Question

Consider a disc of mass 5 kg, radius 2 m, rotating with angular velocity of 10 rad/s about an axis perpendicular to the plane of rotation. An identical disc is kept gently over the rotating disc along the same axis. The energy dissipated so that both the discs continue to rotate together without slipping is \( \_\_\_\_ \) J.

Answer 1

Correct Answer: 250

The problem requires determining the energy lost when a stationary disc is placed upon an identical rotating disc, leading to their eventual synchronized rotation at a common angular velocity.

Concept Used:

This scenario exemplifies an inelastic rotational collision. The fundamental principles applicable are:

1. Conservation of Angular Momentum: In the absence of external torques, the total angular momentum of a system remains constant. When the second disc is superimposed onto the first, the only horizontal forces are frictional and internal to the system, thus exerting no net external torque about the axis of rotation. Consequently, the total angular momentum prior to the interaction equals the total angular momentum after.

\[L_{initial} = L_{final}\]

2. Rotational Kinetic Energy: The rotational kinetic energy (\(K\)) of a body is quantified by its moment of inertia (\(I\)) and its angular velocity (\(\omega\)) using the formula:

\[K = \frac{1}{2}I\omega^2\]

3. Energy Dissipation: In an inelastic collision, the energy dissipated is the difference between the system's initial and final kinetic energies.

\[E_{dissipated} = K_{initial} - K_{final}\]

4. Moment of Inertia of a Disc: The moment of inertia (\(I\)) for a solid disc with mass (\(M\)) and radius (\(R\)), rotating about an axis perpendicular to its plane and through its center, is given by:

\[I = \frac{1}{2}MR^2\]

Step-by-Step Solution:

Step 1: Compute the moment of inertia for a single disc.

Provided data:

• Mass of a disc, \(M = 5 \, \text{kg}\)
• Radius of a disc, \(R = 2 \, \text{m}\)

The moment of inertia (\(I\)) of one disc is calculated as:

\[I = \frac{1}{2}MR^2 = \frac{1}{2}(5 \, \text{kg})(2 \, \text{m})^2 = \frac{1}{2}(5)(4) = 10 \, \text{kg} \cdot \text{m}^2\]

As the discs are identical, their moments of inertia are equal: \(I_1 = I_2 = 10 \, \text{kg} \cdot \text{m}^2\).

Step 2: Determine the initial angular momentum and initial kinetic energy.

Initially, only the first disc rotates at an angular velocity of \( \omega_i = 10 \, \text{rad/s} \), while the second disc is stationary (\( \omega_2 = 0 \)).

Initial angular momentum is:

\[L_{initial} = I_1\omega_i + I_2(0) = (10 \, \text{kg} \cdot \text{m}^2)(10 \, \text{rad/s}) = 100 \, \text{kg} \cdot \text{m}^2/\text{s}\]

Initial kinetic energy is:

\[K_{initial} = \frac{1}{2}I_1\omega_i^2 + 0 = \frac{1}{2}(10)(10)^2 = \frac{1}{2}(10)(100) = 500 \, \text{J}\]

Step 3: Employ conservation of angular momentum to find the final common angular velocity.

Post-interaction, both discs rotate together with a common final angular velocity, \( \omega_f \). The aggregate moment of inertia for the system is \( I_{final} = I_1 + I_2 = 10 + 10 = 20 \, \text{kg} \cdot \text{m}^2 \).

The final angular momentum is:

\[L_{final} = (I_1 + I_2)\omega_f = 20 \omega_f\]

Equating initial and final angular momentum yields:

\[L_{initial} = L_{final} \implies 100 = 20 \omega_f\]\[\omega_f = \frac{100}{20} = 5 \, \text{rad/s}\]

Step 4: Calculate the system's final kinetic energy.

\[K_{final} = \frac{1}{2}(I_1 + I_2)\omega_f^2 = \frac{1}{2}(20)(5)^2\]\[K_{final} = \frac{1}{2}(20)(25) = 10 \times 25 = 250 \, \text{J}\]

Final Computation & Result:

Step 5: Compute the dissipated energy.

The energy dissipated represents the reduction in kinetic energy:

\[E_{dissipated} = K_{initial} - K_{final}\]\[E_{dissipated} = 500 \, \text{J} - 250 \, \text{J} = 250 \, \text{J}\]

The energy dissipated to enable synchronized rotation of both discs is 250 J.