Question

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 w, and on. They are brought into contact
face to face coinciding the axis of rotation. The
expression for loss of energy during this process
[NEET-2014
(1) 1 160 + 02
(3) 1 (0, -02?
(2) 1 1 (0o, - 02/?
(4) 1 (0, - 02/?​

Answer 1

Answer:

I/4(ω1 - ω2)^2.

Explanation:

From the question we get that the angular velocities of both the disc are different, let them be ω1 and ω2 respectively. While the moment of inertia are same for both the disc since the axis passing through the center. So, if we let the moment of inertia be I1 and I2 so, I1=I2.

Now, we know that the kinetic energy loss will  be 1/2(I1*I2)/(I1 + I2)(ω1 - ω2)^2 since, I1=I2. So, 1/2*I^2/2I(ω1 - ω2)^2 which on solving we will get that I/4(ω1 - ω2)^2.

Answer 2

◘ Given data ◘

• Two discs of same moment of inertia rotating about their regular axis passing through the center and perpendicular to the plane of the disc with angular velocity ω₁ and ω₂.
• They are brought into contact face to face coinciding the axis of rotation.

The angular velocity of the first disc = ω₁

The angular velocity of the second disc = ω₂

$\rule{170}2$

◘ Objective ◘

To find the expression for loss coinciding of energy during this process.

$\rule{170}2$

◘ Calculation ◘

According to the given conditions, the angular momentum of system remains constant.

Now, let the angular velocity of the complete system be ω.

• Conservation of angular momentum :-

$\rm{I \omega_1+I \omega_2=\omega(I+I)}\\\\\rm{\longmapsto I( \omega_1+ \omega_2)=2I\omega }\\\\\rm{\longmapsto \omega=\dfrac{I(\omega_1+\omega_2)}{2I} }\\\\\rm{\longmapsto \omega=\dfrac{(\omega_1+\omega_2)}{2} }$

We know,

When an object is rotating about its center of mass, its rotational kinetic energy is

K = ½Iω².

Where,

• K is the kinetic energy

• I is the moment of inertia

• ω is the angular speed

○ Initial kinetic energy :-

$\rm{K_i=\dfrac{1}{2}I\omega_1^2+\dfrac{1}{2}I\omega_2^2 }$

○ Final kinetic energy :-

$\rm{K_f=\dfrac{1}{2}(2I)\omega^2}$

Loss of energy = Δ K = k(i) - k(f)

$\rm{\hookrightarrow \Delta K=\dfrac{1}{2}I\omega_1^2+\dfrac{1}{2}I\omega_2^2-\dfrac{1}{2}(2I)\omega^2 }$

$\rm{\hookrightarrow \Delta K=\dfrac{I}{2}( \omega_1^2+\omega_2^2-2\omega^2)}\\\\\rm{\hookrightarrow \Delta K=\dfrac{I}{2}[\omega_1^2+\omega_2^2-2(\dfrac{\omega_1+\omega_2}{2})^2]}\:\:\:\:\cdots (from\ above)\\\\\rm{\hookrightarrow \Delta K=\dfrac{I}{2}[\omega_1^2+\omega_2^2-2(\dfrac{\omega_1^2+\omega_2^2+2\omega_1 \omega_2}{4})}\\\\\rm{\hookrightarrow \Delta K=\dfrac{I}{2}[\omega_1^2+\omega_2^2-\dfrac{1}{2}(\omega_1^2+\omega_2^2+2\omega_1\omega_2)]}$

$\rm{\hookrightarrow \Delta K=\dfrac{I}{4}(2\omega_1^2-\omega_1^2+2\omega_2^2-\omega_2^2+2\omega_1\omega_2)}\\\\\rm{\hookrightarrow \Delta K=\dfrac{I}{4}(\omega_1^2+\omega_2^2-2\omega_1 \omega_2)}\\\\\boxed{\boxed{\rm{\color{magenta}{\hookrightarrow \Delta K=\dfrac{I}{4}(\omega_1-\omega^2)^2}}}}$

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