Expression for Loss of Energy During Contact of Two Discs – Rankers Physics

Angular Momentum and Conservation of Angular Momentum: Practice Problem & Solution

Two discs of same moment of inertia rotating about their regular axis passing through centre and perpendicular to the plane of disc with angular velocities $\omega_1$ and $\omega_2$. They are brought into contact face to face coinciding the axis of rotation. The expression for loss of energy during this process is: (2017-Delhi)
$\frac{1}{4}I(\omega_1-\omega_2)^2$
$I(\omega_1-\omega_2)^2$
$\frac{1}{8}I(\omega_1-\omega_2)^2$
$\frac{1}{2}I(\omega_1-\omega_2)^2$

Solution Explained:

To solve this problem, we apply the core principles of Angular Momentum and Conservation of Angular Momentum. Understanding the underlying formula is key to arriving at the correct answer below:

By conservation of angular momentum, the final common angular velocity is $\omega = \frac{\omega_1 + \omega_2}{2}$. The loss in rotational kinetic energy is $Delta E = E_i - E_f = \frac{1}{2}I\omega_1^2 + \frac{1}{2}I\omega_2^2 - 2 \cdot \left(\frac{1}{2}I\omega^2\right)$, which simplifies to $\frac{1}{8}I(\omega_1-\omega_2)^2$.

Leave a Reply

Your email address will not be published. Required fields are marked *

Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. Assertion (A): Kinetic energy of a rigid body can be greater than \( \frac{1}{2}mv^2 \), where \( m \) is mass of rigid body & \( v \) is speed of centre of mass of body. Reason (R): Kinetic energy of a particle (point mass) cannot be greater than \( \frac{1}{2}mv^2 \), where \( m \) is mass of particle & \( v \) is speed of particle.
2. A uniform rectangular plate of mass m which is free to rotate about the smooth vertical hinge passing through the centre and perpendicular to the plate, is lying on a smooth horizontal surface. A particle of mass m moving with speed 'u' collides with the plate and sticks to it as shown in figure. The angular velocity of the plate after collision will be
3. When a mass is rotating in a plane about a fixed point, its angular momentum is directed along   (2012 Pre)
4. Assertion (A): By definition, pure rolling of a body occurs when velocity of its point of contact is zero relative to the surface on which it rolls. Reason (R): A body is purely rolling (rolling without slipping). The velocity of point of contact (of body) must be zero with respect to ground.  
5. A solid sphere is rotating freely about its symmetry axis in free space. The radius of the sphere is increased keeping its mass same. Which of the following physical quantities would remain constant for the sphere? (2018)