(A) Initial MI = I

Angular speed \(\displaystyle\omega_{{{i}}}={800}{r}\pm={800}\times{60}\times{2}\pi\) rad. / s

MI of second wheel = 2I

So final MI after coupling = 3I

Final angular speed = \(\displaystyle\omega_{{{f}}}\)

Angular momentum is conserved in the process. So, \(\displaystyle{\left|\omega_{{{i}}}={3}\right|}\omega_{{{f}}}\)

\(\displaystyle\omega_{{{f}}}=\frac{\omega_{{{i}}}}{{3}}={800}\times{60}\times\times{2}\frac{\pi}{{3}}\)

\(\displaystyle={1.005}\times{10}^{{{5}}}\) rad. /s (or) 266.7 rpm

(B) Initial Rotational KE = 0.5 \(\displaystyle{\mid}{\left(\omega_{{{i}}}\right)}^{{{2}}}\)

Final Rotational K.E. = 0.5(3|) \(\displaystyle{\left(\omega_{{{f}}}\right)}^{{{2}}}\)

Fraction of the orginial rotational kinetic energy lostis

\(\displaystyle=\frac{{{\left({R}.{K}.{E}.\right)}_{{{i}}}-{\left({R}.{K}.{E}.\right)}_{{{i}}}}}{{\left({R}.{K}.{E}.\right)}_{{{i}}}}={0.667}\)

Angular speed \(\displaystyle\omega_{{{i}}}={800}{r}\pm={800}\times{60}\times{2}\pi\) rad. / s

MI of second wheel = 2I

So final MI after coupling = 3I

Final angular speed = \(\displaystyle\omega_{{{f}}}\)

Angular momentum is conserved in the process. So, \(\displaystyle{\left|\omega_{{{i}}}={3}\right|}\omega_{{{f}}}\)

\(\displaystyle\omega_{{{f}}}=\frac{\omega_{{{i}}}}{{3}}={800}\times{60}\times\times{2}\frac{\pi}{{3}}\)

\(\displaystyle={1.005}\times{10}^{{{5}}}\) rad. /s (or) 266.7 rpm

(B) Initial Rotational KE = 0.5 \(\displaystyle{\mid}{\left(\omega_{{{i}}}\right)}^{{{2}}}\)

Final Rotational K.E. = 0.5(3|) \(\displaystyle{\left(\omega_{{{f}}}\right)}^{{{2}}}\)

Fraction of the orginial rotational kinetic energy lostis

\(\displaystyle=\frac{{{\left({R}.{K}.{E}.\right)}_{{{i}}}-{\left({R}.{K}.{E}.\right)}_{{{i}}}}}{{\left({R}.{K}.{E}.\right)}_{{{i}}}}={0.667}\)