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 w. They are brought into contact
face to face coinciding the axis of rotation. The
expression for loss of energy during this process
[NEET-2017]
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Answer 1

simple use the law of conservation of momentum

-),let the angular velocity of combined mass be W

initial angular momentum=Iw+Iw

=2Iw

final angular momentum=2IW

intial angular momentum=final angular momentum

2IW=2Iw

W=w

this shows that velocity of combined mass will also be same

intial kinetic energy=1/2×I×w^2+1/2 ×I ×w^2

=Iw^2

final kinetic energy=1/2×2I×W^2

=IW^2

this shows that there is no loss in energy.......

Answer 2

Correct 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 w1, and w2. They are brought into contact face to face coinciding the axis of rotation. The expression for loss of energy during this process

[NEET-2017]

Solution :-

◘ 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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