Question

two discs of same moment of inertia rotating rotating about their regular axis passing through the centre and perpendicular to the plane of the disc with angular velocity w 1 and w 2. they are brought into contact face to face coinciding the axis of rotation
the expression for loss of energy during this process is​

Answer 1

Given :

▪ Two disc having same moment of inertia are rotating about their regular axis passing the centre and perpendicular to the plane.

▪ Angular velocity of disc A = ω_1

▪ Angular velocity of disc B = ω_2

▪ They are brought into contact face to face.

To Find :

▪ Loss of energy during this process.

Thinking Process :

☞ When no external torque acts on system then, angular momentum of system remains constant.

Angular momentum before contact

$\dashrightarrow\sf\:L_i=I_1\omega_1+I_2\omega_2$

Angular momentum after discs brought into contact

$\dashrightarrow\sf\:L_f=I_{net}\omega=(I_1+I_2)\omega$

So, final angular speed of system

$\dashrightarrow\sf\:\omega=\dfrac{I_1\omega_1+I_2\omega_2}{I_1+I_2}$

Now, to calculate loss of energy, we subtract initial and final energies of system.

$\dashrightarrow\sf\:Loss\:of\:energy=\dfrac{1}{2}I{\omega_1}^2+\dfrac{1}{2}I{\omega_2}^2-\dfrac{1}{2}(2I)\omega^2\\ \\ \dashrightarrow\underline{\boxed{\bf{\red{Loss\:of\:energy=\dfrac{1}{4}I(\omega_1-\omega_2)^2}}}}\:\orange{\bigstar}$

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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