling.
Updated On: Jun 26, 2026
- 0.50
- 0.25
- 0.66
- 0.75
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The Correct Option is D
Solution and Explanation
Step 1: Define the rotational inertias and initial angular velocity.
Let the rotational inertia of the first wheel be \[ I \] and its initial angular velocity be \[ \omega. \] The second wheel is initially at rest and has rotational inertia \[ 3I. \] Thus, its initial angular velocity is \[ 0. \]
Step 2: Apply conservation of angular momentum.
Since no external torque acts on the system, angular momentum is conserved.
Initial angular momentum: \[ L_i=I\omega. \] Let the common angular velocity after coupling be \[ \omega_f. \] The total rotational inertia after coupling is \[ I+3I=4I. \] Therefore, \[ I\omega=(4I)\omega_f. \] Hence, \[ \omega_f=\frac{\omega}{4}. \]
Step 3: Calculate the initial rotational kinetic energy.
Initially only the first wheel is rotating.
Therefore, \[ K_i=\frac{1}{2}I\omega^2. \]
Step 4: Calculate the final rotational kinetic energy.
After coupling, \[ K_f=\frac{1}{2}(4I)\left(\frac{\omega}{4}\right)^2. \] Simplifying, \[ K_f=\frac{1}{2}(4I)\frac{\omega^2}{16} \] \[ K_f=\frac{1}{8}I\omega^2. \]
Step 5: Find the fraction of energy remaining.
\[ \frac{K_f}{K_i} = \frac{\frac{1}{8}I\omega^2} {\frac{1}{2}I\omega^2} = \frac{1}{4}. \] Thus, only \[ \frac{1}{4} \] of the original kinetic energy remains.
Step 6: Calculate the fraction of energy lost.
Fraction of kinetic energy lost: \[ 1-\frac{1}{4} = \frac{3}{4}. \] Therefore, \[ \frac{3}{4}=0.75. \]
Step 7: Final conclusion.
Hence, the fraction of the original rotational kinetic energy lost is \[ \boxed{0.75} \]
Let the rotational inertia of the first wheel be \[ I \] and its initial angular velocity be \[ \omega. \] The second wheel is initially at rest and has rotational inertia \[ 3I. \] Thus, its initial angular velocity is \[ 0. \]
Step 2: Apply conservation of angular momentum.
Since no external torque acts on the system, angular momentum is conserved.
Initial angular momentum: \[ L_i=I\omega. \] Let the common angular velocity after coupling be \[ \omega_f. \] The total rotational inertia after coupling is \[ I+3I=4I. \] Therefore, \[ I\omega=(4I)\omega_f. \] Hence, \[ \omega_f=\frac{\omega}{4}. \]
Step 3: Calculate the initial rotational kinetic energy.
Initially only the first wheel is rotating.
Therefore, \[ K_i=\frac{1}{2}I\omega^2. \]
Step 4: Calculate the final rotational kinetic energy.
After coupling, \[ K_f=\frac{1}{2}(4I)\left(\frac{\omega}{4}\right)^2. \] Simplifying, \[ K_f=\frac{1}{2}(4I)\frac{\omega^2}{16} \] \[ K_f=\frac{1}{8}I\omega^2. \]
Step 5: Find the fraction of energy remaining.
\[ \frac{K_f}{K_i} = \frac{\frac{1}{8}I\omega^2} {\frac{1}{2}I\omega^2} = \frac{1}{4}. \] Thus, only \[ \frac{1}{4} \] of the original kinetic energy remains.
Step 6: Calculate the fraction of energy lost.
Fraction of kinetic energy lost: \[ 1-\frac{1}{4} = \frac{3}{4}. \] Therefore, \[ \frac{3}{4}=0.75. \]
Step 7: Final conclusion.
Hence, the fraction of the original rotational kinetic energy lost is \[ \boxed{0.75} \]
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