Question

The angular momentum of a wheel having a rotational inertia of \[ 0.2\,\text{kg m}^2 \] about its symmetric axis decreases from \[ 4 \text{ to } 2\,\text{kg m}^2\text{s}^{-1} \] in \(4\,\text{s}\). The average power of the wheel is:

• A. \(7.5\,\text{W}\)
• B. \(15\,\text{W}\)
• C. \(5\,\text{W}\)
• D. \(12\,\text{W}\)

Hint:

For rotational motion, \[ L=I\omega \] and \[ K=\frac{1}{2}I\omega^2. \] First find angular velocities from angular momentum, then compute rotational kinetic energies.

Answer 1

The Correct Option is A

Step 1: Recall the link between angular momentum and kinetic energy.
Angular momentum is $L = I\omega$, so $\omega = L/I$.
Rotational kinetic energy is $K = \frac{1}{2}I\omega^2 = \frac{L^2}{2I}$.
This formula lets us find kinetic energy directly from $L$ and $I$, without first finding $\omega$.

Step 2: Compute initial kinetic energy.
$L_1 = 4\,\text{kg\,m}^2\text{s}^{-1}$ and $I = 0.2\,\text{kg\,m}^2$:
\[ K_1 = \frac{L_1^2}{2I} = \frac{4^2}{2 \times 0.2} = \frac{16}{0.4} = 40\,\text{J} \]

Step 3: Compute final kinetic energy.
$L_2 = 2\,\text{kg\,m}^2\text{s}^{-1}$:
\[ K_2 = \frac{L_2^2}{2I} = \frac{2^2}{2 \times 0.2} = \frac{4}{0.4} = 10\,\text{J} \]

Step 4: Find the energy lost by the wheel.
\[ \Delta K = K_1 - K_2 = 40 - 10 = 30\,\text{J} \]

Step 5: Calculate average power.
Power is the rate of doing work (or losing energy). The energy was dissipated over $t = 4\,\text{s}$:
\[ P = \frac{\Delta K}{t} = \frac{30}{4} = 7.5\,\text{W} \]

Step 6: State the answer.
\[ \boxed{7.5\,\text{W}} \]