Question

The net work done by forces exerted by the disc on the ball for the duration ball remains on the disc is

• A. $\frac{{{2m{R}^2\omega }_0}^2}{9}$
• B. $\frac{{{m{R}^2\omega }_0}^2}{18}$
• C. $\frac{{{m{R}^2\omega }_0}^2}{6}$
• D. $\frac{{{m{R}^2\omega }_0}^2}{9}$

Answer 1

The correct option is D $\frac{{{m{R}^2\omega }_0}^2}{9}$

By work-energy theorem,

Work done=change in kinetic energy

By conservation of energy,

$\frac{1}{2}I{\omega }_0^2=\frac{1}{2}I{\omega }_1^2+2\times \frac{1}{2}\times \frac{M}{2}{R}^2{\omega }_1^2$

$\frac{1}{2}I{\omega }_0^2$=initial energy

$\frac{1}{2}I{\omega }_1^2$=energy of disc

$2\times \frac{1}{2}\times \frac{M}{2}{R}^2{\omega }_1^2$=energy of two balls

$\frac{1}{2}\frac{M{R}^2}{2}{\omega }_0^2=\frac{1}{2}\times \frac{1}{2}M{R}^2{\omega }_1^2+\frac{1}{2}M{R}^2{\omega }_1^2$

$\frac{1}{2}\frac{M{R}^2}{2R}{\omega }_0^2=\frac{1}{2}M{R}^2{\omega }_1^2(\frac{1}{2}+1)$

${\omega }_1^2=\frac{{\omega }_0^2}{3}$

Work done $=\frac{1}{2}m{R}^2{\omega }_1^2$

$=\frac{1}{2}m{R}^2\frac{{\omega }_0^2}{3}\Rightarrow W=\frac{m{R}^2{\omega }_0^2}{6}$