A flywheel of mass 8 kg and radius 10 cm rotating with a uniform angular speed of 5 rad / sec about its axis of rotation, is subjected to an accelerating torque of 0.01 Nm for 10 seconds. Calculate the change in its angular momentum and change in its kinetic energy.

Answer

To solve the problem step by step, we will calculate the change in angular momentum and the change in kinetic energy of the flywheel.

### Step 1: Calculate the Change in Angular Momentum

The change in angular momentum (ΔL) can be calculated using the formula:
\[ \Delta L = \tau \cdot t \]
where:
- \(\tau\) is the torque applied (0.01 Nm)
- \(t\) is the time duration for which the torque is applied (10 seconds)

Substituting the values:
\[ \Delta L = 0.01 \, \text{Nm} \times 10 \, \text{s} = 0.1 \, \text{kg m}^2/\text{s} \]

### Step 2: Calculate the Moment of Inertia (I)

For a flywheel (considered as a solid disc), the moment of inertia (I) is given by:
\[ I = \frac{1}{2} m r^2 \]
where:
- \(m\) is the mass of the flywheel (8 kg)
- \(r\) is the radius of the flywheel (0.1 m)

Calculating \(I\):
\[ I = \frac{1}{2} \times 8 \, \text{kg} \times (0.1 \, \text{m})^2 = \frac{1}{2} \times 8 \times 0.01 = 0.04 \, \text{kg m}^2 \]

### Step 3: Calculate the Angular Acceleration (α)

Using the relation between torque and angular acceleration:
\[ \tau = I \cdot \alpha \]
we can rearrange this to find \(\alpha\):
\[ \alpha = \frac{\tau}{I} \]

Substituting the values:
\[ \alpha = \frac{0.01 \, \text{Nm}}{0.04 \, \text{kg m}^2} = 0.25 \, \text{rad/s}^2 \]

### Step 4: Calculate the Final Angular Velocity (ω)

Using the equation for angular motion:
\[ \omega_f = \omega_0 + \alpha \cdot t \]
where:
- \(\omega_0\) is the initial angular velocity (5 rad/s)
- \(t\) is the time (10 seconds)

Calculating \(\omega_f\):
\[ \omega_f = 5 \, \text{rad/s} + (0.25 \, \text{rad/s}^2 \times 10 \, \text{s}) = 5 + 2.5 = 7.5 \, \text{rad/s} \]

### Step 5: Calculate the Change in Kinetic Energy (ΔKE)

The change in kinetic energy can be calculated using the formula:
\[ \Delta KE = \frac{1}{2} I \omega_f^2 - \frac{1}{2} I \omega_0^2 \]

Factoring out \(\frac{1}{2} I\):
\[ \Delta KE = \frac{1}{2} I (\omega_f^2 - \omega_0^2) \]

Substituting the values:
\[ \Delta KE = \frac{1}{2} \times 0.04 \, \text{kg m}^2 \times ((7.5)^2 - (5)^2) \]
Calculating the squares:
\[ \Delta KE = \frac{1}{2} \times 0.04 \times (56.25 - 25) = \frac{1}{2} \times 0.04 \times 31.25 \]
\[ \Delta KE = 0.02 \times 31.25 = 0.625 \, \text{J} \]

### Final Answers
- Change in Angular Momentum: \(0.1 \, \text{kg m}^2/\text{s}\)
- Change in Kinetic Energy: \(0.625 \, \text{J}\)