Question

A wheel is rotating freely at angular speed 800 rev/min on ashaft whose rotational inertia is negligible. A secod wheel,initially rest and with twice the rotational inertia of the first,is suddenly coupled to the same shaft.
A) what is the angular speedof the resultant combination of the shaft and the two wheels?
B)What fraction of the orginial rotational kinetic energy is lost?

Answer 1

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