Sigma Percentile
JEE Main 2021
LEVELJEE Advanced

Animated Solution for Physics - Rotational Motion: Two discs have moments of inertia and about their respective axes perpendicular to the plane and passing through the centre. They are rotating with angular speeds, and respectively and are brought into contact face to face with their axes of rotation co-axial. The loss in kinetic energy of the system in the process is given by

Select Answer:

Visualized Solution

  • Two discs rotating independently.
  • and

  • Discs are brought into contact.
  • They rotate together with a common angular velocity .

  • No external torque acts on the system.

  • Kinetic energy is lost due to internal friction.

  • Take common denominator

  • Numerator:
  • Canceling terms leaves:

  • This is a standard result, similar to perfectly inelastic collisions in linear kinematics.

The Sigma Insight: Conservation of Angular Momentum

Solution Diagram

The Setup

A Tale of Two Discs
Imagine you are in a physics laboratory, observing a fascinating experiment. Before you are two solid discs, mounted on the same vertical frictionless axle. However, they are currently separated, spinning completely independently of one another.
The top disc has a moment of inertia and is spinning with an initial angular velocity . The bottom disc is a bit different; it has a moment of inertia and is spinning with its own initial angular velocity . They are like two dancers spinning to their own rhythm, completely unaware of each other.
Suddenly, the support holding the top disc is removed! It drops down, landing squarely on the bottom disc. What happens next is a beautiful display of rotational dynamics.

The Collision

When Worlds Collide
The moment the two discs touch, they don't instantly spin together. Because they are initially rotating at different speeds, their surfaces grind against each other. This grinding is kinetic friction in action. The faster disc tries to speed up the slower one, while the slower disc tries to drag down the faster one.
This internal tug-of-war continues until the friction finally does its job, locking the two discs together. From that moment on, they spin as a single, unified object with a brand new, common angular velocity, which we will call .
But how do we find this new speed? And more importantly, what was the cost of this violent, grinding collision?

The Master Equation

Conservation of Angular Momentum
To find the final speed, we need to look for a conserved quantity. During the collision, the frictional forces acting between the discs are purely internal to our two-disc system. By Newton's Third Law, the torque exerted by the top disc on the bottom disc is exactly equal and opposite to the torque exerted by the bottom disc on the top disc.
Because there is no external twisting force (no outside hand spinning the axle), the net external torque on the system is exactly zero:
When the net external torque is zero, the total angular momentum of the system must remain perfectly constant. This is the Law of Conservation of Angular Momentum.
Let's write down the initial angular momentum. It is simply the sum of the individual momenta of the two discs:
After the collision, the discs are locked together. Their combined moment of inertia is , and they spin at the common speed . So, the final angular momentum is:
Equating the initial and final states, we get our master equation:
Solving for the final common angular velocity , we find:
This formula is incredibly elegant. It is the exact rotational equivalent of finding the final velocity in a perfectly inelastic linear collision!

The Energy Toll

Calculating the Loss
Now we arrive at the core of the problem. While angular momentum was perfectly conserved, kinetic energy was not. The kinetic friction that locked the discs together did negative work, dissipating some of the system's rotational kinetic energy as heat and sound.
To find out exactly how much energy was lost, we need to calculate the difference between the initial kinetic energy and the final kinetic energy:
Let's set up the raw equations. The initial kinetic energy is the sum of the kinetic energies of the two freely spinning discs:
The final kinetic energy is the energy of the combined, locked system spinning at the new speed :
Now, we substitute our expression for into the final kinetic energy equation:

The Algebraic Dance

Simplifying the Mess
At first glance, this equation looks like an algebraic nightmare. But if we take a deep breath and proceed carefully, it will simplify beautifully.
First, notice that the term in the numerator cancels out one of the terms in the denominator of the squared fraction:
Next, we need to combine these terms into a single fraction. We do this by taking a common denominator of :
Now, we carefully expand the terms in the numerator. Watch out for the negative sign!

The Grand Finale

A Beautiful Symmetry
Look closely at the expanded numerator. The magic of algebra is about to happen. The terms cancel each other out perfectly. The terms also cancel each other out perfectly!
We are left with only the cross terms:
Notice that every single remaining term contains the product . Let's factor that out:
Do you recognize the expression inside the parentheses? It is the perfect square expansion of !
Substituting this beautifully simplified numerator back into our full equation, we arrive at our final answer:
This final expression is not just a random jumble of variables; it is a profound physical truth. It perfectly mirrors the energy loss formula for a perfectly inelastic linear collision (), where acts as the "reduced moment of inertia" and is the relative angular velocity. Understanding this deep symmetry will make you a true master of rotational mechanics!

Similar Questions

JEE Main 2019, 10 April Shift-I
LEVELJEE Main

Two coaxial discs, having moments of inertia and are rotating with respective angular velocities and , about their common axis. They are brought in contact with each other and thereafter they rotate with a common angular velocity. If and are the final and initial total energies, then is

(A)
(B)
(C)
(D)
JEE Main 2020, 02 Sep Shift-II
LEVELJEE Main

Two uniform circular discs are rotating independently in the same direction around their common axis passing through their centres. The moment of inertia and angular velocity of the first disc are and respectively, while those for the second one are and , respectively. At some instant they get stuck together and start rotating as a single system about their common axis with some angular speed. The kinetic energy of the combined system is

(A)
(B)
(C)
(D)
JEE Main 2020, 04 Sep Shift-I
LEVELJEE Main

A circular disc of mass and radius is rotating about its axis with angular speed . If another stationary disc having radius and same mass is dropped co-axially on to the rotating disc. Gradually, both discs attain constant angular speed . The energy lost in the process is of the initial energy. Value of is ...........

JEE Main 2020, 05 Sep Shift-I
LEVELJEE Main

A wheel is rotating freely with an angular speed on a shaft. The moment of inertia of the wheel is and the moment of inertia of the shaft is negligible. Another wheel of moment of inertia initially at rest is suddenly coupled to the same shaft. The resultant fractional loss in the kinetic energy of the system is

(A)
(B)
(C)
(D)
JEE Advanced 2013
LEVELJEE Advanced

A uniform circular disc of mass and radius is rotating with an angular velocity of about its own axis, which is vertical. Two uniform circular rings, each of mass and radius , are gently placed symmetrically on the disc in such a manner that they are touching each other along the axis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at rest relative to the disc and the system rotates about the original axis. The new angular velocity (in ) of the system is

JEE Advanced 1983
LEVELJEE Main

A thin circular ring of mass and radius is rotating about its axis with a constant angular velocity . Two objects, each of mass , are attached gently to the opposite ends of a diameter of the ring. The wheel now rotates with an angular velocity

(A)
(B)
(C)
(D)
LEVELJEE Main

Initial angular velocity of a circular disc of mass is . Then, two small spheres of mass are attached gently to two diametrically opposite points on the edge of the disc. What is the final angular velocity of the disc?

(A)
(B)
(C)
(D)
JEE Advanced 1999
LEVELJEE Main

A smooth sphere is moving on a frictionless horizontal plane with angular velocity and centre of mass velocity . It collides elastically and head on with an identical sphere at rest. Neglect friction everywhere. After the collision their angular speeds are and respectively. Then,

(A)
(B)
(C)
(D)
LEVELJEE Main

A thin circular ring of mass and radius is rotating about its axis with a constant angular velocity . Two objects each of mass are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity =

(A)
(B)
(C)
(D)
JEE Main 2021, 18 March Shift-I
LEVELJEE Main

A thin circular ring of mass and radius is rotating about its axis with an angular speed . Two particles having mass each are now attached at diametrically opposite points. The angular speed of the ring will become

(A)
(B)
(C)
(D)