JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Rotational Mechanics

A circular disc of mass M and radius R is rotating about its axis with angular speed

ω_{1}

. If another stationary disc having radius

\frac{R}{2}

and same mass

M

is dropped co-axially on to the rotating disc. Gradually both discs attain constant angular speed

ω_{2}

. The energy lost in the process is

p %

of the initial energy. Value of

p

is ____.

Answer: 20

Solution

No external torque acts while the second disc settles onto the first, so angular momentum is conserved. The first disc has

I_{1} = \frac{1}{2} M R^{2}

and the dropped disc

I_{2} = \frac{1}{2} M (\frac{R}{2})^{2} = \frac{1}{8} M R^{2}

. From

I_{1} ω_{1} = (I_{1} + I_{2}) ω_{2}

ω_{2} = \frac{I _{1}}{I _{1} + I _{2}} ω_{1} = \frac{\frac{1}{2}}{\frac{1}{2} + \frac{1}{8}} ω_{1} = \frac{4}{5} ω_{1}

. The initial energy is

\frac{1}{2} I_{1} ω_{1}^{2} = \frac{M R ^{2} ω _{1}^{2}}{4}

and the final energy is

\frac{1}{2} (I_{1} + I_{2}) ω_{2}^{2} = \frac{1}{2} \cdot \frac{5 M R ^{2}}{8} \cdot \frac{16 ω _{1}^{2}}{25} = \frac{M R ^{2} ω _{1}^{2}}{5}

. The energy lost is

\frac{M R ^{2} ω _{1}^{2}}{4} - \frac{M R ^{2} ω _{1}^{2}}{5} = \frac{M R ^{2} ω _{1}^{2}}{20}

, which as a fraction of the initial energy is

\frac{1/20}{1/4} = \frac{1}{5}

, so

p = 20

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Integer answer