JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Rotational Mechanics

A wheel is rotating freely with an angular speed

ω

on a shaft. The moment of inertia of the wheel is I and the moment of inertia of the shaft is negligible. Another wheel of moment of inertia 3 I initially at rest is suddenly coupled to the same shaft. The resultant fractional loss in the kinetic energy of the system is:

(A) $\frac{3}{4}$ ✓ Correct
(B) $\frac{5}{6}$
(C) $\frac{1}{4}$
(D)0

Answer: (A)

Solution

No external torque acts while the second wheel is coupled, so angular momentum is conserved:

I ω = (I + 3 I) ω^{'}

gives

ω^{'} = \frac{ω}{4}

. The initial kinetic energy is

\frac{1}{2} I ω^{2}

and the final is

\frac{1}{2} (4 I) (\frac{ω}{4})^{2} = \frac{1}{8} I ω^{2}

. The energy lost is

\frac{1}{2} I ω^{2} - \frac{1}{8} I ω^{2} = \frac{3}{8} I ω^{2}

, and as a fraction of the initial energy this is

\frac{3/8}{1/2} = \frac{3}{4}

JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Single correct option

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JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Rotational Mechanics

Solution

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