Question

Two discs of moments of inertia ${I_1}$ and ${I_2}$ about their respective axes (normal to the disc and passing through the centre), and rotating with angular speeds ${\omega _1}$ and ${\omega _2}$ are brought into contact face to face with their axes of rotation coincident. (a) What is the angular speed of the two-disc system? (b) Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy? Take ${\omega _1} \ne {\omega _2}$.

Answer 1

The moment of inertia is a quantity that, like mass, defines the torque required for a given angular acceleration around a rotating axis. It relies on the mass distribution of the body and the axis selected, with greater moments necessitating more torque to affect the rate of rotation.

Complete step by step solution:
The sum of the spin and orbital angular momenta is the total angular momentum. A point particle's orbital angular momentum vector is always parallel and exactly proportional to its orbital angular velocity vector $\omega$, where the proportionality constant relies on both the particle's mass and its distance from the origin.
The total initial angular momentum = ${I_1}{\omega _1} + {I_2}{\omega _2}$
The total moment of inertia about the axis increases as the discs are brought together = ${I_1} + {I_2}$
Let the two-disc system's angular speed be$\omega$
Then there's angular momentum conservation${I_{1}}{\omega _{1}} + {I_2}{\omega _2} = ({I_{1}} + {I_2})\omega$
Thus ${\text{\;\omega = }}\dfrac{{{{\text{I}}_{\text{1}}}{{\text{\omega }}_{\text{1}}}{\text{ + }}{{\text{I}}_{\text{2}}}{{\text{\omega }}_{\text{2}}}}}{{{{\text{I}}_{\text{1}}}{\text{ + }}{{\text{I}}_{\text{2}}}}}$
The kinetic energy of an item is the energy it has owing to its motion in physics. It is the amount of effort required to propel a body of a given mass from rest to a certain velocity. The body retains its kinetic energy after gaining it during acceleration unless its speed changes. When the body decelerates from its current speed to a condition of rest, it does the same amount of effort.
The system's total initial kinetic energy ${E_i} = \dfrac{1}{2}{I_1}{\omega _1}^2 + \dfrac{1}{2}{I_2}{\omega _2}^2$
The system's final kinetic energy$= \dfrac{1}{2}({I_1} + {I_2}){\omega ^2}$
Hence $\dfrac{{\dfrac{1}{2}{{({I_1}{\omega _1} + {I_2}{\omega _2})}^2}}}{{{I_1} + {I_2}}}$
Thus, ${E_{i}} - {E_f} = \dfrac{{{I_1}{I_2}{{({\omega _1} - {\omega _2})}^2}}}{{2({I_1} + {I_2})}} > 0$
The frictional force that occurs when the two discs come into touch with each other is responsible for the loss of KE.

Note:
Examples of how kinetic energy is converted into and out of different types of energy may help you better understand it. A biker, for example, uses chemical energy from meals to accelerate a bicycle to a desired pace. Except to overcome air resistance and friction, this speed can be maintained on a flat surface with no further effort. The chemical energy has been transformed to kinetic energy, or the energy of motion, but the process is inefficient and the rider generates heat as a result.