Consider a Disc of mass 5 ~kg , radius 2 ~m , rotating with angular velocity of 10 ~rad / s about an axis perpendicular to the plane of rotation. An identical disc is kept gently over the rotating disc along the same axis. The energy dissipated so that both the discs continue to rotate together without slipping is \underline\hspace2cm J.

Rotational Motion PYQ — JEE Main 2024 · 30 Jan · Shift 1 · Q87 with Solution

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Correct answer: 250

For a solid disc about its central axis,

I = \frac{1}{2}mR^2

So,

I = \frac{1}{2}(5)(2^2) = \frac{1}{2}(5)(4) = 10\,\text{kg m}^2

Thus, each disc has moment of inertia

I_1 = I_2 = 10\,\text{kg m}^2

Since no external torque acts on the system,

L_i = L_f

Initially only the first disc is rotating:

L_i = I_1\omega_1 = 10 \times 10 = 100

After contact, both rotate together with common angular velocity $\omega$ :

L_f = (I_1 + I_2)\omega = (10+10)\omega = 20\omega

Equating:

100 = 20\omega

\omega = 5\,\text{rad/s}

K_i = \frac{1}{2}I_1\omega_1^2

K_i = \frac{1}{2}(10)(10^2) = 5 \times 100 = 500\,\text{J}

Both discs rotate together:

K_f = \frac{1}{2}(I_1+I_2)\omega^2

K_f = \frac{1}{2}(20)(5^2) = 10 \times 25 = 250\,\text{J}

E_{\text{dissipated}} = K_i - K_f = 500 - 250 = 250\,\text{J}

The energy dissipated is

\boxed{250\,\text{J}}