Two discs of moments of inertia I1 and I2 about their respective axes (normal to the disc and pa

Two discs of moments of inertia I1 and I2 about their...

Key Concepts

Conservation of Angular Momentum

In rotational dynamics, the total angular momentum of a closed system remains constant if no external torques act on the system. This principle is used to determine the final angular velocity when two bodies interact, such as when they are brought into contact and eventually rotate together.

Rotational Kinetic Energy

Rotational kinetic energy is the energy possessed by a rotating object, quantified by the expression ½ I?², where I is the moment of inertia and ? is the angular speed. In processes involving friction or other non-conservative forces, some of this energy can be converted into other forms, resulting in a loss of mechanical energy.

Inelastic Collisions in Rotational Dynamics

When rotating bodies come into contact and eventually move together at a common angular speed, the interaction resembles an inelastic collision. Although angular momentum is conserved during this process, the kinetic energy decreases due to frictional losses, which manifest as a reduction in the system's final rotational kinetic energy.

Moment of Inertia

The moment of inertia is a fundamental concept in rotational motion that quantifies an object's resistance to changes in its rotational speed. It depends on the mass distribution relative to the axis of rotation and plays a crucial role in determining both the rotational kinetic energy and the dynamics of systems undergoing rotational collisions.

Transcript

00:01 So the question asked here is to find the loss of kinetic energy when two disks do two disks with the moment of inertia i1 and i2 and rotating with an angular velocity or angular speed omega -1 omega -2 are brought into contact face -to -face each other and then there will be some kinetic energy losses and we would have to find what is that loss now let's draw that particular figure over here so let this be the disk one sorry put this figure and add this be the first discs and this be the second disk and it is given that both the disks are making in contact each other so i'm just and this is the figure that we got over here so let this be the axis that they are rotating now before the disks are made in contact each other the moment of inertia was i1 and the moment of the angular speed was omega 1 and the moment of inertia of the second disc was i 2 and this angular speed was omega 2 now we can write the initial kinetic energy that before when they when they contacted each other or when they made contact in each other so kinetic energy initial will be equal to half i1 omega 1 square plus half i2 omega 2 square this can be written as half into i 1 omega 1 square plus i 2 omega 2 square so this is the initial kinetic energy now after making them uh contacting each other we can write that the system of the these two disks will be having and a moment of inertia is equal to i1 plus i2 this will be the moment of inertia of the system and we can we have one more thing that after contacting each other this two disks will be moving in a common angular velocity let that be omega why because if one of them is having a higher speed and the other is having a lower speed than the other then after contacting each other the higher speed will be reducing and the smallest will be increasing why because they would have to rotate in a single speed then only the system will be then only the law of physics that is moment of like that is the conservation of momentum will be applied there so we can say that this omega is the final speed of angular speed of this system.

03:09 So according to the law of conservation of momentum, we can write that the initial momentum l initial will be equal to l final.

03:21 So this must happen, which means the angular momentum that is the initial angular momentum was i1 omega 1 plus i2 omega and the final angular momentum will be i of the system that is i1 plus i2 into the common speed that they are rotating that is omega so that we can write the value of omega will be equal to and this omega is the common speed that they're rotating after making them contact each other so i1 omega 1 plus i2 omega 2 divided by i 1 plus i do so this is the one equation that we can write from first i mean from the first law now the first thing that we already found was the kinetic energy that the initial kinetic energy that we found and it is asked to find the loss of kinetic energy so to find the loss of kinetic energy would have to find the final kinetic energy so the kinetic energy final will be equal to their move now they made each other contact each other which means both are rotating in a system like a system so we can try the connectic energy final will be equal to i1 plus i2 will be the combined moment of inertia so the equation was half i omega square so half into instead of i we can write i 1 plus i 2 and omega square instead of omega square we can write this value over here so this will be equal to half into i1 plus i2 into omega square that is i1 omega 1 plus i2 omega 2 divided by i 1 plus i2 and the square of this one so from this we can see that here in this numerator part we have i1 plus i2 and here also we would have i 1 plus i2 whole square so that this i 1 plus i2 square will be there so one i plus i2 we can cancel so this will become as half into i1 omega 1 plus i2 omega 2 square will be only for this now divided by i 1 plus i 2 why because the square is cancelled for this i 1 plus i 2 now this is the value of final kinetic energy now to find the loss of kinetic energy you would have to subtract kinetic energy from initial kinetic energy would have to subtract kinetic energy final so this will we will get this as to the initial kinetic energy was half i1 omega 1 square plus i2 omega 2 square this was the initial kinetic energy minus the final kinetic energy is half i1 omega 1 plus i2 omega 2 whole square divided by i 1 plus i 2 so this one we can write as this i 1 plus i 2 square we can take as common this is equal to and even the half half into i 1 omega 1 square sorry for this yes sorry this we can write it as half it is taken outside so i 1 omega 1 square plus i 2 omega 2 squared minus this half is taken outside so this one we can write i 1 omega 1 sorry i 1 omega 1 plus i 2 omega 2 and this is whole square divided by i 1 plus i 2 so this is the denominator for only this part so i'm just multiplying this denominator with this new so that we can write this determinator as common so this is half so i 1 omega 1 square into i 1 plus i 2 is what we would have to write so this will be equal to this i 1 plus i 2 into i 1 omega 1 square which which will be equal to i 1 square omega 1 square plus i 2 into i 1 omega 1 square plus the next term we can write it as i 1 i 2 omega 2 square plus i 2 square omega 2 square this will be that term minus and this i 1 omega 1 plus i 2 omega 2 whole square we can write it as i 1 omega 1 square this whole square plus i2 omega 2 whole square which is i 2 square omega 2 plus 2 into a b just like a plus b whole b whole square so this will be 2 into i 1 omega 1 into i 2 omega 1 and divided by whole divided by i 1 plus i 2 2 2 this will be the final term...

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