JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Rotational Mechanics

Two coaxial discs, having moments of inertia

I_{1}

and

\frac{I _{1}}{2}

, area rotating with respectively angular velocities

ω_{1}

and

\frac{ω _{1}}{2}

, about their common axes. They are brought in contact with each other and thereafter they rotate with a common angular velocity. If

E_{f}

and

E_{i}

are the final and initial total energies, then

(E_{f} - E_{i})

is:

Answer: (D)

Solution

Angular momentum is conserved when the discs couple. With the second disc having half the moment of inertia and half the angular speed,

I_{1} ω_{1} + \frac{I _{1}}{2} \cdot \frac{ω _{1}}{2} = \frac{3 I _{1}}{2} ω

, giving

ω = \frac{5 ω _{1}}{6}

. The initial kinetic energy is

E_{i} = \frac{1}{2} I_{1} ω_{1}^{2} + \frac{1}{2} \cdot \frac{I _{1}}{2} \cdot \frac{ω _{1}^{2}}{4} = \frac{9 I _{1} ω _{1}^{2}}{16}

and the final energy is

E_{f} = \frac{1}{2} \cdot \frac{3 I _{1}}{2} \cdot \frac{25 ω _{1}^{2}}{36} = \frac{25 I _{1} ω _{1}^{2}}{48}

. The change is

E_{f} - E_{i} = \frac{25 I _{1} ω _{1}^{2}}{48} - \frac{27 I _{1} ω _{1}^{2}}{48} = - \frac{I _{1} ω _{1}^{2}}{24}

, the energy lost to friction.

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Single correct option