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.
Correct Answer: 250
Approach Solution - 1
The problem asks for the energy dissipated when a stationary disc is placed on top of an identical rotating disc, and they eventually rotate together with a common angular velocity.
Concept Used:
This problem is an example of an inelastic rotational collision. The key principles to use are:
1. Conservation of Angular Momentum: In the absence of external torques, the total angular momentum of a system is conserved. When the second disc is gently placed on the first one, the only horizontal forces are internal (friction between the discs), so there is no net external torque about the axis of rotation. The total angular momentum before the collision is equal to the total angular momentum after the collision.
\[ L_{initial} = L_{final} \]2. Rotational Kinetic Energy: The rotational kinetic energy of a body with moment of inertia \(I\) rotating with angular velocity \(\omega\) is given by:
\[ K = \frac{1}{2}I\omega^2 \]3. Energy Dissipation: The energy dissipated in an inelastic collision is the difference between the initial kinetic energy and the final kinetic energy of the system.
\[ E_{dissipated} = K_{initial} - K_{final} \]4. Moment of Inertia of a Disc: The moment of inertia of a solid disc of mass \(M\) and radius \(R\) about an axis perpendicular to its plane and passing through its center is:
\[ I = \frac{1}{2}MR^2 \]Step-by-Step Solution:
Step 1: Calculate the moment of inertia of a single disc.
Given values:
- 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:
\[ I = \frac{1}{2}MR^2 = \frac{1}{2}(5 \, \text{kg})(2 \, \text{m})^2 = \frac{1}{2}(5)(4) = 10 \, \text{kg} \cdot \text{m}^2 \]Since the two discs are identical, they both have the same moment of inertia, \(I_1 = I_2 = 10 \, \text{kg} \cdot \text{m}^2\).
Step 2: Calculate the initial angular momentum and initial kinetic energy.
Initially, only the first disc is rotating with angular velocity \( \omega_i = 10 \, \text{rad/s} \). The second disc is at rest (\( \omega_2 = 0 \)).
Initial angular momentum:
\[ L_{initial} = I_1\omega_i + I_2(0) = (10 \, \text{kg} \cdot \text{m}^2)(10 \, \text{rad/s}) = 100 \, \text{kg} \cdot \text{m}^2/\text{s} \]Initial kinetic energy:
\[ K_{initial} = \frac{1}{2}I_1\omega_i^2 + 0 = \frac{1}{2}(10)(10)^2 = \frac{1}{2}(10)(100) = 500 \, \text{J} \]Step 3: Apply conservation of angular momentum to find the final common angular velocity.
After the collision, both discs rotate together with a common final angular velocity, \( \omega_f \). The total moment of inertia of the combined 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:
\[ L_{initial} = L_{final} \implies 100 = 20 \omega_f \] \[ \omega_f = \frac{100}{20} = 5 \, \text{rad/s} \]Step 4: Calculate the final kinetic energy of the system.
\[ K_{final} = \frac{1}{2}(I_1 + I_2)\omega_f^2 = \frac{1}{2}(20)(5)^2 \] \[ K_{final} = \frac{1}{2}(20)(25) = 10 \times 25 = 250 \, \text{J} \]Final Computation & Result:
Step 5: Calculate the energy dissipated.
The energy dissipated is the loss in kinetic energy:
\[ E_{dissipated} = K_{initial} - K_{final} \] \[ E_{dissipated} = 500 \, \text{J} - 250 \, \text{J} = 250 \, \text{J} \]The energy dissipated so that both the discs continue to rotate together is 250 J.
Approach Solution -2
Step 1. Calculate Initial Moment of Inertia and Kinetic Energy:
For the first disc:
\( I_1 = \frac{1}{2}MR^2 = \frac{1}{2} \times 5 \times (2)^2 = 10 \, \text{kg} \cdot \text{m}^2 \)
Initial angular velocity \( \omega = 10 \, \text{rad/s} \). Initial kinetic energy:
\( E_i = \frac{1}{2}I_1 \omega^2 = \frac{1}{2} \times 10 \times (10)^2 = 500 \, \text{J} \)
Step 2. Final Moment of Inertia and Angular Velocity:
When the second disc is placed on top, the combined moment of inertia becomes:
\( I_f = I_1 + I_2 = 10 + 10 = 20 \, \text{kg} \cdot \text{m}^2 \)
Using conservation of angular momentum:
\( I_1 \omega = I_f \omega_f \Rightarrow 10 \times 10 = 20 \times \omega_f \)
\( \omega_f = 5 \, \text{rad/s} \)
Step 3. Calculate Final Kinetic Energy:
\( E_f = \frac{1}{2} I_f \omega_f^2 = \frac{1}{2} \times 20 \times (5)^2 = 250 \, \text{J} \)
Step 4. Energy Dissipated:
Energy dissipated \( \Delta E = E_i - E_f \):
\( \Delta E = 500 - 250 = 250 \, \text{J} \)
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